Student’s Book Numerical Functional Analysis Editor: Stefan Engblom Fall 2014 Preface This little book collects some of the student’s output during the course Numerical Functional Analysis, which was given for the first time in the fall 2014 at the Department of Information technology, Uppsala university. The course consisted of a total of five lectures, the first three of which went over Metric spaces, Normed spaces, and Inner product spaces, thus following closely the first three chapters of Kreyszig’s book Introductory functional analysis with applications. Two more lectures were devoted to the five ‘Big’ theorems of functional analysis as presented by the students themselves: the Hahn-Banach theorem, the Uniform boundedness theorem, the Open mapping theorem, the Closed graph theorem, and the Banach fixed point theorem. The final part of the course consisted of short essays on various topics, typically connecting Numerical analysis and Functional analysis in one way or the other. The essays were improved by double open review among the students themselves after which the final version entered this collection. A selection of student’s solution to book exercises has also been included: the selection was very loosely based on the difficulty of the exercise itself and on the solution quality. Stefan Engblom Uppsala, January 2015 Contents I Student’s essays 3 Dual-consistent approximations of PDEs and super-convergent functional output Martin Almquist 4 Lax equivalence theorem Siyang Wang 15 The Schauder fixed point theorem Hannes Frenander 20 Existence of a weak solution to the Stokes equations Hanna Holmgren 24 The Lax-Milgram theorem Andrea Alessandro Ruggiu 31 A time dependent mapping from Cartesian coordinate systems into curvilinear coordinate systems and the geometric conservation law Samira Nikkar 37 The Babuˇ ska inf-sup condition Simon Sticko 42 Uniformly best wavenumber approximations by spatial central difference operators Viktor Linders 47 The Ekeland variational principle with applications Markus Wahlsten 54 Construction of Sobolev spaces Cristina La Cognata 59 1 The Arzel´ a -Ascoli theorem Tomas Lundquist 65 A completeness theorem for non-self-adjoint eigenvalue problems Saleh RezaeiRavesh 69 The metric Lax and applications Cheng Gong 77 Fixed-point proof of Lax-Milgram theorem Fatemeh Ghasemi 82 II 87 Solutions to exercises 1 Metric spaces 88 2 Normed spaces 96 3 Inner product spaces 112 2 Part I Student’s essays 3 Dual-consistent approximations of PDEs and super-convergent functional output Martin Almquist∗ December 22, 2014 Abstract Finite-difference operators satisfying the summation-by-parts (SBP) rule can be used to obtain high-order accurate time-stable schemes for partial differential equations (PDEs), when the boundary and interface conditions are imposed weakly by the simultaneous approximation term method (SAT). The most common SBP-SAT discretizations are accurate of order 2p in the interior and order p close to the boundaries, which yields a global accuracy of p + 1 for first order PDEs. However, Berg and Nordstr¨om [2012] and Hicken and Zingg [2014] have shown that any linear functional computed from the time-dependent numerical solution will be accurate of order 2p if the spatial discretization is dual-consistent. This text discusses the concepts of dual problem and dual consistency. We walk through some of the definitions and results in Berg and Nordstr¨om [2012] and Hicken and Zingg [2014] while emphasizing the underlying functional analysis considerations. 1 Dual problems and dual consistency We introduce the concepts of dual problem and dual consistency by considering linear partial differential equations with homogeneous boundary conditions. Let L be a linear differential operator of order m on a domain Ω and consider the problem Lu − f = 0, x ∈ Ω, (1.1) ∗ Division of Scientific Computing, Department of Information Technology, Uppsala university, SE-751 05 Uppsala, Sweden. [email protected] 1 subject to homogeneous boundary conditions. f ∈ L2 (Ω) is independent of the solution u ∈ H k (Ω) ⊂ L2 (Ω). Here H k (Ω) = W k,2 (Ω) is the Sobolev space consisting of functions whose weak derivatives up to order k (for some integer k) exist and are square-integrable on Ω. We assume that m ≤ k. For v, w ∈ L2 (Ω) we use the notation Z (v, w) := vw dx. (1.2) Ω for the standard L2 inner product. We now introduce the formal adjoint of the operator L. Definition 1.1. Let u ∈ H k (Ω). The formal adjoint of a linear differential operator L is the operator L† such that (φ, Lu) = (L† φ, u) The word formal in the definition emphasizes that we are not yet specifying the domain D(L† ) of the adjoint operator. The reason for this definition is that L is not bounded on L2 (Ω), in general. Hence we need to distinguish between the formal adjoint and the Hilbert-adjoint (which concerns bounded operators) even though L2 (Ω) is a Hilbert space. Note also that the definition of L† is abstract; an explicit expression for L† φ can be obtained via (repeated) integration by parts on (φ, Lu). We now consider a bounded linear functional of the solution u, J (u) = (z, u) (1.3) for some z ∈ L2 (Ω). Note that since L2 (Ω) is a Hilbert space, the representation (1.3) includes all linear functionals, by Riesz’s representation theorem. To derive the adjoint equation of (1.1) with respect to the functional J , we take the inner product of (1.1) with a function φ ∈ D(L† ) and express J as J (u) = (z, u) = (z, u) − (φ, Lu − f ) = (φ, f ) − (L† φ − z, u). (1.4) We now obtain the adjoint or dual equation by seeking φ such that J becomes independent of u, which requires L† φ − z = 0, yielding J = (φ, f ). 2 (1.5) Remark. To simplify the analysis we have disregarded the boundary conditions. In general, the dual problem depends on the type of boundary conditions in the primal problem, but not on the particular boundary data. Boundary conditions for the dual problem can be constructed as the minimal set of homogeneous conditions which, together with the homogeneous primal boundary conditions, make all boundary terms resulting from the integration by parts procedure vanish. We outline the procedure for a model problem in Section 2.3. Definition 1.2. The continuous dual problem associated with the primal problem (1.1) and the functional J is L† φ − z = 0 subject to the dual boundary conditions. The solution φ is called the dual variable. Note that the forcing function f and functional representation z in the primal problem have switched roles in the dual problem. We now turn to the discrete case. Let Lh uh − f = 0 (1.6) be a consistent discretization of (1.1) including boundary conditions, where uh ∈ Rn may hold basis function coefficients and/or collocation values. Lh is a bounded linear operator which can be represented by matrix multiplication, Lh : Rn −→ Rn . Let (·, ·)h denote a discrete inner product on Rn × Rn such that (Rn , (·, ·)h ) is a Hilbert space. We can then use the Hilbert-adjoint: Definition 1.3. The Hilbert-adjoint of Lh is the operator L†h : Rn −→ Rn such that (φh , Lh uh )h = (L†h φh , uh )h for all φh , uh ∈ Rn . Let Jh (uh ) = (z, uh )h be an approximation of the functional J . We now derive the discrete dual problem by taking the inner product of the discretization (1.6) with φh ∈ Rn and expressing Jh as Jh (uh ) = (z, uh )h − (φh , Lh uh − fh )h = (φh , fh )h − (L†h φh − z, uh )h . (1.7) We obtain the discrete dual problem by seeking φh such that Jh is independent of uh , which requires L†h φh − z = 0 . (1.8) 3 Definition 1.4. The discrete dual problem associated with (1.6) and the functional Jh is L†h φh − z = 0 Definition 1.5. A discretization (1.6) of a continuous problem (1.1) is called dual consistent if the discrete dual problem (1.8) is a consistent discretization of the continuous dual problem (1.5). 1.1 Time-dependent problems We now extend the concepts of dual problems and dual consistency to timedependent problems. Consider the problem ut + Lu − f = 0, x ∈ Ω, t > 0 (1.9) subject to homogeneous boundary and initial conditions. Let d u dt h + Lh uh − f = 0, t > 0 (1.10) be a semi-descretization of (1.9), including the boundary conditions. Definition 1.6. The semi-discretization (1.10) is called spatially dual-consistent if Lh uh − f = 0 is a dual-consistent discretization of the steady adjoint problem, i.e., the dual problem corresponding to (1.9) with ut = 0. 2 SBP-SAT discretizations We begin this section by introducing notation and some basic concepts related to SBP-SAT discretizations, and then proceed to consider stable and dual-consistent SBP-SAT discretizations. 2.1 Definitions Consider as an example the advection equation on 0 ≤ x ≤ 1, ut + ux = f u(0, t) = g(t). 4 (2.1) We introduce the N + 1 equidistant grid points xj = jh, j = 0, 1, . . . , N, h= 1 N (2.2) and denote the semi-discrete solution vector corresponding to grid size h by uh , where uh = [uh,0 , uh,1 , . . . , uh,N ]T ∈ RN +1 and uh,j (t) approximates u(xj , t). The restriction [f (x0 ), f (x1 ), . . . , f (xN )]T of f onto the grid will also be denoted by f , with no risk for confusion. Spatial derivatives will be approximated by first-derivative SBP operators: Definition 2.1. A matrix D is called a first-derivative SBP operator if D can be written as D = H −1 Q where H = H T is positive definite, and thus defines a norm, and Q satisfies Q + QT = diag[−1, 0, . . . , 0, 1]. In this text we will restrict ourselves to diagonal norm matrices H. In that case, D consists of a 2pth-order accurate centered difference stencil in the interior and pth-order accurate one-sided stencils near the boundaries. The global accuracy for first order problems can then be shown to be p + 1 Sv¨ard and Nordstr¨om [2006]. Let u, v ∈ RN +1 . We define the discrete inner product and norm as kuk2h = (u, u)h . (u, v)h = uT Hv, (2.3) The inner product space (RN +1 , (·, ·)h ) is complete and thus a Hilbert space. Now consider a discretization matrix Lh . We can compute the Hilbert-adjoint L†h with respect to the inner product (2.3) explicitly: Lemma 2.1. Let Lh be an (N + 1) × (N + 1) matrix with real entries. Then the Hilbert adjoint is L†h = H −1 LTh H (2.4) Proof. Starting from Definition 1.3, we have (φh , Lh uh )h = (L†h φh , uh )h ∀φh , uh ∈ Rn φTh HLh uh = (L†h φh )T Huh ∀φh , uh ∈ Rn φTh HLh uh = φTh (L†h )T Huh ∀φh , uh ∈ Rn HLh = (L†h )T H (L†h )T = HLh H −1 L†h = H −1 LTh H 5 ⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒ 2.2 Stable SBP-SAT discretizations The SBP-SAT semi-discretization of (2.1) can be written d uh + Duh = f + τ H −1 e0 (eT0 uh − g) dt (2.5) where e0 = [1, 0, . . . , 0]T and τ is a penalty parameter, which will be chosen such that the scheme (2.5) is stable. Multiplying (2.5) by uTh H from the left and setting f = 0, g = 0 yields d kuh k2h = (1 + 2τ )u2h,0 − u2h,N , dt (2.6) from which it is clear that τ ≤ − 21 leads to a stable scheme. Thus, the penalty parameter τ is allowed to vary in a semi-infinite range. This flexibility creates a possibility to impose additional conditions on the scheme, such as dual consistency. 2.3 Dual-consistent SBP-SAT discretizations Consider again the advection equation (2.1) on 0 ≤ x ≤ 1, ut + ux = f u(0, t) = 0. (2.7) with some initial condition and an associated linear functional Z 1 J (u) = zu dx. (2.8) 0 Note that J is a time-dependent functional. The dual problem is obtained by setting ut = 0 and finding φ such that J = (φ, f ). Using integration by parts and inserting the boundary condition in (2.7), we obtain Z 1 Z 1 J (u) = J (u) − φ(ux − f ) dx = −φ(1, t)u(1, t) + (z + φx )u dx + (φ, f ), 0 0 and hence the dual problem is −φx = z φ(1, t) = 0 . (2.9) Note that the sign has changed and the boundary condition is located at the opposite boundary, compared to the primal problem. We are now ready to state the main result in this text: 6 Theorem 2.2. Let d uh + Lh uh = f (2.10) dt be a stable and spatially dual consistent SBP-SAT approximation of the continuous problem ut + Lu = f. (2.11) A linear functional J (u) = (z, u) (2.12) Jh (uh ) = (z, uh )h (2.13) Jh (uh ) = J (u) + O(h2p ) (2.14) approximated by satisfies Proof. The proof can be found in Berg and Nordstr¨om [2012]. We now investigate whether the scheme (2.5) can be made spatially dualconsistent. Assuming homogeneous boundary conditions (g = 0) and neglecting the time-derivative, the scheme can be written as Lh uh = f (2.15) Lh = D − τ H −1 e0 eT0 . (2.16) where By definition the scheme is spatially dual-consistent if the discrete dual problem L†h φh − z = 0 (2.17) is a consistent discretization of (2.9). Using Lemma 2.1 and the SBP property, we have L†h φh −z = H −1 LTh Hφh −z = H −1 −Q + eN eTN − (τ + 1)e0 eT0 φh −z (2.18) which is a consistent discretization of (2.9) if and only if τ = −1. For τ 6= −1 the scheme imposes a boundary condition at x = 0 which does not exist in the continuous equation. Note that the requirement τ = −1 does not contradict the stability requirement τ ≤ − 21 . Thus, we expect convergence of order 2p for any linear functional of the solution if we choose τ = −1. 7 3 Numerical experiments For the interested reader, we include a set of numerical experiments which validate the theoretical results. We solve the advection equation using the scheme (2.5) and consider the functional Z 1 cos(4πx)u dx. (3.1) J (u) = 0 Using the technique of manufactured solutions, we choose the exact solution u = cos(4πx) sin(2πt), (3.2) which corresponds to the forcing function f = 2π cos(4πx) cos(2πt) − 4π sin(4πx) sin(2πt). (3.3) The analytic solution is used as initial and boundary data. With the solution (3.2), the exact time-dependent functional becomes J = 1 sin 2πt. 2 (3.4) We use an SBP-SAT discretization of internal order 6 (2p = 6) and hence expect 4th order convergence in the l2 -norm for any τ ≤ − 21 . We expect the functional error to converge as (at least) h4 for τ ≤ − 12 . For the dualconsistent choice τ = −1, we expect 6th order convergence in the functional. This is verified by Figures 3.1-3.3, which show the convergence rates for τ = −0.5, τ = −1 and τ = −1.5, respectively. The l2 -error converges as h4 regardless of τ , whereas the functional error converges as h6 for τ = −1 and as h4 otherwise. Figure 3.4 shows both the l2 -error and the functional error as functions of τ , using N = 256. As expected, the functional error is minimized by the dual-consistent choice τ = −1. It is interesting to note that also the l2 -error is minimized by τ ≈ −1 in this case, although none of the theory in this text leads us to expect that. However, the l2 -error is much less sensitive to the choice of τ than the functional error, which decreases by approximately 4 magnitudes in the vicinity of τ = −1. References J. Berg and J. Nordstr¨om. Superconvergent functional output for time-dependent problems using finite differences on summation-by-parts 8 10 0 10 -2 10 -4 10 -6 10 -8 l 2 -error 4th order reference functional error 6th order reference 10 -10 1 10 10 2 10 3 N Figure 3.1: Convergence in k·kl2 and functional for τ = −0.5; dualinconsistent discretization. 10 0 10 -5 10 -10 l 2 -error 4th order reference functional error 6th order reference 10 -15 1 10 10 2 10 3 N Figure 3.2: Convergence in k·kl2 and functional for τ = −1; dual-consistent discretization. 9 10 0 10 -2 10 -4 10 -6 10 -8 l 2 -error 4th order reference functional error 6th order reference 10 -10 1 10 10 2 10 3 N Figure 3.3: Convergence in k·kl2 and functional for τ = −1.5; dualinconsistent discretization. 10 -6 10 functional error l 2 -error 10 -4 -5 10 -6 -2 -1.5 -1 10 -8 10 -10 10 -12 -2 -0.5 τ -1.5 -1 -0.5 τ (a) (b) Figure 3.4: The (a) l2 -error ku−uh kl2 and (b) functional error, as functions of the penalty parameter τ . Only τ = −1 yields a dual-consistent discretization. 10 form. Journal of Computational Physics, 231(20):6846–6860, Aug. 2012. ISSN 00219991. doi: 10.1016/j.jcp.2012.06.032. URL http://www. sciencedirect.com/science/article/pii/S0021999112003476. J. Hicken and D. Zingg. Dual consistency and functional accuracy: a finitedifference perspective. Journal of Computational Physics, 256:161–182, Jan. 2014. ISSN 00219991. doi: 10.1016/j.jcp.2013.08.014. URL http:// www.sciencedirect.com/science/article/pii/S0021999113005524. M. Sv¨ard and J. Nordstr¨om. On the order of accuracy for difference approximations of initial-boundary value problems. J. Comput. Physics, 218: 333–352, October 2006. 11 Lax Equivalence Theorem Siyang Wang∗ December 26, 2014 Abstract Convergence of numerical methods for solving differential equations is of great importance, but difficult to prove in a direct way. The Lax equivalence theorem links consistency, stability and convergence. This enables us to prove convergence by consistency and stability analysis. In this essay, the Lax equivalence theorem is proved from the functional analysis point of view. A differential equation example is also presented. 1 Introduction The Lax equivalence theorem is originally presented in Lax and Richtmyer [1956]. It applies to an initial value problem ut = f, 0 ≤ t ≤ tf , u = u0 , t = 0. (1.1) (1.2) The theorem reads Given a properly posed initial value problem 1.1, 1.2 and a finite difference approximation C(∆t) to it that satisfies the consistency condition, stability is a necessary and sufficient condition that C(∆t) be a convergent approximation.1 Remark. The term properly posed is equivalent to wellposed, which is used more often nowadays. That is, a unique solution U exists, and a small perturbation of the data leads to a small perturbation of the solution. ∗ Division of Scientific Computing, Department of Information Technology, Uppsala University, SE-751 05 Uppsala, Sweden. [email protected] 1 This theorem is copied entirely from Lax and Richtmyer [1956]. 1 2 The theorem We prove the Lax equivalence theorem from the functional analysis point of view. We reformulate the mathematical problem to a more general one. Find x ∈ X such that T x = y, (2.1) where y ∈ Y , T : X → Y is a linear operator from a normed space X onto a normed space Y . We assume that the above problem is wellposed, that is, T −1 : Y → D(T ) is continuous. This ensures that a small perturbation in the data y leads to a small perturbation in the solution x. Since T is linear, T −1 is also linear. The spaces X and Y are equipped with appropriate norms. The mathematical problem 2.1 is solved numerically as Tn xn = yn , (2.2) where Tn : Xn → Yn , xn ∈ Xn and yn ∈ Yn . The spaces Xn and Yn are equipped with appropriate norms. Here, n can be considered as the resolution of 2.2. As n → ∞, we hope for xn → x. In practice (think of 2.1 as a differential equation), the true solution x is a continuous function, while the numerical solution xn is discrete. In order to compare xn with x to check convergence, we define the following two operators RnX : X → Xn , RnY : Y → Yn . These two operators project x and y (in X and Y ) to RnX x and RnY y (in Xn and Yn ). We make the following definition • The numerical method 2.2 is consistent with 2.1 if Tn RnX z → RnY T z, for any z ∈ D(T ). • The numerical method 2.2 is stable if kTn−1 k ≤ K, for all n, where K is a constant independent of n. • The numerical method is convergent if Tn−1 RnY y → RnX T −1 y, for every y ∈ Y . 2 The Lax equivalence theorem from the functional analysis point of view reads: If the numerical method is consistent, it is convergent if and only if it is stable. Proof. 1. consistent + stable ⇒ convergent: kTn−1 RnY y − RnX T −1 yk =kTn−1 RnY T x − RnX xk =kTn−1 (RnY T x − Tn RnX x)k ≤kTn−1 kkRnY T x − Tn RnX xk ≤K kRnY T x − Tn RnX xk | {z } →0 by consistency Tn−1 RnY y RnX T −1 y. Therefore, → The method is convergent. Note that this proof does not require deep functional analysis knowledge. 2. consistent + convergent ⇒ stable: From convergence, we know that the sequence (Tn−1 RnY y) is bounded [Kreyszig, 1978, Lemma 1.4–2], that is kTn−1 RnY yk ≤ cy , for every y ∈ Y and n = 1, 2, .... By uniform boundedness theorem [Kreyszig, 1978, Lemma 4.7–3], the sequence of the norms kTn−1 RnY k is bounded independent of n. That is, there exists a constant β such that kTn−1 RnY k ≤ β. In order to proceed, we have to make the following assumption (*): For every wn ∈ Yn , kwn k ≤ 1, there exists w ∈ Y, kwk ≤ α, such that RnY w = wn , where α is a constant independent of n. This is a plausible assumption in most cases in practice, since RnY is just a restriction operator bringing a continuous function to its corresponding discrete one. It then follows kTn−1 RnY wk kTn−1 wn k kTn−1 RnY k = sup = sup ≤β kwk kwk kwk6=0 kwk6=0 The norm of Tn−1 can be written as kTn−1 k = supkzk=1 kTn−1 zk. Choose wn such that kwn k = 1 and kTn−1 k = kTn−1 wn k. With this wn , we still have kwk ≤ α. Therefore, kTn−1 k ≤ αβ. This proves that the numerical method is stable. 3 Remark. There are interesting examples Sanz-Serna and Palencia [1985] where the assumption (*) does not hold, so that a consistent and convergent method is not stable. Also, the uniform boundedness theorem requires that Yn is complete, i.e. a Banach space. This is not assumed or guaranteed, and might be a strong assumption in practice. Rosinger even claims that the Lax equivalence theorem is wrong Rosinger [2005]. 3 Application We choose the operator T in 2.1 as a differential operator d/dt. Then we have a differential equation of the type xt = y, 0 ≤ t ≤ tf , (3.1) with the initial condition x(0) = c. Comparing with the mathematical problem 2.1, the space X and Y can be chosen as {x ∈ C 2 [0, tf ] | x(0) = c} equipped with the maximum norm in [0, tf ]. We introduce the time discretization tj = jk, j = 0, 1, · · · , N, where the step size k = tf /N . The corresponding discrete solutions are x0 , x1 , · · · , xN . We equip the discrete space with the maximum norm. The finite difference scheme can be written as xn+1 = P (k)xn , n = 0, 1, · · · , N − 1, where P (k) is the finite difference operator taking the discrete solution from the current time step to the next time step. The recursion relation can be written as x1 P (k)x0 1 −P (k) x2 0 1 x3 0 −P (k) 1 . = .. .. .. .. . . . . xN 0 −P (k) 1 {z } | Q(k) Stability requires that kQ−1 (k)k ≤ cQ , where cQ is independent of k. Due to 4 the special structure of Q(k), its inverse can be computed exactly, yielding 1 P (k) 1 2 1 Q−1 (k) = P (k) P (k) . .. . . . . . . . . . . N −1 2 P (k) · · · P (k) P (k) 1 Therefore, kQ−1 (k)k = |P (k)| < 1. PN −1 i=0 |P i (k)| = 1−|P (k)|N . 1−|P (k)| Stability is ensured if Remark. By taking y = −λx, we get the familiar test equation, which is often used to analyze the stability properties of numerical methods for time integration. For forward Euler method, P (k) = 1 − kλ, and stability requires |1 − kλ| < 1. This is the same result as if we would do the standard stability analysis. References E. Kreyszig. Introductory functional analysis with applications. John Wiley & Sons, 1978. P. D. Lax and R. D. Richtmyer. Survey of the stability of linear finite difference equations. Comm. Pure Appl. Math., IX:267–293, 1956. E. E. Rosinger. What is wrong with the Lax-Richtmyer fundamental theorem of linear numerical analysis? Technical report, 2005. J. M. Sanz-Serna and C. Palencia. A general equivalence theorem in the theory of discretization methods. Math. Comp., 45(171):143–152, 1985. 5 The Schauder fixed point theorem Hannes Frenander∗ January 5, 2015 Abstract The Brouwer and Schauder fixed point theorems are presented; the latter with a complete proof. An application of the Schauder fixed point theorem regarding solutions to partial diﬀerential equations has also been investigated. 1 Brouwer fixed point theorem For completeness, we start by introducing the Brouwer fixed point theorem (BFPT) since it is essential for understanding the proof of our main theorem: the Schauder fixed point theorem (SFPT). However, we neglect the proof since it would obscure the main points of this essay. The theorem is typically used in numerical analysis to prove existence of initial boundary value problems, which will be demonstrated in the end of this essay. In general, the BFPT states that every continous mapping that maps a convex set into itself has at least one fixed point. That is: Theorem 1.1 (Brouwer fixed point theorem). Let X be a compact and convex set and f : X → X a continuous map. Then there is a x0 ∈ X s.t f (x0 ) = x0 . To illustrate, lets take a simple example by considering the domain X = [0, 1] ⊂ R, i.e. the set of all real numbers between zero and one. This set is compact and convex. BFPT now states that every mapping f (x) that remains in this domain has at least one fixed point. For example, f (x) = x2 has a fixed point at x0 = 0, 1, f (x) = sin(x) has a fixed at x0 = 0 and so on. ∗ Division of Computational Mathematics, Department of Mathematics, Link¨oping university, SE-58183 Link¨oping, Sweden. [email protected] 1 2 Shauder fixed point theorem The SFPT was proved 1930 by Juliusz Schauder, but only for special cases. It remained for Robert Cauty to prove it in its final form in 2001. The theorem is a generalization of BFPT into vector spaces and claims that every continuous mapping from a convex vector space into itself has at least one fixed point. That is: Theorem 2.1 (Shauder fixed point theorem). Let X be a normed vector space and K ⊂ X a convex and compact subset of X. Then every continuous map f : K → K has a fixed point. Proof. Since K is compact, it has a finite subcover. That is: there is a finite sequence xi such that the open balls Bϵ (xi ) cover the set K, where ϵ > 0 is the radius of the balls. Lets define the continuous functions gi (x) as { ϵ − ||x − xi || if ||x − xi || ≤ ϵ gi (x) = 0 if ||x − xi || ≥ ϵ. ∑ Since gi (x) ≥ 0 and i gi (x) > 0 for all x ∈ K, we can define another continuous function in K: ∑ gi (x)xi g(x) = ∑i . i gi (x) 1 We can easily∑verify that g is a map from ∑ K into the convex hull K0 of K. That is, g = i αi xi where αi ≥ 0 and i αi = 1. One can also observe that ||g(x) − x|| ≤ ϵ for all x ∈ K. The continuous map H = g ◦ f maps K0 into itself. Since K0 is compact, the BFPT tells us that there is a z ∈ K0 such that H(z) = g(f (z)) = z and therefore ||f (z) − z|| = ||f (z) − g(f (z))|| ≤ ϵ. Since ϵ can be arbitrarily small, there is for all m ∈ N a zm ∈ K such that ||f (zm ) − zm || ≤ 1 1 . m The ∑ convex hull K0∑of K is the smallest convex set that contains K. In this case, the set { i αi xi |αi > 0, i αi = 1} is a convex hull of K. 2 Remember that the theorem demands K to be compact, so there is a sub sequence (zm )k such that f ((zm )k ) → x0 for some x0 ∈ K. Hence, we have ||(zm )k − x0 || = ||(zm )k − f ((zm )k ) + f ((zm )k ) − x0 || ≤ ||(zm )k − f ((zm )k )|| + ||f ((zm )k ) − x0 || 1 ≤ + ||f ((zm )k ) − x0 || → 0 mk which means (zm )k → x0 . Since f is continuous, we also have f ((zm )k ) → f (x0 ). This means f (x0 ) = x0 , i.e. x0 ∈ K is a fixed point. 3 An application The Schauder fixed point theorem is commonly used to prove existence of solutions to partial diﬀerential equations (Pouso [2012],Cao et al. [2014],Agarwal et al. [2013]), and we will in this section consider an example given by Pouso [2012]. For clarity, we skip some details of the proof and the reader is referred to Pouso [2012] for the full version. Consider the problem xtt = f (t, x) x(0, t) = x(1, t) = 0 t ∈ I = [0, 1] (3.1) under the Carath´eodory’s conditions Pouso [2012]. Hence, in (3.1), the function f (t, x) is allowed to be discontinuous. We now prove that (3.1) has a solution. First, we demand that there is a M , such that |f (t, x)| ≤ M (t), and consider the domain ∫ t 1 K = {x ∈ C (I), x(0) = x(1) = 0, |xt (t) − xt (s)| ≤ M (r)dr} s which can be proved to be compact and convex subset of the Banach space C 1 (I) if we define the norm ||x|| = maxt∈I |x(t)| + maxt∈I |xt (t)|. Consider now the operator T , defined by ∫ t T x(t) = G(t, s)f (s, x(s))ds 0 3 where G(t, s) is the Green’s function to (3.1). T maps K into itself if all the above conditions are fulfilled. Therefore, according the SFPT, T has a fixed point: T x(t) = x(t) which is then the solution to (3.1). References R. Agarwal, S. Arshad, and D. O’regan. A Schauder fixed point theorem in semilinear spaces and applications. Fixed Point Theory and Applications, 306, 2013. Z. Cao, C. Yuan, and X. Li. Applications of Schauders fixed point theorem to semipositone singular diﬀerential equations. Journal of Applied Mathematics, 2014, 2014. H. Pouso. Schauder’s fixed-point theorem: new applications and a new version for discontinous operators. Boundary Value Problems, 92, 2012. 4 Existence of a weak solution to the Stokes equations Application of the Babuška-Brezzi inf-sup theorem to the Stokes equations Hanna Holmgren∗ January 5, 2015 Abstract The dynamics of incompressible viscous fluids are described by the Stokes equations if stationary laminar flow conditions with low Reynolds numbers are assumed, see Figure 0.1 for an example. The existence of a unique solution to the weak formulation of the Stokes equations is proved in this essay. First the Two Hilbert Spaces H10 (Ω) and L20 (Ω) are introduced and the weak formulation of the Stokes equations is derived. Then, the BabuškaBrezzi inf-sup theorem is applied to the corresponding variational problem to show existence and uniqueness of a weak solution. Figure 0.1: Laminar flow conditions on the River Derwent in North Yorkshire. 1 Division of Scientific Computing, Department of Information Technology, Uppsala university, SE-751 05 Uppsala, Sweden. [email protected] ∗ 1 1 The Stokes Equations The dynamics of incompressible fluids are generally described by the Navier-Stokes equations. These equations are derived by considering the conservation of mass, momentum and energy of a fluid flow. If the flow is “slow and calm” it is called a laminar flow and an example is illustrated in Figure 0.1. For steady laminar flows which have a small Reynolds number (intertial forces are small compared with viscous forces) the Navier-Stokes equations are reduced to the Stokes equations. In the Stokes equations all time dependent terms as well as the term describing advection have been neglected. This essay aims at proving the existence of a unique solution to the weak formulation of the Stokes equations. Assume an incompressible viscous fluid with kinematic viscosity ν > 0 is filling up a domain Ω ⊂ R3 with boundary Γ, then the Stokes system of equations is given by −ν∆u + ∇p = f, in Ω ∇ · u = 0, in Ω u = 0, on Γ (1.1) (1.2) (1.3) where f is a given volume force, u is the unknown vector describing the velocity of the fluid and p is the unknown pressure. The Two Hilbert Spaces H10(Ω) and L20(Ω) 2 Before deriving the weak formulation of the Stokes system (1.1)−(1.3) we first need to introduce two Hilbert spaces related to the two unknowns u and p. We assume the velocity u to be in the Hilbert space H10 (Ω) (here Ω ⊂ R3 ), which is the space of all vector fields v = (v1 , v2 , v3 ) with components vi in the space H01 defined by H01 (Ω) = {v ∈ H 1 (Ω) : v|Γ = 0}. (2.1) Remember that the space H 1 (Ω) is equipped with the norm defined by kvk2H 1 (Ω) = kvk2L2 (Ω) + k∇vk2L2 (Ω) . The Hilbert space(s) H10 (Ω) is equipped with the inner product and seminorm defined by (u,v) H10 (Ω) = Z Ω ∇u : ∇v dx |v|2H10 (Ω) = (v,v)H10 (Ω) 1 Picture from http://evidence.environment-agency.gov.uk 2 (2.2) (2.3) where Z Ω ∇u : ∇v dx = Z Ω ∇u1 · ∇v1 dx + Z Ω ∇u2 · ∇v2 dx + Z Ω ∇u3 · ∇v3 dx. Further, we assume the pressure p to be in the Hilbert space L20 (Ω) defined by L20 (Ω) 2 = {q ∈ L (Ω) : Z Ω q dx = 0} (2.4) and equipped with the usual inner product and norm of the space L2 (Ω). 3 The Weak Form of Stokes Equations Now, the weak formulation is derived by first multiplying the momentum equation (1.1) by a test vector v ∈ H10 (Ω) and then integrating by parts which gives the following weak formulation of the momentum equation ν Z Ω ∇u : ∇v dx − Z Ω (∇ · v)p dx = Z Ω f · v dx, where all boundary terms from the integration by parts vanishes since v|Γ = 0. Further, multiplying the incompressibility constraint (1.2) by a test function q ∈ L20 (Ω) and integrating yields Z Ω (∇ · u)q dx = 0. Summarizing, the variational formulation derived from the Stokes system of equations reads: Find the solution pair (u,p) ∈ H10 (Ω) × L20 (Ω) such that a(u,v) + b(v,p) = l(v) ∀v ∈ H10 (Ω) b(u,q) = 0 ∀q ∈ L20 (Ω), (3.1) (3.2) where the following linear forms have been introduced: a(u,v) = ν Z b(u,q) = − l(v) = Z ZΩ Ω ∇u : ∇v dx, (3.3) (∇ · u)q dx, (3.4) Ω f · v dx. 3 (3.5) 4 The Babuška-Brezzi inf-sup theorem For the abstract variational problem: Find u ∈ V , such that a(u,v) = l(v), ∀v ∈ V, where V is a Hilbert space with inner product (·,·), a(u,v) is a coercive continuous bilinear form on V and l(v) is a continuous linear form on V , the Lax-Milgram lemma can be used to prove the existence and uniqueness of a solution [1] . Remember that a bilinear form a(·,·) is said to be V -coercive if there is a positive constant m such that mkvk2V ≤ a(v,v) ∀v ∈ V. However, in the case of the weak formulation of the Stokes equations (3.1)-(3.2), the existence of a unique solution does not follow from the Lax-Milgram lemma alone, since it is impossible to establish coercivity for the bilinear form b(·, ·) in (3.4) [2]. In this case we need the Babuška-Brezzi inf-sup theorem [1]: Theorem 1: The Babuška-Brezzi inf-sup theorem. Let V and M be two Hilbert spaces, and let a(·,·) : V × V −→ R and b(·,·) : V × M −→ R be two continuous bilinear forms with the following properties: There exists a positive constant α such that a(v,v) ≥ αkvk2V for all v ∈ U0 = {v ∈ V : b(v,q) = 0 ∀q ∈ M }, i.e. a(·,·) is U0 -coercive, and there exists a positive constant β such that |b(v,q)| ≥ β, q∈M,q6=0 v∈V,v6=0 kvkV kqkM inf sup where the latter is called the Babuška-Brezzi inf-sup condition. Finally, let l : V −→ R and g : M −→ R be two continuous linear forms. Then the variational problem: Find (u,p) ∈ V × M such that a(u,v) + b(v,p) = l(v) b(u,q) = g(q) has one and only one solution. 4 ∀v ∈ V, ∀q ∈ M, 5 Existence and uniqueness of a weak solution We use the Babuška-Brezzi inf-sup theorem to prove existence and uniqueness of a solution to the variational problem (3.1)-(3.2) derived from the Stokes equations. Let V = H10 (Ω) and M = L20 (Ω), where V and M are the two spaces from the Babuška-Brezzi inf-sup theorem. First, we prove that a(·,·) is U0 -coercive: 1. a(·,·) is H10 (Ω)-coercive since a(v,v) = ν|v|2H10 (Ω) ∀v ∈ H10 (Ω). 2. Then a(·,·) is also coercive on U0 = {v ∈ H10 (Ω) : b(v,q) = 0 ∀q ∈ L20 (Ω)} since the above equality holds for all v ∈ H10 (Ω). Then, for proving the second part of the Babuška-Brezzi inf-sup theorem (i.e. that b(·,·) fulfills the Babuška-Brezzi inf-sup condition) we need the following theorem [1]: Theorem 2. Let Ω be a domain in RN . Then the injective continuous linear operator div : (ker div)⊥ −→ L20 (Ω) is surjective and has a continuous inverse. Here “div” is another notation used for the divergence, i.e. divv = ∇ · v for any v. The space (ker div) is defined by (ker div) = {v ∈ H10 (Ω) : divv = 0 in L20 (Ω)}. Now, we prove that b(·,·) fulfills the Babuška-Brezzi inf-sup condition: 1. Since div : (ker div)⊥ −→ L20 (Ω) is surjective (by Theorem 2), there exists a unique vector field w ∈(ker div)⊥ ⊂ H10 (Ω) such that divw = q in L20 (Ω), and besides, since div : (ker div)⊥ −→ L20 (Ω) has a continuous inverse (also by Theorem 2), there exists a constant C such that |w|H10 (Ω) ≤ CkqkL20 (Ω) 5 ∀q ∈ L20 (Ω). 2. Now we can prove the Babuška-Brezzi inf-sup condition holds for b(·,·), for each non-zero q ∈ L20 (Ω): R |b(v,q)| | Ω (∇ · v)q dx| sup = sup v∈H10 (Ω),v6=0 |v|H10 (Ω) kqkL20 (Ω) v∈H10 (Ω),v6=0 |v|H10 (Ω) kqkL20 (Ω) h ∇ · v = divv i R | Ω (divv)q dx| = sup v∈H10 (Ω),v6=0 |v|H10 (Ω) kqkL20 (Ω) h i since w ∈ (ker div)⊥ ⊂ H10 (Ω) we can choose v = w R | (divw)q dx| ≥ Ω |w|H10 (Ω) kqkL20 (Ω) h divw = q i R q 2 dx| = |w|H10 (Ω) kqkL20 (Ω) | Ω hR = q 2 dx = kqk2L2 (Ω) 0 i kqkL20 (Ω) |w|H10 (Ω) h ≥ Ω |w|H10 (Ω) ≤ CkqkL20 (Ω) i 1 =β C So, by the Babuška-Brezzi inf-sup theorem, there exists a unique solution pair (u,p) ∈ H10 (Ω) × L20 (Ω) to the variational problem (3.1)-(3.2) derived from the Stokes equations. 6 References [1] Philippe G. Ciarlet, Linear and Nonlinear Functional Analysis with Applications. Society for Industrial and Applied Mathematics, Philadelphia, 2013. [2] Mats G. Larson, Fredrik Bengtzon, The Finite Element Method - Theory, Implementation and Applications. Springer, Umeå, Sweden, 2012. 7 The Lax–Milgram Theorem Andrea Alessandro Ruggiu∗ January 7, 2015 Abstract In this essay we state and prove the Lax–Milgram Theorem, a result which claims well–posedness of variational problems. This theorem is here proven through the classical approach, that is to say in this work the Banach Fixed Point Theorem is not taken into account. 1 Introduction Various problems in science can be formulated through PDEs, which necessarly don’t have a classic solution. Solving a problem weakly means find a solution for the problem, called weak solution, whose derivatives could also not exist in a classical sense. Not rarely they are the only solutions that is possible to find. In particular, it’s possible to formulate any differential problem as a variational problem of the following kind given F ∈ H ∗ , with H ∗ dual space of H, find u ∈ H such that a (u, v) = F (v) , ∀v ∈ H. (1) A common example is the Poisson problem −∆u = f, x ∈ Ω u = 0, x ∈ ∂Ω (2) that can be formulated as a variational problem through the integration by parts. ∗ Computational Mathematics, Department of Mathematics, Link¨ oping university, SE581 83 Link¨oping, Sweden. [email protected] 1 Indeed, multiplying the equation by v ∈ H01 (Ω) and integrating by parts leads to Z Z (3) ∇u · ∇v dx = f v dx, ∀v ∈ H01 (Ω) . Ω Ω The solution of (3), u ∈ 2 H01 (Ω), is said weak solution for (2). Background theorems In this section we recall two important theorems that we want to use in order to prove the Lax–Milgram Theorem. Both of them were studied and analysed in the Numerical Functional Analysis course. The first one is the Hilbert Projection Theorem. Theorem 2.1. Let us consider a Hilbert space H, a non–empty subspace V ⊂ H closed in H and x ∈ H. Then ∃!v = PV (x) ∈ V such that kx − vk = inf v∈V kx − vk = d ≥ 0. Moreover • PV (x) = x ⇐⇒ x∈V, • QV (x) = x − PV (x) ∈ V ⊥ . The second theorem that we want to use is the Riesz Representation Theorem Theorem 2.2. Let H be a Hilbert space. Then ∀L ∈ H ∗ ∃!u ∈ H s.t. L (v) = (u, v) ∀v ∈ H and moreover kukH = kLkH ∗ . 3 Lax–Milgram Theorem In this section we prove the following statement Theorem 3.1. Let H be a Hilbert space. If a (·, ·) is a bilinear form such that • ∃M > 0 such that |a (u, v)| ≤ M kuk kvk, • ∃α > 0 such that a (u, u) ≥ α kuk2 , 2 and F ∈ H ∗ is linear and bounded, then ∃!u solution for (1) and kuk ≤ 1 kF k∗ , α (4) where k·k∗ is the norm associated to H ∗ . Proof. By Theorem 2.2 ∃!z ∈ H s.t. F (v) = (z, v) , ∀v ∈ H and kF k∗ = kzk. But a (u, ·) : V 7→ a (u, v) is a linear operator in H ∗ , then, again by Theorem 2.2 we know that ∃!h ∈ H s.t. a (u, v) = (h, v) , ∀v ∈ H with u fixed. Thus we can call h = A (u). Notice that if u is fixed, then h exists and it is unique. Therefore (1) can be rewritten as (A (u) , v) = (z, v) , ∀v ∈ H, which implies A (u) = z, A : H → H. In order to show the uniqueness of u we have to prove that A is invertible. This proof requires 5 steps • A is linear. As stated before, by Theorem 2.2 (A (u) , v) = a (u, v) , ∀v ∈ H. Then we can use the bilinearity of a (·, ·) in order to prove the linearity of A (A (λu1 + µu2 ) , v) = a (λu1 + µu2 , v) = λa (u1 , v) + µa (u2 , v) = = λ (A (u1 ) , v) + µ (A (u2 ) , v) = = (λA (u1 ) + µA (u2 ) , v) , ∀v ∈ H. This implies that A (λu1 + µu2 ) = λA (u1 ) + µA (u2 ), that is to say A is linear. For this reason from now on we write Au instead of A (u). • A is bounded. We can write kAuk2 = (Au, Au) = a (u, Au) ≤ M kuk kAuk . (5) If kAuk = 0, then the boundedness follows trivially. Therefore we assume kAuk = 6 0: dividing by kAuk both the left and the right–hand side of (5), the boundedness of A follows. 3 • A is injective. This statement is equivalent to nul (A) = {0}, where nul is the null space of A. Using the hypotheses and Cauchy–Schwarz inequality in a (u, u) = (Au, u) leads to α kuk2 ≤ a (u, u) = (Au, u) ≤ kAuk kuk . This implies that if u 6= 0 (trivially an element of nul (A)) then kAuk ≥ α kuk, that is to say kuk ≤ 1 kAuk , α ∀u ∈ H. (6) Therefore if we consider u ∈ nul (A) (kAuk = 0), then u = 0. This means that the only element in the nullspace of A is 0, then A is injective. • The image of A is closed. Let yn ∈ Im (A) an arbitrary convergent sequence in H, let’s say yn → y. The statement is proven if, and only if, y ∈ Im (A). But since yn ∈ Im (A) ⇐⇒ ∃xn s.t. yn = Axn we can also prove the statement by showing the existence of x s.t. y = Ax. As a first step we can prove that xn is convergent. Using (6) and the linearity of A we can write 0 ≤ kxn − xm k ≤ 1 1 1 kA (xn − xm )k = kAxn − Axm k = kyn − ym k . α α α But since yn is convergent in H, then it is also Cauchy. In particular, this implies by the above equation that xn is Cauchy in H. The completeness of H leads to the convergence of xn to a certain x ∈ H. By hypothesis it is known that Axn = yn → y. (7) But, on the other hand, we know by the continuity of A (it is linear and bounded) that Axn → Ax. (8) By (7), (8) and the uniqueness of the limit we have y = Ax, which prove the closedness of Im (A) since x is such that y = Ax ∈ Im (A). 4 • A is surjective, that is to say Im (A) = H. We want to prove this statement by contradiction: let us assume that Im (A) ( H. Then ∃q0 ∈ / Im (A) and applying Theorem 2.1 (notice that Im (A) is closed by the previous point) leads to QIm(A) (q0 ) = q0 − PIm(A) (q0 ) ∈ Im (A)⊥ . If we define q0 − PIm(A) (q0 ) q= q0 − PIm(A) (q0 ) , then q ∈ Im (A)⊥ and kqk = 1. Then we can write 0 = (Aq, q) = a (q, q) ≥ α kqk2 = α > 0. Since we have a contradiction, then is impossible to have Im (A) ( H and therefore A is surjective. Finally, we have proven that A is invertible by proving that A is injective and surjective. In particular Au = z has the unique solution u = A−1 z and this means that there exists a unique solution for (1). In order to prove the remaining part of the statement we use (6) and Theorem 2.2 1 1 1 kuk ≤ kAuk = kzk = kF k∗ . α α α 4 Example of application Let’s take into account the problem (2) and its weak formulation given by (3). We can write Z Z a (u, v) ∇u · ∇v dx, F (v) = f v dx. (9) Ω Ω With this notation we have written (3) as a problem of the same kind of (1) with H ≡ H01 (Ω). It can be shown that a suitable norm for H01 (Ω) is given by k·kH 1 (Ω) = k∇·kL2 (Ω) 0 and, equivalently, we can use as inner product in H01 (Ω) (u, v)H 1 (Ω) = (∇u, ∇v)L2 (Ω) . 0 5 We can easily check that |a (u, v)| ≤ k∇ukL2 (Ω) k∇vkL2 (Ω) = kukH 1 (Ω) kvkH 1 (Ω) 0 and 0 a (u, u) = k∇uk2L2 (Ω) = kuk2H 1 (Ω) . 0 Moreover F is linear and it is bounded, indeed Z f v dx ≤ kf kL2 (Ω) kvkL2 (Ω) ≤ cp kf kL2 (Ω) kvkH 1 (Ω) , F (v) = 0 Ω where we have used the Cauchy–Schwarz and the Poincar´ e inequalities with constant cp . Then the Lax–Milgram theorem holds with α = M = 1 and therefore (2) admits a unique solution u ∈ H01 (Ω) that is such that kuk ≤ cp kf kL2 (Ω) . References [1] Evans L.C., Partial Differential Equations, American Mathematical Society, PP.297-299, 1997. [2] Salsa S., Equazioni a derivate parziali. Metodi, modelli e applicazioni, Springer, Chapter 6, 2nd edition, 2010. 6 A time dependent mapping from Cartesian coordinate systems into curvilinear coordinate systems and the geometric conservation law Samira Nikkar∗ January 9, 2015 Abstract A time-dependent transformation that maps a moving domain in the physical space (in terms of the Cartesian coordinates) into a fixed domain in the computational space (in terms of the curvilinear coordinates) is considered. The continuous transformation is usually non-singular, which means that the Geometric Conservation Law (GCL) holds exactly. In the numerical set up, the differential operators and the metric formulations must satisfy some certain criteria in order for the Numerical GCL (NGCL) to be fulfilled. With the use of the Summation-by-Parts (SBP) operators for the space and time discretization, we can prove that even for the discrete mapping, the NGCL is also guaranteed. 1 The continuous problem Consider a time-dependent transformation from the Cartesian coordinates into curvilinear coordinates, which results in a fixed spatial domain, as x = x(τ, ξ, η), y = y(τ, ξ, η), t = τ, ξ = ξ(t, x, y), η = η(t, x, y), τ = t. (1.1) such that 0 ≤ ξ ≤ 1, 0 ≤ η ≤ 1, 0 ≤ τ ≤ T , see Figure 1. The Jacobian ∗ Division of scientific computing, Department of Mathematics, Linkoping university, Sweden. [email protected] 1 y η c b c' d' d Ω n a x ab bc cd da a' b' ξ a'b' : South (s) b'c' : East (e) c'd' : North (n) d'a' : West (w) Figure 1.1: A schematic of the moving and fixed domains and boundary definitions. matrix of the transformation is xξ yξ 0 [J] = xη yη 0 , xτ y τ 1 (1.2) Jξt = xη yτ − xτ yη , Jξx = yη , Jξy = −xη Jηt = yξ xτ − xξ yτ , Jηx = −yξ , Jηy = xξ , (1.3) The relation between [J], and its inverse, which transforms the derivatives back to the Cartesian coordinates leads to the metric relations in which J = xξ yη −xη yξ > 0 is the determinant of [J]. If we want to preserve any conservation property while transforming the physical domain into the computational domain, the transformation and the metric relations must satisfy the GCL Farhat et al. (2001); Sj¨ogreen et al. (2014), summarized as the following Jτ + (Jξt )ξ + (Jηt )η = 0, (Jξx )ξ + (Jηx )η = 0, (Jξy )ξ + (Jηy )η = 0. (1.4) In practice (1.4) holds for any non-singular transformation. Now let us investigate the requirements for the discrete transformation such that the numerical version of (4) also holds. 2 2 The discrete problem The spatial computational domain Ω is a square in ξ, η coordinates, see Figure 1.1, and discretized using N and M nodes in the direction of ξ and η respectively. In time we use L time levels from 0 to T. The fully-discrete grid vector is a column vector of size LM N organized as follows Z0 .. . Z= [Zk ] .. . ZL Z0 Z0 .. .. . . ; [Zk ] = [Zi ] ; [Zi ] = [Zj ] , k .. .. . . ZM ki ZN k (2.1) where Zkij = Z(τk , ξi , ηj ) and Z ∈ {x, y}. The first derivative xξ is approximated by Dξ x, where Dξ is a so-called SBP operator of the form Dξ = Pξ−1 Qξ , (2.2) and x = [x0 , x1 , · · · , xN ]T is the x coordinate of the grid points. Pξ is a symmetric positive definite matrix, and Q is an almost skew-symmetric matrix that satisfies Qξ + QTξ = E1 −E0 = B = diag(−1, 0, ..., 0, 1). (2.3) In (2.3), E0 = diag(1, 0, ..., 0) and E1 = diag(0, ..., 0, 1). The η and τ directions are discretized in the same way. A finite difference approximation including the time discretization Nordstr¨om and Lundquist (2013), on SBP-SAT form, is constructed by extending the one-dimensional SBP operators in a tensor product fashion as Dτ = Pτ−1 Qτ ⊗ Iξ ⊗ Iη , Dξ = Iτ ⊗ Pξ−1 Qξ ⊗ Iη , Dη = Iτ ⊗ Iξ ⊗ Pη−1 Qη (2.4) where ⊗ represents the Kronecker product Loan (2000). Note that in (2.4) we have used the same names for the differential operators in multi dimensional cases as the operators in the one dimensional case. All matrices in the first position are of size L×L, the second position N×N , the third position M×M . I denotes the identity matrix with a size consistent with its position in the Kronecker product. 3 The Kronecker product is bilinear and associative. For square matrices the following rules exist Nordstr¨om and Berg (2013), (A⊗B)(C ⊗D) = (AC ⊗BD), (A⊗B)−1 = A−1 ⊗B −1 , (A⊗B)T = AT ⊗B T . (2.5) For later reference we need Lemma 2.1. The difference operators in (2.4) commute. Proof. The properties (2.5) of the Kronecker product lead to Dτ Dξ = (Pτ−1 Qτ ⊗ Iξ ⊗ Iη )(Iτ ⊗ Pξ−1 Qξ ⊗ Iη ) = Pτ−1 Qτ ⊗ Pξ−1 Qξ ⊗ Iη = (Iτ ⊗ Pξ−1 Qξ ⊗ Iη )(Pτ−1 Qτ ⊗ Iξ ⊗ Iη ) = Dξ Dτ . The proof is analogous for the other coordinate combinations. We show Lemma 2.2. The Numerical Geometric Conservation Law (NGCL) holds: Jτ + (Jξt )ξ + (Jηt )η = 0, (Jξx )ξ + (Jηx )η = 0, (Jξy )ξ + (Jηy )η = 0. (2.6) Proof. Consider the following definitions, Jτ (Jξ t )ξ (Jη t )η (Jξx )ξ (Jξy )ξ (Jηx )η (Jηy )η = = = = = = = diag[Dτ (Dη M (1) − Dξ M (2) )] diag[Dξ (Dτ M (2) − Dη M (3) )] diag[Dη (Dξ M (3) − Dτ M (1) )] diag[Dξ (Dη y)] diag[−Dξ (Dη x)] diag[−Dη (Dξ y)] diag[Dη (Dξ x)] (2.7) in which x and y are the discrete Cartesian coordinates defined in (2.1). Also M (1) = diag(y)(Dξ x), M (2) = diag(y)(Dη x) and M (3) = diag(y)(Dτ x). By Lemma 2.1 we find that the NGCL holds exactly. 3 Summary and conclusions We have considered a time-dependent curvilinear coordinates. By using SBP operators in space and time and a special definitions for the metric coefficients the Geometric Conservation Law is proven to hold numerically. 4 References C. Farhat, P. Geuzaine, C. Grandmont, The discrete geometric conservation law and the nonlinear stability of ALE schemes for the solution of flow problems on moving grids, Journal of Computational Physics 174 (2001) 669–694. B. Sj¨ogreen, H. C. Yee, M. Vinokur, On high order finite-difference metric discretizations satisfying GCL on moving and deforming grids, Journal of Computational Physics 265 (2014) 211–220. J. Nordstr¨om, T. Lundquist, Summation-by-parts in time, Journal of Computational Physics 251 (2013) 487–499. C. F. V. Loan, The ubiquitous Kronecker product, Journal of Computational and Applied Mathematics 123 (2000) 85–100. J. Nordstr¨om, J. Berg, Conjugate heat transfer for the unsteady compressible Navier-Stokes equations using a multi-block coupling, Computers & Fluids 72 (2013) 20–29. 5 The Babuˇska Inf-Sup Condition Simon Sticko∗ January 12, 2015 Abstract A general weak formulation, typically arising from a linear timeindependent partial differential equation, is discussed. The Babuˇska inf-sup condition is shown to guarantee that the weak formulation is a well-posed problem. 1 Introduction The finite element method applied to a linear time-independent partial differential equation typically leads to a weak formulation of the following form: find u ∈ U such that a(u, v) = l(v) ∀v ∈ V. (1.1) Here U and V are Hilbert spaces, a(·, ·) : U × V → R a bounded bilinear form: |a(u, v)| ≤ αC kukU kvkV , αC ∈ (0, ∞) and l(v) : V → R is a bounded linear functional. Our goal is to find a condition for when (1.1) is a well-posed problem. That is, the problem should fulfill the following requirements: (E) There exist a solution u. (U) The solution is unique. (C) The solution depends continuously on initial data. ∗ Division of Scientific Computing, Department of Information Technology, Uppsala university, SE-751 05 Uppsala, Sweden. [email protected] 1 2 Theory In order to understand if the problem in (1.1) is well posed or not we would like to reformulate it into something which is a bit more familiar. In the following we denote the scalar product in V and U by h·, ·iV and h·, ·iU respectively. Since U and V are Hilbert spaces a(·, ·) has a Ritz-representation A : U → V such that: a(u, v) = hAu, viV , where A is a bounded linear operator with the same norm as a(·, ·). Also l(·) has a Ritz-representations f ∈ V : l(v) = hf, viV . Using this we can now express (1.1) as hAu − f, viV = 0 ∀v ∈ V. Since this should hold for all v ∈ V it holds in particular for v = Au − f . So this leads to the equation Au = f. (2.1) This formulation is equivalent to (1.1), but it’s easier to understand what is required for the problem to be well posed. In order for a solution of (2.1) to exist for an arbitrary f ∈ V we must require that there is at least one u ∈ U such that Au = f . Thus A must be a surjective mapping. If we want uniqueness we can not have two different elements u1 , u2 ∈ U such that Au1 = f and Au2 = f . Thus we must also require A to be injective. Finally, consider two different right-hand sides f1 , f2 , such that Au1 = f1 and Au2 = f2 . Since A is a linear operator we obtain A(u1 − u2 ) = f1 − f2 , which leads to ku1 − u2 kU ≤ A−1 V kf1 − f2 kV . Thus if we want a small change in right hand side to generate a small change in the solution we must require that A−1 is a bounded operator. So in summary we have the following requirements for our problem to be well posed: 2 (E) A is surjective. (U) A is injective. (C) A−1 is a bounded operator. Our goal is to show that A fulfills these three requirements. This was shown by Babuˇska [1971] and can be summarized as: Theorem 2.1 (Babuˇska-Lax-Milgram). The weak formulation in (1.1) is well posed if there exists αV , αU > 0 such that a(·, ·) fulfills: αV kukU ≤ sup v∈V αU kvkV ≤ sup u∈U |a(u, v)| ∀u ∈ U kvkV |a(u, v)| ∀v ∈ V. kukU (2.2) (2.3) Proof. The three requirements (E), (U ) and (C) are proved separately in Lemma 2.2, 2.3 and 2.4. Remark. The conditions (2.2) and (2.3) are frequently referred to as inf-sup conditions. The reason for this is that they can be formulated as: there exist αU , αV > 0 such that |a(u, v)| ≥ αV inf sup u∈U v∈V kuk kvk U V inf sup v∈V u∈U |a(u, v)| ≥ αU . kukU kvkV Lemma 2.2 ((U ) Injectivity). If a(·, ·) satisfies (2.2) its Ritz-representation is injective. Proof. By the definition of the Ritz-representation and the Cauchy-Schwartz inequality we have | hAu, viV | kAukV kvkV |a(u, v)| = ≤ = kAukV . kvkV kvkV kvkV By taking the supremum and utilizing (2.2) we obtain: αV kukU ≤ sup v∈V |a(u, v)| ≤ kAukV . kvkV 3 (2.4) This gives us that Au = 0 =⇒ u = 0 which implies injectivity since Au1 − Au2 = 0 =⇒ A(u1 − u2 ) = 0 =⇒ u1 − u2 = 0. Lemma 2.3 ((E) Surjectivity). If a(·, ·) satisfies (2.2) and (2.3) its Ritzrepresentation is surjective. Proof. Consider the range of A: R(A), this is a subspace of V . In order to show that this is a complete subspace let {yj }∞ j=1 be a Cauchy sequence in R(A). For every > 0 there exists N such that kyi − yj kV ≤ ∀i, j ≥ N. But since yi ∈ R(A) we have yi = Axi with xi ∈ U so we get kAxi − Axj kV ≤ ∀i, j ≥ N, which by linearity and (2.4) gives us: αV kxi − xj kU ≤ kA (xi − xj )kV ≤ ∀i, j ≥ N. This shows us that {xj }∞ j=1 is a Cauchy sequence, which converges since U is a Hilbert space. Denote the limit by x and let y = Ax. Since we originally assumed that A is a bounded operator we now see that yj converges to y: ky − yj kV = kAx − Axj kV ≤ αC kx − xj kU → 0 j → ∞. Thus the conclusion is that R(A) is a complete subspace of V . This can only happen if R(A) is also closed. Since R(A) is a closed subspace of V we can decomposed V as a direct product space: M V = R(A) R(A)⊥ . (2.5) But consider now a point v ∈ R(A)⊥ . From (2.3) we get αU kvkV ≤ sup u∈U | hAu, vi | = 0, kukU since Au ∈ R(A) and v ∈ R(A)⊥ . So this implies v = 0, which means that R(A)⊥ = {0} since v was arbitrary. By (2.5) we must then have R(A) = V , so that A is surjective. 4 Lemma 2.4 ((C) Bounded Ritz-Inverse). Given that a(·, ·) fulfills (2.2), the inverse of the Ritz-projection fulfills −1 A ≤ 1 , U αV and thus is a bounded operator. Proof. By rewriting (2.4) we obtain kukU ≤ 1 kAukV . αV Since we now know that A is a bijection we knot that for each u ∈ U there exist an unique f ∈ V such that u = A−1 f . Inserting this gives us −1 A f ≤ 1 kf k . V U αV Since this holds for all f ∈ V we obtain −1 −1 A = sup kA f kU ≤ 1 . V kf kV αV f ∈V References I. Babuˇska. Error-bounds for finite element method. Numerische Mathematik, 16:322–333, 1971. ISSN 0029599X. doi: 10.1007/BF02165003. 5 Uniformly best wavenumber approximations by spatial central difference operators Viktor Linders∗ January 12, 2015 Abstract We show that the problem of finding uniformly best wavenumber approximations by central difference schemes is equivalent to approximating a continuous function from a finite dimensional subspace. A characterisation theorem for best approximations is proven. 1 Introduction and motivation Consider a central difference approximation of a first derivative ux of some function u(x, t) at the point x = xi , ∂u ∂x i p + O(∆x2p ) = 1 X (p) c (ui+k − ui−k ). ∆x k=1 k (1.1) (p) Here ∆x is the spatial step size of the discretisation and ck are the coefficients of the 2pth order classical central difference scheme. It is well known that the numerical dispersion relation of this scheme is ξ¯c = 2 p X (p) ck sin (kξ). (1.2) k=1 where ξ = κ∆x is the normalised wavenumber. The subscript c serves as a reminder that (1.2) corresponds to a classical stencil. Here we have introduced an overbar to signify that the wavenumber is a numerical approximation. ∗ Division of Computational Mathematics, Department of Mathematics, Link¨ oping University, SE-581 83 Link¨oping, Sweden. [email protected] 1 Noting that the smallest wavelength the scheme can resolve is λmin = 2∆x, the largest wavenumber is κmax = 2π/λmin = π/∆x so that we have |ξ| ≤ π. Typically, waves with high frequencies and wavenumbers require small spatial increments, ∆x, in order to be properly resolved. Over sizeable intervals, ¯ may come to dominate the error in the the so called dispersion error, |ξ − ξ|, approximation, which gravely restricts the ∆x that may be used. Problems of this type are commonly encountered in computational fluid dynamics, aeroacoustics, electromagnetism, elasticity, seismology, and other fields where energy is propagated. Wave properties encoded within the dispersion relation include phase velocity, group velocity and dissipation. It is therefore of interest to develop difference schemes that preserve the analytic dispersion relation of the governing equations for a wide range of spatial increments. 2 Problem formulation Let us perturb the stencil (1.1) by adding n new coefficients, without increasing the accuracy. Thus, the scheme we consider has the form p+n 1 X ∂u 2p ak (ui+k − ui−k ). (2.1) + O(∆x ) = ∂x i ∆x k=1 Such a scheme uses (p + n) points on either side of xi to approximate the derivative, however the formal order of accuracy remains O(∆x2p ). In view of (1.2) the numerical wavenumber corresponding to this scheme is ξ¯ = 2 p+n X ak sin (kξ). (2.2) k=1 Our goal may thus be formulated in terms of the following problem: Choose ak , k = 1, . . . , p + n, such that the error function ¯∞ kEk∞ := kξ − ξk is minimised, ensuring that the stencil has formal accuracy O(∆x2p ). (p) (2.3) Note that if ap+j = 0, ∀j ≥ 1, we have ak = ck . By linearity we may thus write n X (p) (j) ak = c k + αk ap+j , k = 1, . . . , p (2.4) j=1 2 (j) where the parameters αk are independent of ap+j . We summarise in the following proposition: Proposition 2.1. The numerical wavenumber of the 2pth order central difference stencil (2.1) can be written as ξ¯ = ξ¯c + n X ap+j φj (ξ) j=1 where ξ¯c is the wavenumber apprpximation of the classical central difference stencil and p X (j) φj (ξ) = 2 sin ((p + j)ξ) + 2 αk sin (kξ). k=1 (j) The coefficients αk are independent of ap+j for all k = 1, . . . , p and j = 1, . . . , n. Proof. Inserting (2.4) into (2.2) and collecting terms multiplying ap+j immediately gives the desired result. The set {φj } is linearly independent and thus span some n-dimensional vector space, which we will denote by Ξn . We may thus consider each φj as a basis function of Ξn . P Let ψ(ξ, a) = nj=1 ap+j φj (ξ). We see from Proposition 2.1 that the problem (2.3) may equivalently be written find a ∈ Rn that minimises kE(ξ, a)k∞ = kEc (ξ) − ψ(ξ, a)k∞ , (2.5) where Ec (ξ) = ξ − ξ¯c is the error of the classical stencil and ψ(ξ, a) ∈ Ξn . 3 Approximation theory For convenience we use the following definition: Definition 3.1. If a vector a solves problem (2.5), then ψ(ξ, a) is called a best approximation. Finding and characterising best approximations are central problems in the field of approximation theory. It is well-known that a solution to problem (2.5) exists Riesz [1918]. It is in place to introduce some relevant concepts: 3 Definition 3.2. A set of functions, {θi }, i = 1, . . . , n forms a Chebyshev set on the interval [0, ξmax ] if any non-trivial linear combination has at most (n − 1) zeros in [0, ξmax ]. We will also need the following definition: Definition 3.3. Let M (a) denote the set of points {ξ1 , ξ2 , . . . } at which |E(ξ, a)| = kE(ξ, a)k∞ . Then E(ξ, a) is said to alternate n times on [0, ξmax ] if M (a) contains (n + 1) points 0 ≤ ξ1 < ξ2 < · · · < ξn+1 ≤ ξmax such that E(ξi , a) = −E(ξi+1 , a), i = 1, 2, . . . , n. The set {ξi } is referred to as an alternating set. The usefulness of the above definitions becomes apparent when we consider the following well-known extension of the Chebyshev alternation theorem: Theorem 3.1. Let {θi (ξ)}, i = 1, . . . , n be a Chebyshev set on [0, ξmax ]. Then a solves problem (2.5) if and only if there is an alternating set of (n + 1) points in [0, ξmax ]. Such a solution exists and is unique. For a proof, see e.g. Watson [1980]. Theorem 3.1 identifies best approximations as those whose error oscillates between a positive and a negative fixed value. The main task left is thus to show that the set {φj } constitutes a Chebyshev set. 4 Characterisation of best approximations The following observation is useful. The proof is omitted for brevity. Lemma 4.1. Consider the function f (p) (x, a) = 1 − 2 p+n X ak kTk (x) k=1 where ak is defined as before and Tk (x) is the k th order Chebyshev polynomial of the first kind, uniquely defined through the relation Tk (cos (φ)) = cos (kφ). Then f (p) has a root at x = 1 of multiplicity p. We are in position to prove our main result: Theorem 4.2. The set {φj }, j = 1, . . . , n defined in Proposition 2.1 forms a Chebyshev set on the semi-open interval (0, ξmax ]. 4 Proof. Recall the form of the basis functions, φj (ξ) = 2 sin ((p + j)ξ) + 2 p X (j) αk sin (kξ). k=1 Note first that φj (ξ) is a trigonometric polynomial. Thus the number of zeros of any non-trivial combination, ψ(ξ, a) is finite in a bounded interval. The function f (p) (x, a) is a polynomial in x of degree p + n with a root of multiplicity p at x = 1, i.e. f (p) (x, a) = (1 − x)p Pn (x, a), where Pn (x, a) is some polynomial of degree n that depends linearly on a. Let x = cos (ξ). We have f (p) (cos (ξ), a) = 1 − 2 p+n X k=1 ak kTk (cos (ξ)) = 1 − 2 p+n X ak k cos (kξ) = k=1 dE . dξ Thus, E(ξ, a) has an extreme point at ξ = 0 and at most n other extreme points ξ1 ≤ ξ2 ≤ · · · ≤ ξn in [0, ξmax ]. If ξ1 = 0 the stencil would be of higher order than O(∆x2p ) so we may assume that 0 < ξ1 . Consider now another error function, E(ξ, b). Linearity gives ∂ (E(ξ, a) − E(ξ, b)) = (1 − cos (ξ))p [Pn (cos (ξ), a) − Pn (cos (ξ), b)] ∂ξ = (1 − cos (ξ))p Pn (cos (ξ), a − b) but also ∂ (E(ξ, a) − E(ξ, b)) ∂ = [(Ec (ξ) − ψ(ξ, a)) − (Ec (ξ) − ψ(ξ, b))] ∂ξ ∂ξ ∂ = ψ(ξ, b − a), ∂ξ where, as before, ψ is a linear combination of the basis functions, {φj }. The vectors a and b are general so we may write c = b − a for any vector c ∈ Rn . Thus ψ(ξ, c) has at most n extrema in the open interval (0, ξmax ]. Obviously there is at most one zero between any two consecutive extreme points. Thus ψ(ξ, c) has at most n − 1 zeros in (0, ξmax ]. This completes the proof. 5 Corollary 4.3. Let {φj (ξ)}, j = 1, . . . , n be defined as in Proposition 2.1. Then a solves problem (2.5) if and only if there exists an alternating set of (n + 1) points in (0, ξmax ]. Such a solution exists and is unique. Proof. Theorem 4.2 shows that for any δ satisfying 0 < δ < ξmax , {φj } is a Chebyshev set on [δ, ξmax ]. Thus Theorem 3.1 applies. Letting δ → 0 and noting that at ξ = 0 the approximation is exact yields the desired result. 5 Example We conclude with a brief example: Figure 5.1 shows the error of a stencil optimised with respect to the max-norm. The stencil is second order accurate and is 31 points wide. Here we have set ξmax = π/2. In this region the dispersion error is bounded from above by kEk∞ ∼ O(10−12 ). Clearly the alternation property holds as expected from Theorem 3.1. 1.5 kEk∞ = 1.337520e − 12 ×10 -12 1 E 0.5 0 -0.5 -1 -1.5 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 ξ Figure 5.1: Optimised stencil for the case p = 1 and n = 14, i.e. the stencil has formal accuracy O(∆x2 ) and is 31 points wide. 6 References ¨ F. Riesz. Uber lineare functionalgleichungen. Acta Mathematica, 41:71–98, 1918. G. A. Watson. Approximation Theory and Numerical Methods. John Wiley & Sons Ltd., 1980. 7 The Ekeland Variational Principle With Applications Markus Wahlsten ∗ January 12, 2015 Abstract We state and prove the Ekeland Variational Principle and exemplify its usefulness with some applications. 1 Introduction The Ekeland Variational Principle is an important tool in areas such as nonlinear analysis and optimization theory. The principle is often used to establish existence of solutions for optimization problems where the Weierstrass’ theorem does not apply. Further, it’s proven to be a valuable tool when studying partial differential equations Ekeland [1974]. 2 The Ekeland Variational Principle Before we state The Ekeland Variational Principle we recall the Cantor’s intersection theorem and some definitions. Theorem 2.1. (Cantor’s intersection theorem) Let {Sk } be a sequence of non-empty, closed and bounded sets satisfying S0 ⊇ S1 ⊇ · · · Sk ⊇ Sk+1 · · · , then ∞ \ k Sk ! ∗ 6= ∅. (2.1) (2.2) Department of Mathematics, Computational Mathematics, Link¨ oping University, SE581 83 Link¨oping, Sweden [email protected] 1 Definition 2.1. (Upper and lower semicontinuity) Let f be a function and x0 a point such that f : R → R and x0 ∈ R. The functiom f is said to be upper semicontinuous at the point x0 if f (x0 ) ≥ lim sup f (x) (2.3) f (x0 ) ≤ lim inf f (x) (2.4) x→x0 or lower semicontinuous if x→x0 Theorem 2.2. (The Ekeland Variational Principle) Let X be a complete metric space and φ : X → R an upper semi-continuous map that is bounded from above. Then, for any > 0 there exists a y ∈ X such that, φ(y) > φ(x) − d(x, y), ∀x ∈ X\{y} (2.5) Remark. The theorem can be modified by replacing the words ”upper” with ”lower” when also ”above” is replaced by ”below” and finally in (2.5) the ”−” and ”>” is replaced by ”+” and ”<” respectively. Proof. By the fact that sup φ(X) < ∞, ∃z0 ∈ X such that, φ(z0 ) > sup φ(X) − . (2.6) ∞ If we now construct two sequences {zi }∞ i=0 and {Si }i=0 inductively in the following manner 1. Assume zi is given 2. Define Si := {x ∈ X : φ(zi ) ≤ φ(x) − d(x, zi )} 3. Pick any element zi+1 ∈ Si such that 1 φ(zi+1 ) ≥ sup φ(Si ) − i+1 (2.7) Since φ is upper semi-continuous, the map φ(·) − d(·, zi ) is also upper semicontinuous, which leads to the fact that Si is closed. Next, we need to prove that Si+1 ⊆ Si for i = 0, 1, . . . . First, we pick an arbitrary i ∈ Z+ and let x ∈ Si+1 . Then, by using the triangle inequality, and the facts that x ∈ Si+1 and zi+1 ∈ Si , we obtain the relation φ(x) − d(x, zi ) ≥ φ(x) − d(x, zi+1 ) − d(zi+1 , zi ) ≥ φ(zi+1 ) − d(zi+1 , zi ) ≥ φ(zi ) which means x ∈ Si and since x was arbitrarily chosen Si+1 ⊆ Si . 2 (2.8) Further, in order to apply the Cantor Frechet Intersection Theorem we also need to prove that δ(Si ) = sup d(x, y) ≤ x,y∈Si 2 . i (2.9) By picking an i ∈ Z+ , and an x ∈ Si+1 , which also is x ∈ Si which we know from the proof above. By using this fact and the inequality in (2.7) we get, d(x, zi ) ≤ φ(x) − φ(zi ) ≤ sup φ(Si−1 ) − φ(zi ) ≤ 1i (2.10) which proves the desired result. By applyingT the Cantor Frechet Intersection Theorem we conclude that the intersection +∞ i=0 Si = {y} for some y ∈ X. To conclude our proof we need to show that the element y satisfies (2.7). This is done by assuming the contrary, let x ∈ X\{y}, hence we have φ(x) − d(x, y) ≥ φ(y), (2.11) φ(x) − d(x, y) ≥ φ(y) ≥ φ(zi ) + d(x, zi ) (2.12) then since (2.12) is satisfied for all i = 0, 1, . . . we have φ(x) ≥ φ(zi ) + (d(x, y) + d(y, zi )) ≥ φ(zi ) + d(x, zi ) (2.13) is true for all i = 0, 1, . . . , hence x ∈ hence a contradiction. 3 T+∞ i=0 (2.13) Si , which means x = y, Applications In this section several useful applications of the Ekeland Variational Principle is presented. Theorem 3.1. Caristi’s Fixed Point Theorem Let ψ be a self-map on a complete metric space X. If d(x, ψ(x)) ≤ φ(x) − φ(ψ(x)), ∀x ∈ X (3.1) for some lower semi-continuous φ ∈ RX that is bounded from below, then ψ has a fixed point in X. 3 Proof. From the Ekeland Variational Principle, we know that there exists some y ∈ X such that, φ(y) < φ(x) + d(x, y), ∀x ∈ X\{y}. (3.2) However, by (3.1) we have that φ(y) ≥ φ(ψ(y)) + d(y, ψ(y)). (3.3) So ψ(y) ∈ / X\{y}, that is ψ(y) = y. Definition 3.1. (Palais-Smale condition) We say that a C 1 functional φ : X → R satisfies the Palais-Smale condition if every sequence {un }∞ n=1 ∈ X which satisfies kψ(un )k ≤ const, and ψ(un ) → 0 in X possesses a convergent subsequence. (3.4) Theorem 3.2. Let ψ : X → R be a C 1 functional where X is a Hilbert space. Let S be a closed and convex subset of X. Suppose that • K ≡ I − ψ 0 which maps S into S • ψ is bounded from below in S • ψ satisfies definition 3.1 in S Then there exists a u0 ∈ S such that ψ 0 (u0 ) = 0 and inf ψ(u) = ψ(u0 ) u∈S Proof. We start by applying the Ekeland Variational Principle to ψ. Given there exists a u ∈ C such that ψ(u ) ≤ inf ψ(u) + and u∈S ψ(u ) ≤ ψ(u) + ku − u k , ∀u ∈ S (3.5) If we now let u = (1 + t)u + tKu , where 0 < t < 1. Next, we use Taylors formula to expand ψ(u + t(Ku − u )) about u and obtain, 2 t kψ 0 (u )k ≤ kψ 0 (u )k + O(t), (3.6) which implies kψ 0 (u )k < . Finally we use the property (3.1). 4 Summary The Ekeland Variational Principle has been stated and proved in section 2 followed by some useful applications of the theorem in section 3. 4 References I. Ekeland. On the variational principle. Journal of Mathematical Analysis and applications, 47:324–353, 1974. 5 Construction of Sobolev spaces Cristina La Cognata∗ January 15, 2015 Abstract Sobolev spaces are Banach spaces that are characterized by the definition of ”weak” derivatives. The functions in Sobolev spaces and their weak derivatives lie in Lp spaces. Introduction This essay is an introduction to the theory of “Sobolev” spaces. They are powerful tools from Functional Analysis for defining the solutions of Partial Differential Equations (PDE). In particular, they are used in those classes of problems in which smoothness and continuity are too strong requirements. In a certain sense, functions that belong to Sobolev spaces represent a good compromise as they have some, but not all, smoothness properties [Evans, 2002]. We start with the definition of the concept of weak derivatives and then we will apply it to the construction of the Sobolev spaces. We conclude with an overview of some of the most interesting properties concerning these spaces. 1 Weak derivatives A kth-order PDE can be symbolically represented by the following expression F (Dk u(x), Dk−1 u(x), ..., Du(x), u(x), x) = 0, x ∈ U, (1.1) where k ≥ 1 is a fixed integer, U is an open set in Rn and u : U → R is the unknown. The equation (1.1) involves an unknown function u : U → R, ∗ Division of Computational Mathematics, MAI, Link¨ oping University, SE-581 83 Link¨oping, Sweden. [email protected] 1 of variable x ∈ U , and some of its partial derivatives. By solving a PDE of type (1.1) we mean (ideally) finding a simple (in the best case), explicit expression for such a u, or at least guaranteeing its existence together with some properties. Let us ignore for the moment the strict concept of a wellposed PDE problem. What do we require for u to be a “solution” of (1.1)? It is natural to desire having a solution to be at least k times continuously differentiable (u ∈ C k ). A solution of this type is called “classical” solution. Unfortunately there exist certain PDEs (or most of them) which cannot be solved in a classical sense. For instance, in general the conservation law of the form ut + F (u)x = 0, x ∈ U, (1.2) has no classical solutions Evans [2002],R.LeVeque [1992]. Then we need to allow a more general or weaker class of functions to be our solutions. Let us denote by Cc∞ (U ) the spaces of infinitely many times differentiable functions φ : U → R, with compact support on U . We will refer to a function in Cc∞ (U ) as a test function. Furthermore, we will widely use Lp spaces and their norms. For a a reader which is not familiar with these type of spaces we remand to Evans [2002], R.LeVeque [1992]. Definition. Suppose that u, v ∈ L1loc (U )1 and α is a multi-index. We say that v is the αth-weak derivative of u if Z Z α |α| uD φ dx = (−1) vφ dx (1.3) U U for all test functions φ ∈ Cc∞ (U ). In other words, a weak derivative is a generalization of the concept of the derivative of a function (strong derivative) for functions not assumed differentiable, but only integrable in the Lebesgue sense. Note that, since u is not necessarily a differentiable function, the locally summability assumption is needed to ensure the existence of the right-hand side of (1.3). The notion of weak derivatives originated with the works of J. Leray (19061998) and later from S. Sobolev (1908-1989) when he came to deal with non-continuous functions with almost everywhere existing derivatives. Lemma 1.1. The αth-weak derivative of u is uniquely defined almost everywhere2 1 2 see Appendix see Appendix 2 Proof. Let us assume that v, w ∈ L1loc (U ) both satisfy Z Z Z α |α| |α| uD φ dx = (−1) vφ dx = (−1) wφ dx, U for all test functions φ ∈ U Cc∞ (U ). Z U U Then (v − w)φ dx = 0, for all test functions φ ∈ Cc∞ (U ). Hence, (v − w) = 0 a.e. Example. Let n = 1, U = (a, b) and consider u, v defined as follows x a < x ≤ (a + b)/2 u(x) = 1 (a + b)/2 ≤ x < b 1 a < x ≤ (a + b)/2 v(x) = 0 (a + b)/2 ≤ x < b. We can show that v is the weak derivative of u by choosing any test function φ ∈ Cc∞ (U ) and calculate Rb 0 R (a+b)/2 0 Rb uφ dx = a xφ dx + (a+b)/2 φ0 dx a R (a+b)/2 = − a φ dx + [φ((a + b)/2) − φ((a + b)/2)] ((a + b)/2) = Rb = − a vφ dx. as required. 2 Sobolev spaces We now define a function space which contains elements with weak derivatives of various orders from Lp spaces. Let us fix 1 ≤ p ≤ ∞ and let k be a nonnegative integer. Definition (Sobolev spaces). The Sobolev space W k,p (U ) consists of all locally summable functions u : U → R such that for each multi-index α with |α| ≤ k, the weak derivative Dα u exists and lie in Lp (U ). Definition (Sobolev norms). W k,p (U ) is a normed space equipped with the norm defined as 1/p R P α p (1 ≤ p ≤ ∞) |α|≤k U |D u| dx kukW k,p (U ) = , (2.1) P α ess sup |D u| (p = ∞) U |α|≤k where u ∈ W k,p (U ). For the definition of ess sup f see 3 3 We finally introduce the following notation for convergence: k,p Definition (Sobolev convergence). We say that {um }∞ (U ) m=1 , um ∈ W k,p converge to u ∈ W (U ) provided that lim kum − ukW k,p (U ) = 0. m→∞ Next we list some elementary properties Sobolev functions. The proofs can be found in Evans [2002]. Note that the following properties are trivially true for functions in C k (U ) spaces. For Sobolev functions, instead, we must think about these rules as in the weak definition. Theorem 2.1. Assume u, v ∈ W k,p (U ), |α| ≤ k. Then 1. Dβ (Dα u) = Dα (Dβ u) = Dα+β u for all |α| + |β| ≤ k. 2. For each λ, µ ∈ R, λu + µv ∈ W k,p (U ) and λDα u + µDα , |α| ≤ k. 3. if V is an open set in U , then U ∈ W k,p (V ). Some of these properties are needed to prove that the Sobolev spaces are Banach spaces. Theorem 2.2 (Sobolev spaces as a function spaces). For each k = 1, ..., and 1 ≤ p ≤ ∞, the Sobolev space W k,p (U ) is a Banach space. Proof. First we check that (2.1) is a norm. It is easy to check from the definition that kλukW k,p (U ) = |λ|kukW k,p (U ) kukW k,p (U ) = 0 and iff u=0 a.e. (2.2) Next we assume that u, v ∈ W k,p (U ) with 1 ≤ p < ∞. Then ku + vkW k,p (U ) = P |α|≤k kD u+D α vkpLp (U ) 1/p p 1/p α α p (U ) + kD vkLp (U ) kD uk L |α|≤k P 1/p P 1/p p p α α ≤ kD uk + kD vk p p |α|≤k |α|≤k L (U ) L (U ) = kukW k,p (U ) + kvkW k,p (U ) . ≤ P α Note that in the second inequality we have used Minkowski’s relation defined in 3. Next we prove that W k,p (U ) is complete by showing that any Cauchy sequence um ∈ W k,p (U ) converges. 4 1. For any |α| ≤ k, Dα um is a Cauchy sequence in Lp (U ). Since Lp (U ) is complete, the existence of u and uα , ∀α, such that um → u and Dα um → uα in Lp (U ), (2.3) is guaranteed. 2. Consider any test function φ ∈ Cc∞ (U ). Then we have R R α uD φ dx = lim u Dα φ dx m→∞ U U m R = limm→∞R(−1)|α| U Dα um φ dx = (−1)|α| U uα φ dx. Thus, uα = Dα u, |α| ≤ k. 3. From the two previous steps we find that um → u in Lp (U ) and the uα are the weak derivatives of u. Therefore um → u in W k,p (U ). Thus W k,p (U ) is a Banach space. We conclude by noting that in the special case p = 2, the Sobolev space has a more interesting structure. Definition (Hilbert-Sobolev spaces). We define the Hilbert-Sobolev space as H k (U ) = W k,2 (U ). The space H k (U ) is a Hilbert space endowed with the inner product XZ (u, v)H k (U ) = Dα uDα v dx. |α|≤k U Note that H 0 (U ) = L2 (U ). For the curious reader, more details and results can be found in Brezis [1983]. 3 Appendix Definition (Almost everywhere). If a property holds everywhere in Rn , except for a measurable set of Lebesgue measure zero, we say that the property is true almost everywhere (a.e.). Definition (ess sup). If a real function f is measurable, the essential supremum of f is defined as ess sup f = inf {µ ∈ R s.t. |f > µ| = 0} . 5 Definition (L1loc functions). L1loc = {u : U → R s.t. u ∈ Lp (V ) for each V ⊂⊂ U } . Definition (Minkowski’s inequality). Assume 1 ≤ p ≤ ∞ and u, v ∈ Lp (U ). Then ku + vkLp (U ) ≤ kukLp (U ) + kvkLp (U ) . The Minkowski inequality is the triangle inequality in Lp (U ). References H. Brezis. Analyse Fonctionnelle. Masson, 1983. L. Evans. Partial Differential Equations. AMS, 2002. R.LeVeque. Numerical Methods for Conservation Laws. Birkh¨auser, 1992. 6 The Arzel´a-Ascoli theorem Tomas Lundquist∗ January 15, 2015 Abstract We discuss the Arzel´ a-Ascola theorem from a functional analysis point of view, and give examples of applications within the field, in particular relating to the theory of compact embeddings in function spaces. 1 Introduction The Arzel´a-Ascola theorem is of fundamental importance in both real analysis and functional analysis, as well as in topology. In functional analysis, it is particularly useful in function space theory, providing a constructive criterion for compactness of subsets in C(X), the set of complex valued continuous functions on a metric space X. In its original form, the theorem was formulated for sequences of continuous functions on compact intervals of the real line, giving a criterion for the existence of uniformly bounded subsequences. Later it was extended to compact metric spaces then further to compact Hausdorﬀ topological spaces. Here we will focus on the functional analysis version of the theorem. In section 2 we present the theorem and discuss some of its implications. Section 3 is a brief introduction to the theory of compact embeddings in continuous function spaces, to which the theorem is closely related. 2 Statement of the theorem The Arzel´a-Ascoli theorem gives a necessary and suﬃcient condition for compactness of continuous function spaces. Before we proceed with stating the theorem, we need the following definition. ∗ Division of Computational Mathematics, Department of Mathematics, Link¨oping university, SE-581 83 Link¨oping, Sweden. [email protected] 1 Definition 2.1. A set F of functions between two metric spaces X and Y is said to be equicontinuous if for each ϵ > 0, ∃ a δ > 0 such that d(f (x0 ), f (x1 )) < ϵ for all f ∈ F , and x0 , x1 ∈ X with d(x0 , x1 ) < δ. The parameter δ may depend on x0 but not on f . Note the contrast between equicontinuity and uniform continuity. In the former case, ϵ is not allowed to depend on f , whereas in the latter case, ϵ is not allowed to depend on x0 . The metric space version of the Arzel´a-Ascoli theorem now follows below. Theorem 2.1. Let X be a compact metric space, and denote with C(X) the space of real (or complex) valued continuous functions on X with respect to the supremum norm. Then a subset M of C(X) is relatively compact in C(X) if and only if M is equicontinuous and bounded. Proof. A proof can be found e.g. in Nagy [2006]. ¯ of We recall that relatively compact in this case means that the closure M ¯ M is compact in C(X); that is, each sequence in M has a convergent subsequence. The Arzela-Ascola theorem is used in the proof of many important theoretical results. One example is Peano’s theorem of local existence for continuous ordinary diﬀerential equations. Another is to prove that an operator between Banach spaces is compact if and only if the adjoint operator is compact (see e.g. Garrett [2012]). 3 Application: compact embeddings of continuous functions The concept of compact embeddings between function spaces is important not the least in the study of numerical methods for partial diﬀerential equations. For example, the famous Sobolev embedding theorems are examples of results relating to compact embeddings. These results are important when applying Sobolev spaces to partial diﬀerential equations. For a discussion of this, see e.g. Dlotko [2014]. An embedding E : X ← Y from a subset X ⊂ Y into Y is simply the identity operator. As an example, consider the space C 1 (Ω) of C 1 −diﬀerentiable functions on the domain Ω ∈ Cn , equipped with the norm ∥f ∥C 1 = sup|f | + sup|∇f |. With this definition, the embedding E(C 1 (Ω)) into C(Ω) is the space formed by the set C 1 (Ω) and standard supremum norm. As we shall later see, this embedding is indeed compact. 2 Definition 3.1. Let (X, ∥·∥X ) and (Y, ∥·∥Y ) be two normed spaces, where X is a subset of Y . We say that X is compactly embedded in Y if the following holds: 1. ∃ c > 0 : ∥x∥Y ≤ c∥x∥X ∀x ∈ X. (uniform continuity) 2. The embedding of X into Y is a compact operator. (compactness) We recall that an operator is compact if it maps bounded sets to relatively compact sets. The Arzel´a-Ascoli theorem provides a constructive way to show compactness of embeddings. Too see this, we continue with the example above. Thus, consider a bounded subset M of the space C 1 (Ω). The boundedness of sup|f | and sup|∇f | then implies equicontinuity of the set (this part is left as an exercise). From the Arzel´a-Ascoli theorem, it also follows that M is relatively compact in C(Ω). i.e. the embedding E(M) into C(Ω) is compact. Moreover, uniform continuity follows directly from the definitions of the norms involved: ∥f ∥C(Ω) = sup|f | ≤ sup|f | + sup|∇f | = ∥f ∥C 1 (Ω) . Thus the space C 1 (Ω) is compactly embedded in C(Ω) with respect to the norm given above. Analogous results to the one given above can be derived using the Arzol´aAscoli theorem for a variety of continuous function spaces. In particular the Sobolev embedding theorems, of fundamental importance for Sobolev space theory, can be proven using similar arguments. The most famous of these result states that the embedding H m+1 (Ω) in H m (Ω) is compact, where H m (Ω) is the Sobolev space with square integrable weak derivatives up to order m. Note that H 0 (Ω) = L2 (Ω), so that all Sobolev spaces are compactly embedded in the space of square integrable functions. A full proof of this, generalized to Lp (Ω), is given e.g. in Adams [1975]. 4 Summary We have reviewed the metric space formulation of the Arzol´a-Ascoli theorem, and shown its usefulness as a constructive criterion for compactness of continuous function spaces. An example of application is given by the famous Sobolev embedding theorems. References R. Adams. Sobolev spaces. Academic Press, New York, 1975. 3 T. Dlotko. Sobolev spaces and embedding theorems. www.icmc.usp.br/~ andcarva/sobolew.pdf, 2014. P. Technical report, Garrett. Compact operators on banach spaces: Fredholm-riesz. Technical report, http://www.math.umn.edu/~ garrett/m/fun/fredholm-riesz.pdf, 2012. G. Nagy. A functional analysis point of view of the arzela-ascoli theorem. Real Analysis Exchange, 32:583–586, 2006. 4 A Completeness Theorem for Non-Self-Adjoint Eigenvalue Problems Saleh RezaeiRavesh∗ January 16, 2015 Abstract An important completeness theorem for a general class of nonself-adjoint eigenvalue problems is stated. The corresponding proof is reproduced in connection with basic material in spectral theory and functional analysis. 1 Introduction In many applications in hydrodynamic stability analysis we are confronted with eigenvalue problems in the form (L − λM )ϕ = 0 (1.1) where L and M are differential operators and λ ∈ σ(L, M ). The associated boundary conditions for (1.1) are ϕ(0) = Dϕ(0) = ϕ(1) = Dϕ(1) = 0, where, D := d/dy is the differentiation operator. One of the famous examples is the Orr-Sommerfeld equation which is derived in linearized stability analysis of parallel flows (e.g., cf. Drazin and Reid [1981]). The general idea for studying the spectrum of (1.1) is to write L := Ls +B where Ls and M are self-adjoint operators and B is called perturbation. In Section 2, some of the definitions and theorems required in this document are expressed. Then, theorem 3.1 shows how we can find a self-adjoint extension operator G for M −1 Ls in such a way that two problems Gϕ = λϕ and Ls ϕ = λM ϕ become equivalent. Subsequently, theorem 3.2 proves the completeness of the generalized eigenvectors of (1.1). Finally, an application of the completeness theorem is briefly explained. ∗ Division of Scientific Computing, Department of Information Technology, Uppsala university, SE-751 05 Uppsala, Sweden. [email protected] 1 2 Spectral Theory Definition 2.1 (Resolvent Set & Spectrum of T ) Let X 6= {0} be a complex normed space and T : D(T ) → X 1 a linear operator with D(T ) ⊂ X. Then, (a) The resolvent set ρ(T ) of T is the set of all regular values λ of T . If T : X → X is a bounded linear operator and X is a Banach space, then, ρ(T ) = {λ ∈ |(T − λI) is one-to-one and onto} Evans [2010] (Any λ ∈ σ(T ) is called a spectral value of T (cf. 7.2-1 Kreyszig [1989])), (b) The spectrum of T is the complement of ρ(T ); i.e., σ(T ) = − ρ(T ) (cf. 7.2-1 Kreyszig [1989]). Definition 2.2 (Resolvent Operator) If Tλ := (T − λI) has an inverse, denoted by Rλ (T ) := Tλ−1 = (T − λI)−1 , we call it the resolvent operator of T (cf. 7.2 Kreyszig [1989]). Definition 2.3 (Point, Continuous, Residual Spectra) The point or discrete spectrum σp (T ) is the set such that Rλ (T ) does not exist. λ ∈ σp (T ) is called an eigenvalue of T . The continuous spectrum σc (T ) is the set such that Rλ (T ), exists and has a domain dense in X, but is unbounded. The residual spectrum σr (T ) is the set such that Rλ (T ), (a) exists (may be bounded or not) but its domain is not dense in X (cf. 7.2-1 Kreyszig [1989]). C C Theorem 2.1. (Distance from σ(T )) If T ∈ B(X, X), where X is complex Banach space and λ ∈ ρ(T ), then (cf. 7.5-3 Kreyszig [1989]) kRλ (T )k ≥ 1 , d(λ, σ(T )) d(λ, σ(T )) := inf |λ − s| s∈σ(T ) (2.1) Definition 2.4 (Eigenvalue & Eigenvector) We say that λ ∈ σ(T ) is an eigenvalue of T provided N (T − λI) 6= {0}. We write σp (T ) to denote the collection of eigenvalues of T ; clearly, σp (T ) is a point spectrum. If λ is an eigenvalue and ϕ 6= 0 satisfies T ϕ = λϕ, we say ϕ is an associated eigenvector (cf. D.3 Evans [2010]). Definition 2.5 (Generalized Eigenvector of (L, M )) Let L and M be linear operators with D(L) ⊂ D(M ) dense in a Hilbert space H. We say that ψ ∈ H is a generalized eigenvector of (L, M ) if and only if for some λ and some integer p > 1, there exist p non-zero vectors ψ1 , ψ2 , · · · , ψp = ψ such that Lψi = λM ψi + M ψi−1 , i = 1, · · · , p, where ψ0 = 0. By allowing p = 1, eigenvectors become included in generalized eigenvectors (DiPrima and Habetler [1969]). 1 In this manuscript, D(T ), R(T ), and N (T ) respectively denote domain, range and null space of operator T . 2 Definition 2.6 (Completeness) The generalized eigenvectors of a linear operator with domain dense in H are said to span H if the following is true. Let ϕk1 , ϕk2 , · · · , ϕknk be the eigenvectors (and possibly generalized eigenvectors) corresponding to the eigenvalues in rk−1 < |λ| < rk for k > 1 and |λ| < r1 for k = 1; then each ϕ ∈ H has a representation ϕ= ni ∞ X X aij ϕij (2.2) i=1 j=1 with convergence in the norm of H (DiPrima and Habetler [1969]). Theorem 2.2. (Friedrichs Extension) A positive-bounded below (hence positive definite) operator with domain dense in a Hilbert space can be extended to a self-adjoint operator which possesses an inverse defined on the entire space (cf. p 11 Mikhlin [1965]). Theorem 2.3. (Naimark’s Theorem) If T is a closed linear operator on a dense subspace of a Hilbert space H with norm k.k whose resolvent (T − λI)−1 is compact for some λ = λ0 , and if there exists a sequence of circles Ck , |λ| = rk such that (a) there are no eigenvalues of T on any Ck , (b) limk→∞ rk = ∞, and (c) limk→∞ supλ∈Ck k(T − λI)−1 k = 0, then the generalized eigenvectors of T span H in the sense of (2.2) (e.g., see DiPrima and Habetler [1969]). 3 Completeness Theorem Theorem 3.1. (Equivalent EVP’s) Let Ls and M be linear operators with D(Ls ) ⊂ D(M ) dense in a Hilbert space H and such that, 1. M and Ls are positive-bounded below with R(M ) = R(Ls ), 2. G−1 is compact and R(G−1 ) ⊂ D(M ) where G−1 is the inverse of the self-adjoint extension of M −1 Ls in HM .2 Then, the two problems Gϕ = λϕ and Ls ϕ = λM ϕ are equivalent (DiPrima and Habetler [1969]). Proof: I Step 1. (Construct G−1 ): M is psoitive-bounded below, i.e., ∀x ∈ D(M ) ∃α > 0 : hM x, xi ≥ αkxk2 2 A new Hilbert space HM which is embedded in H can be constructed by completing the pre-Hilbert space HM (D(M ), [., .]), where [f, g] = hf, M gi for f, g ∈ D(M ) is the 1 corresponding inner product and the norm is kf kHM = [f, f ] 2 (cf. 3 Mikhlin [1965]). 3 Combining this with the Schwarz inequality results in, αkxk2 ≤ hM x, xi ≤ kM xkkxk Note that y = M x for x and y in D(M ) and R(M ) of M , respectively; hence, kM −1 yk ≤ 1 kyk α Therefore, M −1 exists and is bounded. Furthermore, M −1 Ls is positivebounded below in HM ; since ∀x ∈ D(Ls ) ⊂ D(M ), [x, M −1 Ls x] = hx, M M −1 Ls xi = hx, Ls xi ≥ βkxk2 , β>0 where the last inequality holds since Ls is positive-bounded below in H. According to Theorem 2.2, M −1 Ls has a self-adjoint extension (call it G) that has an inverse G−1 as mentioned in hypothesis 2. I Step 2. (Find the Spectrum of G): Since G−1 is compact, then it has a point spectrum, σp (G) (cf. 8.3-1, 8.3-3 Kreyszig [1989]). Furthermore, the set of eigenvalues G is countable (may be finite and even empty) and λ = 0 is the only possible point of accumulation of that set. In general, on D(G) ⊂ X which is a complex normed space two cases can be considered; if X is finite dimensional, then G has a matrix representation and clearly 0 may or may not belong to σ(G) = σp (G); that is, if dimX < ∞ we may have 0 ∈ / σ(G); then 0 ∈ ρ(G). However, if dimX = ∞, then we must have 0 ∈ σ(G) (cf. pp. 420 and 433 Kreyszig [1989]). I Step 3. (Equivalent Spectra): We show that σ(G) = σp (G) = σp (Ls , M ). 3.a) Suppose λ ∈ σ(G); then, there exists ϕ ∈ D(G) such that Gϕ = λϕ. Since from hypothesis 2, D(G) = R(G−1 ) ⊂ D(M ), then we have ϕ ∈ D(M ) and hence Gϕ ∈ D(M ). But, since R(M ) = R(Ls ), M −1 Ls is onto D(M ). This means that there exists ψ ∈ D(Ls ) such that M −1 Ls ψ = Gϕ, and since G = M −1 Ls on D(Ls ) and G−1 exists, we have ψ = ϕ. As a result, M −1 Ls ψ = M −1 Ls ϕ = λϕ ⇒ Ls ϕ = λM ϕ ⇒ λ ∈ σp (Ls , M ) 3.b) Conversely, consider λ ∈ σp (Ls , M ) and correspondingly, Ls ϕ = λM ϕ, ϕ 6= 0. Then immediately we have M −1 Ls ϕ = λϕ an hence Gϕ = λϕ and λ ∈ σ(G). Consequently, σp (Ls , M ) ∈ σ(G) and an eigenvector of G is an eigenvector of (Ls , M ). This completes the proof of equivalency of the Gϕ = λϕ and Ls ϕ = λM ϕ . Theorem 3.2. (Completeness) Let L = Ls + B, and let Ls , B, and M be linear operators with D(Ls ) = D(B) ⊂ D(M ) dense in a Hilbert space H, such that, 4 1. Ls and M satisfy conditions of theorem 3.1, 2. M −1 B is bounded in HM and D(Ls ) is dense in HM , 3. for some λ, R(Ls + B + λM ) = R(M ) and (G + M −1 B + λI) has an inverse, 4. there exists a sequence of concentric circles {Ck } with the radii {rk } such that, (a) limk→∞ rk = ∞, (b) d(Ck , σp (Ls , M )) > 0 for all k, (c) limk→∞ d(Ck , σp (Ls , M )) = ∞. Then, 1. the eigenvalues of (L, M ) lie within circles of radii kM −1 BkHM about those of (Ls , M ), and, 2. the generalized eigenvectors of (L, M ) span HM in the sense of (2.2) (DiPrima and Habetler [1969]). Proof: I Step 1. (Define T and Show it is Closed) Consider T := G + M −1 B where M −1 B is a unique bounded extension from D(B) to D(G). Now, we show that T satisfies the condition of Naimark’s theorem 2.3. According to the proof of theorem 3.1, G is self-adjoint and therefore closed. In addition by hypothesis 2., M −1 B is bounded and when we add it to the closed operator G, we end up with closed operator T . I Step 2. (Show that ∃λ0 : Rλ (T ) = (T − λ0 I)−1 is Compact): Let’s choose λ0 such that d(λ0 , σ(G)) > kM −1 BkHM . This is always possible; since according to hypothesis 4.(c), it is enough to choose λ0 on Ck for sufficiently large k. G−1 is compact and G is closed, so (G − λI)−1 is compact for all λ ∈ ρ(G) (cf. 5.7 Engel and Nagel [2006]), and hence for λ = λ0 . Now, we can show that (T − λ0 I)−1 is compact. we have, (T − λ0 I)−1 = (G + M −1 B − λ0 I)−1 = [I + (G − λ0 I)−1 M −1 B]−1 (G − λ0 I)−1 By using theorem 2.1, we can show that [I +(G−λ0 I)−1 M −1 B]−1 is bounded: kI+(G−λ0 I)−1 M −1 BkHM ≥ 1−k(G−λ0 I)−1 kHM kM −1 BkHM ≥ 1− kM −1 BkHM >0 d(λ0 , σ(G)) Therefore, (T − λ0 I)−1 which is the product of a bounded operator and a compact operator, is compact (cf. 8.3-2 Kreyszig [1989]). I Step 3. (Obtain Conclusion 1.): 3.a) Since (T − λ0 I)−1 is compact and T is closed, then (T − λI)−1 is compact for all λ ∈ ρ(T ) (see 5.7 Engel and Nagel [2006]). Therefore, σ(T ) is a point spectrum such that each λ ∈ σ(T ) = σp (T ) has finite multiplicity (cf. 8.3-1 Kreyszig [1989]). On the other hand, if λ is such that d(λ, σ(G)) > kM −1 BkHM then necessarily λ ∈ ρ(T ). Consequently, for λ ∈ σ(T ), we have d(λ, σ(G)) < kM −1 BkHM . In other 5 words, σ(T ) lies within the union of a set of circles of radii kM −1 BkHM and center σ(G) = σp (Ls , M ). 3.b) By a similar discussion made in Step 3. of the proof of theorem 3.1, it can be shown that σ(T ) = σp (M −1 L). For this, replace Ls by L + λM ; similarly, existence of G−1 is replaced by existence of (G + M −1 B + λI)−1 for λ mentioned in hypothesis 3. Since it is possible that G + M −1 B does not have an inverse, it is required that such λ exists. Hence, σp (M −1 L) lies within the union of a set of circles of radii kM −1 BkHM centered about σp (M −1 Ls ). I Step 4. Obtain Conclusion 2.: According to hypothesis 4.(c) we can choose a k0 such that for all integers k ≥ k0 , d(Ck , σp (Ls , M )) = d(Ck , σp (G)) > kM −1 BkHM . Then for such a k and λ ∈ Ck , we have k(T − λI)ϕkHM = k(G − λI)ϕ + M −1 BϕkHM ≥ k(G − λI)ϕkHM − kM −1 BϕkHM ≥ {d(λ, σp (Ls , M )) − kM −1 BkHM }kϕkHM Furthermore, k(T − λI)ϕkHM ≤ k(T − λI)kHM kϕkHM . Combining this with the latest inequality and then taking supremum of both sides results in, sup k(T − λI)−1 kHM ≤ λ∈Ck 1 d(λ, σp (Ls , M )) − kM −1 BkHM Clearly, by hypothesis 4.(c), limk→∞ supλ∈Ck k(T − λI)−1 kHM = 0. Therefore all the conditions of Naimark’s theorem are satisfied by T for circles Ck with k ≥ k0 . Consequently, the generalized eigenvectors of T , and hence those of (L, M ) span HM in the sense of (2.2) . 4 Application To show an application of theorem 3.2, consider the Orr-Sommerfeld equation rewritten in the form of equation (1.1); then λ = iαcRe Ls = (D − α ) , B = −iαRe[U (D2 − α2 ) − U 00 ] M = −(D2 − α2 ) 2 2 2 (4.1) (4.2) (4.3) Here, U (y) is the base flow profile and U 00 = d2 U/dy 2 . Moreover, α and Re are wavenumber and Reynolds number, respectively. In hydrodynamic temporal stability analysis we are interested in finding the phase speed c appearing in λ for a set of input parameters. In association with the completeness theorem mentioned above, consider H = L2 (Ω) = H 0 (Ω), HM = H01 (Ω), 6 where Ω = [0, 1] Then, through a detailed process, one can show that differential operators in the Orr-Sommerfeld equation satisfy all hypotheses of theorems 3.1 and 3.2. Here as a step of that process, the construction of concentric circles is presented. Consider σp (Ls , M ) = {µn } ∪ {ϑn } where the µn are the ordered positive eigenvalues of the problem Ls ϕ1 = µM ϕ1 , ϕ1 (0) = ϕ01 (0) = ϕ1 (1) = ϕ01 (1) = 0 (repeated eigenvalues, if any occurs, are listed once) and the numbers νn are the ordered positive eigenvalues of the problem M ϕ2 = ϑϕ2 , ϕ2 (0) = ϕ2 (1) = 0 For the operators in the Orr-Sommerfeld equation, it can be shown that as n → ∞, µn+1 − µn → ∞ and ϑn+1 − ϑn → ∞. Thus, the sequence {Ck }∞ k=1 is a set of concentric circles with center at the origin and with radii rk as follows. For n ≥ 1 consider the interval (µn , µn+1 ). Then, (a) if there is no ϑl ∈ (µn , µn+1 ), take rn = 21 (µn + µn+1 ); (b) if there is only one ϑl ∈ (µn , µn+1 ), then when ϑl ∈ (µn , 21 (µn +µn+1 )) take 3 1 rn = 14 µn + 34 µn+1 and when ϑl ∈ ( 21 (µn + µn+1 ), µn+1 ) take rP n = 4 µn + 4 µn+1 ; m 1 (c) if there are several (say m) ϑl in (µn , µn+1 ), take rn = m i=1 ϑi (DiPrima and Habetler [1969]). As a result of theorem 3.2, for a bounded flow, the spectrum of the OrrSommerfeld equation consists of an infinite number of discrete eigenvalues and the associated eigenvectors span HM = H01 (Ω) in the sense of (2.2). By this completeness, an arbitrary initial disturbance which satisfies the physical boundary conditions can be expanded in terms of these eigenfunctions. This technique is very useful in studying different hydrodynamic instability mechanisms. References R. C. DiPrima and G. J. Habetler. The completeness theorem for nonselfadjoint eigenvalue problems in hydrodynamic stability. Archieve for Rational Mechanics and Analysis, 34(3):218–227, 1969. P. G. Drazin and W. H. Reid. Hydrodynamic Stability. Cambridge Univ. Press, 1981. K.-J. Engel and R. Nagel. A Short Course on Operator Semigroups. Springer, 2006. 7 L. C. Evans. Partial Differential Equations. American Mathematical Society, 2nd edition, 2010. E. Kreyszig. Indtroductory Functional Analysis with Applications. John Wiley & Sons, 1989. S. G. Mikhlin. The Problem of the Minimum of a Quadrature Functional. Holden-Day INC., 1965. 8 The Metric Lax and Applications∗ Cheng Gong† January 16, 2015 Abstract We define consistency, convergence, and stability of numerical methods in the setting of metric spaces. The Lax equivalence theorem is formulated and a proof is given in this setting. We also indicate its use in some applications. 1 Introduction The Lax (or Lax-Richtmyer) equivalent theorem Lax and Richtmyer [1956] is the fundamental theorem in numerical analysis. It was established by Peter Lax and Robert D. Richtmyer in 1956. The theorem states that “a consistent finite difference method for a well-posed linear initial value problem is convergent if and only if it is stable”. The convergence of a numerical method is a measurement of the differences between the numerical solution and the analytic solution from the mathematical problem. In general, the analytic solutions are usually not accessible and the convergence is difficult to be observed directly. However, the consistency and stability can be evaluated from the properties of the numerical method, e.g. error estimate, Fourier analysis and Von Neumann analysis Gustafsson et al. [2013]. Thus, the Lax equivalent theorem provides a practical way to prove the convergence of a numerical method from stability. This principle can be formulated and proved in metric spaces. ∗ This essay is based on the notes from Stefan Engblom. [email protected] Division of Scientific Computing, Department of Information Technology, Uppsala university, SE-751 05 Uppsala, Sweden. [email protected] † 1 2 Definition ˜ be two metric spaces and we define the Let X = (X, d) and Y = (Y, d) mathematical problem as: find x ∈ X, such that T x = y, (2.1) where T : X −→ Y is an operator defined by the problem and y ∈ Y is given. Definition 2.1. The mathematical problem in (2.1) is well-posed if T −1 exists and is continuous in some neighborhood containing y. Then, the solution of a well-posed problem with the same form as in (2.1) is given by the inverse operator of T as x = T −1 y. (2.2) The problem in (2.1) is approximated by a sequence of numerical problems: find xn ∈ X such that Tn xn = y, (2.3) where Tn : X −→ Y is an operator of the corresponding numerical method and y ∈ Y is the same right-hand-side as in (2.1). The family of equations in (2.3) is called the numerical approximations for the problem in (2.1) [Hunter and Nachtergaele, 2001] and n refers to the indices of the sequence such that Tn −→ T as n −→ ∞. Definition 2.2. Let D(T ) be the domain of T . The numerical method in (2.3) is consistent, if ∀x ∈ D(T ), where D(T ) ⊂ X is the domain of T in X, then Tn x −→ T x as n −→ ∞. Definition 2.3. The numerical method in (2.3) is stable if ∀n ∈ N, Tn−1 exists and is continuous in some neighborhood containing y. Definition 2.4. The numerical method in (2.3) is convergent if xn −→ x as n −→ ∞. For a stable numerical method, the solution is given as xn = Tn−1 y by the inverse operator of Tn−1 . If the numerical method is convergent, the numerical solution will converge to the analytic solution of the mathematical problem. 2 3 The Lax Equivalent Theorem A general form of Lax equivalent theorem is given in Theorem 3.1. Theorem 3.1. A consistent method applied to a well-posed problem is convergent if and only if it is stable. By introducing the definitions from Section 2, the Lax equivalent theorem is formulated in Theorem 3.2. Theorem 3.2. Define a mathematical problem as (2.1) and the numerical approximation as (2.3). Suppose T −1 is continuous and Tn x −→ T x for any x in the domain of T . Then, Tn−1 y −→ x as n −→ ∞ if and only if Tn−1 is continuous for all n. Proof. [Engblom, 2014] ‘=⇒’ By the consistency of Tn , for any given δ, ∃N ∈ N such that, for all n > N ˜ n x, T x) < δ. d(T (3.1) From the assumption in (2.1) and (2.3), T x = y = Tn xn , (3.2) together with the well-posedness of the problem, ˜ n x, Tn xn ) < δ. d(T (3.3) By the stability of the numerical method, Tn−1 is continuous at y for all n. Denote y1 = Tn x, then ∀ > 0 and for all y1 satisfying (3.3), ∃δ > 0, such that d(Tn−1 y1 , Tn−1 y) < , (3.4) which is equivalent to d(x, xn ) < . (3.5) Therefore, xn −→ x as n −→ ∞ and the numerical method is convergent. ‘⇐=’ If the numerical method in (2.3) is not stable, then ∃ > 0, for ∀δ > 0, ˜ 1 , y) < δ, we have ∀y1 ∈ Y satisfying d(y d(Tn−1 y1 , Tn−1 y) > . (3.6) From the consistency, there exists an N ∈ N, ∀n > N , ˜ n x, T x) < δ. d(T 3 (3.7) If we take y1 = Tn x, then (3.6) is written as d(Tn−1 Tn x, Tn−1 Tn xn ) > , (3.8) d(x, xn ) > , (3.9) by (3.2) and consider the well-posedness of the problem that T −1 exists in some neighborhood of y, (3.10) d(T −1 y, Tn−1 y) > , which indicates that the numerical method is not convergent. Therefore, a consistent and convergent numerical method is stable. 4 Applications In a metric space, the metric can be used to define convergence, continuity and compactness. In section 3, the metric form of Lax equivalent theorem is shown in Theorem 3.2. Usually, this theorem is applied to the problems defined in normed spaces, or even in Banach spaces, to prove the numerical method is convergent by using the stability. There are many linear and nonlinear examples for this forward direction of Lax equivalent theorem, e.g. [Sanz-Serna and Palencia, 1985, Palencia and Sanz-Serna, 1984]. On the other hand, to get stability from convergence is not very practical in some finite difference method since the convergence always relates to the analytic solution of the problem which is not accessible in many cases. However, this direction of the Lax equivalent theorem can be use to show a consistent numerical method does not converge from the condition that it is unstable. An interesting example is to show that Newton-Cotes formulas for numerical integration do not converge [Engels, 1980] by knowing that Newton-Cotes formulas are numerically unstable [Krommer and Ueberhuber, 1998]. The importance of the Lax equivalent theorem is that the convergence of a numerical method is difficult to be established while it is usually desired. The consistency is straightforward to be verified and the stability which concerns the ‘local’ properties of numerical method is much easier to show than the convergence. Moreover, the metric Lax equivalent theorem only requires the operator for the numerical method to be defined on a metric space which may provide a wider use than the classical one for numerical analysis. However, the operators Tn and T are defined on the same metric space which could cause some troubles for the applications in the future. 4 References S. Engblom. Notes of lax equivalent theorem, 2014. H. Engels. Numerical quadrature and cubature. 1980. B. Gustafsson, H.-O. Kreiss, and J. Oliger. Time-dependent problems and difference methods, volume 121. John Wiley & Sons, 2013. J. K. Hunter and B. Nachtergaele. Applied analysis. World Scientific, 2001. A. R. Krommer and C. W. Ueberhuber. Computational integration, volume 53. Siam, 1998. P. D. Lax and R. D. Richtmyer. Survey of the stability of linear finite difference equations. Communications on Pure and Applied Mathematics, 9(2): 267–293, 1956. C. Palencia and J. Sanz-Serna. An extension of the lax-richtmyer theory. Numerische Mathematik, 44(2):279–283, 1984. J. Sanz-Serna and C. Palencia. A general equivalence theorem in the theory of discretization methods. Mathematics of computation, 45(171):143–152, 1985. 5 Fixed-point proof of Lax-Milgram theorem Fatemeh Ghasemi∗ January 16, 2015 Abstract Lax-Milgram theorem gives conditions under which a bilinear form can be inverted to show existence and uniqueness of weak solution of a given partial differential equation (PDE). There are different ways for proving this theorem. Here we use the well-known Banach fixed-point theorem. 1 Introduction Riesz representation theorem is of the fundamental theorem about Hilbert space that is the basis for the existence and uniqueness for (symmetric) variational problems. For using this theorem, the variational problems needs to be symmetric. But a wide variety of variational problems are not symmetric and this means that this theorem is not applicable. Fortunately, there is an analogous result, the Lax-Milgram theorem, that does apply to nonsymmetric problems. The Lax-Milgram lemma has played a critical role over the last half century by establishing existence and uniqueness of weak solutions of operator equations Au = f where A is a continuous and coercive linear operator from a Hilbert space H into its topological dual H 0 . The result has had far-reaching impact on the analytic and numerical study of elliptic and parabolic partial differential equations; boundedness and coercivity of the weak operators are all that need be verified. While in 1954 the paper (1) of Lax and Milgram provided a nonconstructive proof of the result, constructive proofs appeared over the next several ∗ Division of Computational Mathematics, Department of Mathematics, Link¨ oping University, SE-581 83 Link¨oping, Sweden . [email protected] 1 years. In 1960, Zaran- tanello provided a generalization of Lax-Milgram to strongly monotone nonlinear operators via Banach fixed points. A similar fixed point proof in the linear case was given by Lions and Stampacchia in a 1967 paper (2) on variational inequalities. In both cases, the mapping shown to be contractive involves applying the Riesz map for H to a residual Av − f ∈ H for some v ∈ H. Also in 1967, Petryshyn gave a constructive proof of the result based on so-called upper and lower semi-orthogonal bases for subspaces of H. In this essay, after introducing some preliminaries in the first section, we prove Lax-Milgram theorem by using Banach fixed-point theorem. 2 Preliminaries Let H denote a real Hilbert space equipped with inner product (., .)H and 0 0 associated norm k.k. H denotes the topological dual with h., .i the H × H duality pairing. In this essay we use Banach fixed-point theorem and Riesz Representation theorem in order to prove Lax-Milgram Theorem. In the following we review these two theorems. Theorem 2.1. (Banach fixed-point theorem) Let (X, d) be a non-empty complete metric space, and T : X → X a map such that d(T x, T y) ≤ kd(x, y) for some real positive k < 1. Then T has a unique fixed point, i.e. there exists a unique x ∈ X such that T x = x; furthermore for any initial x0 , the iterates T n x0 converge to the fixed point x. Theorem 2.2. (Riesz Representation theorem) For every bounded linear functional f ∈ H there exists a unique element u ∈ H such that f (v) = (u, v), 3 ∀v ∈ H. Lax-Milgram Theorem Theorem 3.1. Suppose that B : H × H → R is a bilinear form that is 1. bounded, i.e. there exists an α > 0 such that |B(u, v)| ≤ αkukkvk, and 2 ∀u, v ∈ H, 2. coercive, i.e. there exists a β such that B(u, u) ≥ βkuk2 , ∀u ∈ H. 0 Then for any f ∈ H there exists a unique uf ∈ H such that B(uf , v) = f (v), ∀v ∈ H. (3.1) Furthermore kuf k ≤ β −1 kf kH 0 , (3.2) where kf kH 0 = sup v∈H,kvk=1 |hf, vi|. in particular uf depends continuously on f , i.e. kuf − ug k ≤ β −1 kf − gk. Proof. The uniqueness is straightforward. Suppose there exist two solutions u, u¯ then B(u, v) = B(¯ u, v) = f (v), ∀v ∈ H. Since B is bilinear, we have B(u − u¯, v) = 0 for all v ∈ H. Let v = u − u¯. Using the coercivity of B leads to βku − u¯k2 ≤ B(u − u¯, u − u¯) = 0, i.e. u = u¯. Let v = uf in (3.1) to get B(uf , uf ) = f (uf ) By definition of k.kH 0 we know that f (uf ) ≤ kf kH 0 . By using this inequality and coercivity of B, the bound (3.2) is derived. The continuity result follows by considering B(uf , v) − B(ug , v) = B(uf − ug , v) = (f − g, v), and setting v = uf − ug . So only existence requires any work. Fix u ∈ H and consider the map v → B(u, v). We claim that this defines a bounded linear functional on H. Since B is bilinear and bounded, it is clearly linear and bounded. It follows from the Riesz Representation Theorem that there exists a w ∈ H such that (w, u) = B(u, v), 3 ∀v ∈ H. We define Au = w. By this definition we have (Au, u) = B(u, v), ∀v ∈ H. and claim that this definition yields a bounded linear operator from H into itself. Indeed, for every v ∈ H (A(α1 u1 + α2 u2 ), v) =B(α1 u1 + α2 u2 , v) =α1 B(u1 , v) + α2 B(u2 , v) =α1 (Au1 , v) + α2 (Au2 , v) =(α1 Au1 + α2 Au2 , v) Since this holds for every v ∈ H, it follows that A(α1 u1 + α2 u2 ) = α1 Au1 + α2 Au2 , and this implies that A is linear. To show that A is bounded, note that kAuk2 = (Au, Au) = B(u, Au) ≤ αkukkAuk, hence kAuk ≤ αkuk and A is bounded. Using the Riesz Representation, we know that there exists a ϕ ∈ H such that (ϕ, v) = f (v), ∀v ∈ H. We can rewrite our equation in the following way (Au, v) = (ϕ, v), ∀v ∈ H. This means that u satisfies (3.1) if and only if Au = ϕ. Now we have to show that this equation has a solution. In this step of proof we use Banach fixed-point theorem. Clearly, for any λ > 0 Au = ϕ ⇔ u = u − (Au − ϕ). Let T u = u − λ(Au − ϕ). So we are investigating a fixed point for T. Since H is a Hilbert space, it is enough to show that T is a contraction. We have kT u − T vk2 =k(u − v) − λA(u − v)k2 =ku − vk2 − 2λ(A(u − v), u − v) + λ2 kA(u − v)k2 =ku − vk2 − 2λB(u − v, u − v) + λ2 kA(u − v)k2 ≤ku − vk2 − 2λβku − vk2 + λ2 α2 ku − vk2 =(1 − 2λβ + λ2 α2 )ku − vk2 , where we have used the coercivity of B and the fact that A is bounded. It follows that if we choose sufficiently small λ, T is a contraction. It therefore has a unique fixed point, which provides our solution. 4 References [1] P. D. Lax and A. N. Milgram, Parabolic equations , in Contributions to the theory of partial differential equations, Annals of Mathematics Studies, no. 33, Princeton University Press, Princeton, N. J., 1954, pp. 167-190. [2] J.-L. Lions and G. Stampacchia, Variational inequalities ,Comm. Pure Appl. Math. Comm. Pure Appl. Math., 20 (1967), pp. 493-519. 5 Part II Solutions to exercises 87 Chapter 1 Metric spaces 88 Ex. 1.2.4 Consider the sequence 1 , n ≥ 2. log (n) Clearly this sequence converges to zero. However, log (n) grows slower than any polynomial sequence so for each 1 ≤ p < ∞ there exists an Np ∈ N such that ξnp ≥ 1/n for all n ≥ Np . Therefore, since the harmonic series diverges, the sequence ξn is not in any space lp . ξn = Ex. 1.3.8 Show that the closure B(x0 ; r) of an open all B(x0 ; r) in a metric space can differ from the ˜ 0 ; r). closed ball B(x Solution: Let x0 = (0, 0) and r = 1. Define a metric space (X, d), where X = {(x1 , x2 ) | x21 + x22 ≤ 1, x1 and x2 are real and nonnegative} ∪ {(x1 , x2 ) = (cos(1.25π), sin(1.25π))}, and d is the Euclidean distance. The points in set X are shown in Figure 1.1. According 1.5 1 0.5 0 −0.5 −1 −1.5 −1.5 −1 −0.5 0 0.5 1 1.5 Figure 1.1: Points in X to the definitions, B(x0 ; r) = {(x1 , x2 ) ∈ X | x21 + x22 < 1}, ˜ 0 ; r) = {x ∈ X | d(x, x0 ) ≤ 1} = X, B(x B(x0 ; r) = {(x1 , x2 ) ∈ X | (x1 , x2 ) 6= (cos(1.25π), sin(1.25π))}. 89 ˜ 0 ; r). Therefore, B(x0 ; r) can differ from B(x Ex. 1.3.8 Let X be a set with at least two elements and let the metric space (X, d) have the discrete metric ( 0 x=y d(x, y) = . 1 x 6= y ˜ 0 , 1) be the closed ball around x0 ∈ X Let B(x ˜ 0 , 1) = {x ∈ X : d(x, x0 ) ≤ 1}. B(x By the definition of the metric this means ˜ 0 , 1) = X. B(x Let now B(x0 , 1) = {x ∈ X : d(x, x0 ) < 1} be the open ball around x0 . Which means B(x0 , 1) = {x0 }. Let now x˜ ∈ X be an accumulation point of B(x0 , 1). This means that d(˜ x, x0 ) < , ∀ > 0 which implies x˜ = x0 thus the closure of the open ball is B(x0 , 1) = {x0 }, which is not equal to the closed ball. Ex. 1.3.8 Consider the discrete metric on R2 , X = (R2 , d) where B(x0 ; 1) = X and B(x0 ; 1) = {x0 }, clearly differing. Ex. 1.3.12 Let us consider the family of characteristic functions X = x (t) = χ(a,c) (t) , c ∈ (a, b) 90 that is non–countable because (a, b) is a non–countable set. Notice that d (x, y) = 1, x 6= y, x, y ∈ X. Now, let us assume that there exists a dense set in B [a, b], call it G: then there exist as 1 many non–empty balls B x0 , 2 , x0 ∈ X as the elements of X. On the other hand, we know that 1 1 0 0 B x0 , ∩ B x0 , = ∅, x0 6= x0 2 2 and therefore G is non–countable as well. Thus, it does not exist a countable dense set in B [a, b], since we have proved that every dense set in X is non–countable. Ex. 1.3.12 Consider the subset M ⊂ B[a, b] consisting of the functions δx = 1 at x and zero everywhere else on [a, b]. Then M is uncountable, and sup |δx − δy | = 1 whenever x 6= y. Consider now all open balls B(δx ; 1/2). These are disjoint, and so if a subset S ⊂ B[a, b] is dense, then it must intersect each ball in a different point. This renders S uncountable. Since S is general this shows that B[a, b] is not separable. Ex. 1.4.2 If xnk → x, then for every ε > 0 there is an N1 = N1 (ε) such that d (xnk , x) < ε 2 ∀nk > N1 and, since xn is Cauchy we know that there is an N2 = N2 (ε) such that d (xn , xm ) < ε 2 ∀n, m > N2 . Hence by the triangular inequality we obtain for n, nk > max {N1 , N2 } d (xn , x) ≤ d (xn , xnk ) + d (xnk , x) < and the claim follows. 91 ε ε + =ε 2 2 Ex. 1.4.2 For the subsequence we have d(xnk , x) −→ 0, and since (xn ) is Cauchy, Using the triangle inequality we get d(xn , xnk ) −→ 0. d(xn , x) ≤ d(xn , xnk ) + d(xnk , x) −→ 0 which proves that (xn ) is convergent with the limit x. Ex. 1.4.8 If d1 and d2 are metrics on the same set X and there are positive numbers a and b such that for all x, y ∈ X, ad1 (x, y) ≤ d2 (x, y) ≤ bd1 (x, y), (1.0.1) show that the Cauchy sequences in (X, d1 ) and (X, d2 ) are the same. Solution: We use the relation (1.0.1) to show (xn ) is Cauchy in (X, d1 ) ⇐⇒ (xn ) is Cauchy in (X, d2 ) . First assume that (xn ) is Cauchy in (X, d1 ). That is, for any ε > 0 there is an N such that d1 (xn , xm ) < ε b ∀m, n > N. Using (1.0.1) we obtain d2 (xn , xm ) ≤ bd1 (xn , xm ) < ε ∀m, n > N. which proves that (xn ) is Cauchy also in (X, d2 ). Now assume that (xn ) is Cauchy in (X, d2 ). That is, for any ε > 0 there is an N such that d2 (xn , xm ) < aε ∀m, n > N. Using (1.0.1) we obtain 1 d1 (xn , xm ) ≤ d2 (xn , xm ) < ε ∀m, n > N. a which proves that (xn ) is Cauchy also in (X, d1 ). Thus, (xn ) is Cauchy in (X, d1 ) iff (xn ) is Cauchy in (X, d2 ), i.e., the Cauchy sequences in (X, d1 ) and (X, d2 ) are the same. 92 Ex. 1.5.6 Show that the set of all real numbers constitutes an incomplete metric space if we choose d(x, y) = |arctan x − arctan y|. Solution: Let X = R. We show that (X, d) is incomplete by constructing a Cauchy sequence in X which does not converge. Consider the sequence (xn ), where xn = n. Clearly, (xn ) does not converge. Since π , 2 lim arctan x = x→∞ for any ε > 0 we can find N = N (ε) such that π ε arctan n − < 2 2 ∀n > N. Hence, π π d(xn , xm ) = |arctan xn − arctan xm | = arctan n − − arctan m + 2 2 π π ε ε ≤ arctan n − + arctan m − < + = ε ∀n, m > N 2 2 2 2 which shows that (xn ) is Cauchy. Ex. 1.5.8 Let y1 , y2 ∈ Y . Since Y is a subspace it is enough to show that Y is closed. Let {yn }∞ n=1 be a sequence in Y with yn → y ∈ C[a, b], this means ∀ > 0 ∃N : sup |y(t) − yn (t)| < |. t∈[a,b] But by definition |y(a) − yn (a)| ≤ sup |y(t) − yn (t)| t∈[a,b] so we get and by the same argument also yn (a) → y(a), yn (b) → y(b). But since each yn ∈ Y we have yn (a) = yn (b) so we get |y(a) − y(b)| ≤ |y(a) − yn (a)| + |y(b) − yn (b)| → 0 which shows that y(a) = y(b). This means that y ∈ Y and thus is a closed subspace and complete. 93 Ex. 1.5.9 Assume: that (xm ) is the sequence of continuous functions on [a, b] that uniformly converges on [a, b]; Prove: that the limit function x is continuous on [a, b]. Solution: Since (xm (t)) is converging to x, ∀ > 0 ∃N (), ∀t ∈ [a, b], |xm (t) − x(t)| < 3 (1.0.2) On the other hand, (xm (t)) is continuous at any to ∈ [a, b]; so, we have, ∀ > 0 ∃δ > 0, |t − to | < δ ⇒ |xm (t) − xm (to )| < 3 (1.0.3) Now, we want to show that x(t) is continuous on [a, b]; therefore, ∀ > 0 ∃δ1 > 0, |t − to | < δ1 ⇒ |x(t) − x(to )| < (1.0.4) For this purpose, we can find the following relation, using (1) and (2): |x(t) − x(to )| = |x(t) ± xm (t) ± xm (to )| = = |(x(t) − xm (t)) + (xm (t) − xm (to )) + (xm (to ) − x(to ))| ≤ |x(t) − xm (t)| + |xm (t) − xm (to )| + |xm (to ) − x(to )| + + = ≤ 3 3 3 which is equivalent to (3). Ex. 1.6.14 The only question here is whether d(x, y) = 0 ⇒ x = y or not. i) Let d(x, y) = 0 and let h(t) = x(t) − y(t) which implies Z a b |h(t)|dt = 0. Assume that there exist t0 ∈ [a, b] such that |h(t0 )| = 6 0. Since x and y are continuous h is continuous. So let 0 < = |h(t0 )| and let δ be so small that if |t − t0 | < δ we have |h(t) − h(t0 )| < . 94 The reverse triangle inequality gives us: |h(t0 )| − |h(t)| ≤ |h(t) − h(t0 )| < . Since = |h(t0 )| this now implies |h(t)| > 0, ∀t ∈ [t0 − δ, t0 + δ]. But since: Z a this implies b |h(t)|dt ≥ Z a Z t0 +δ t0 −δ |h(t)|dt b |h(t)|dt > 0 which is a contradiction. So no t0 with |h(t0 )| = 6 0 can exist, thus x(t) = y(t). ii) For Riemann-integrable functions d(x, y) = 0 does not imply x = y. An example is: let tm = (a + b)/2 and ( 0 t < tm x(t) = 1 t ≥ tm ( 0 t ≤ tm y(t) = 1 t > tm We have x(tm ) 6= y(tm ) but d(x, y) = 0. 95 Chapter 2 Normed spaces 96 Ex. 2.3.10 Suppose that the Banach space (X, k · k) has a Schauder basis {en }∞ n=1 so that for any x ∈ X there is a unique sequence of scalars, α1 , α2 , . . . such that x= ∞ X αk ek . k=1 We will for simplicity assume that αk ∈ R, though generalisation to complex numbers is straight forward. Given any > 0 there is an N such that kx − N X αk ek k < . 2 k=1 Now, recall that Q is dense in R, meaning that for any αk ∈ R we can find a βk ∈ Q such that |αk − βk | < /2k+1 . By the triangle inequality we have N X N X N X N X N X kx − βk ek k ≤ kx − αk e k k + k αk ek − βk ek k < + < . 2 k=1 2k+1 k=1 k=1 k=1 k=1 Thus the Schauder basis constitutes a basis for a dense subset of X, which is countable since βk ∈ Q for all k. Consequently, X is separable. Ex. 2.3.10 Let Y be the set, N X Y ={ αj ej : αj ∈ R, N ∈ N+ }. j=1 Let x ∈ X, since we have a Schauder-basis there exist {αj }∞ j=1 such that kx − xn k → 0 where xn := n X αj ej . j=0 Thus for every x ∈ X we have a sequence in Y converging to x. This shows that Y¯ = X. 97 Let now instead xn = set Pn j=1 αj ej ∈ Y be an arbitrary element in Y and let M be the N X M ={ βj ej : bj ∈ Q, N ∈ N+ }, j=1 k which is countable. Since Q is dense in R there exist a sequence {βjk }∞ k=1 with βj ∈ Q such that |αj − βjk | ≤ , ∀k ≥ K, Cn where C = max kej k . j∈{1,...,n}] Let xkn = Pn j=1 βjk ej ∈ M , we get: n n X X xn − xkn = (αj − βjk )ej |αj − βjk | kej k , ≤ j=1 j=1 so for k ≥ K we get n n X X xn − xkn ≤ kej k ≤ 1 ≤ . Cn n j=1 j=1 Showing that xkn → xn when k → ∞. Thus for every xn ∈ Y we can find a sequence in M ¯ = Y and in total converging to xn . So M ¯) = M ¯ X = Y¯ = (M showing that M is dense in X so that X is separable. Ex. 2.4.4 Show that equivalent norms on a vector space X induce the same topology for X. Solution: Assume that k·k1 and k·k2 are equivalent norms on X. Let T1 be the topology for X induced by k·k1 and let T2 be the topology for X induced by k·k2 . In other words, T1 and T2 are the collections of all open subsets of X in the sense of k·k1 and k·k2 , respectively. We show that T1 = T2 by showing that any subset of X is open in the sense of k·k1 iff it is open in the sense of k·k2 . Since k·k1 and k·k2 are equivalent, we have a k·k1 ≤ k·k2 ≤ b k·k1 (2.0.1) for some a, b > 0. Let Bi (x; r) denote the open ball of radius r around x in the sense of k·ki . Now assume that M is an open subset of X in the sense of k·k1 . By the definition of an open set, M contains an open ball around each of its points, i.e., for any x ∈ M ∃ε > 0 98 such that B1 (x; ε) ⊂ M . Because of (2.0.1) we have B2 (x; aε) ⊂ B1 (x; ε), which means that M contains an open ball in the sense of k·k2 around each of its points. This proves that any M ⊂ X is open in the sense of k·k2 if it is open in the sense of k·k1 . By reversing the above argument, we can show that any M ⊂ X is open in the sense of k·k1 if it is open in the sense of k·k2 . Ex. 2.4.4 Suppose that the norms kk0 and kk1 are equivalent. Then there are constants c1 and c2 with 0 ≤ c1 ≤ c2 such that c1 k x k0 ≤k x k1 ≤ c2 k x k0 for all x ∈ X. Let U be a subset of X which is open with respect to the topology on X induced by the norm kk1 . For u ∈ U there is an such that {x ∈ X k u − x k1 < } ⊂ U . But since the norms are equivalent {x ∈ X k u − x k1 < } ⊂ {x ∈ X k u − x k0 < /c2 }. This means that U is open with respect to the topology induced by norm kk0 . By the same argument, we can show that any open subset of X with respect to the topology induced by the norm kk0 is open with respect to the topology induced by kk1 . Ex. 2.5.4 Show that for an infinite subset M in the space s to be compact, it is necessary that there are numbers γ1 , γ2 , . . . such that for all x = (ξk (x)) ∈ M we have |ξk (x)| ≤ γk . Solution: The metric on the space s is d(x, y) = ∞ X 1 |ξk − ηk | 2k 1 + |ξk − ηk | k=1 where x = (ξj ), y = (ηj ). Assume that not all (ξk ) in M are bounded by |ξk | ≤ γk . Then there is at least one k0 such that |ξk0| is unbounded. Now consider a sequence of elements (m) (m) in M , x1 , x2 , . . ., where xm = ξk . We choose the sequence such that ξk0 = m. We then have, for m 6= n, ∞ (m) (n) X 1 |m − n| 1 1 1 |ξk − ξk | ≥ k0 ≥ k0 . |xm − xn | = (m) (n) k 2 1 + |ξk − ξk | 2 1 + |m − n| 2 2 k=1 Clearly (xm ) can not have a convergent subsequence. Thus, M is not compact. Ex. 2.5.10 Let X and Y be metric spaces, X compact, and T : X → Y bijective and continuous. Show that T is a homeomorphism. 99 In the proof, I use the result from Prob. 14 in Section 1.3: a mapping T : X → Y is continuous if and only if the inverse image of any closed set M ⊂ Y is a closed set in X. This result is equivalent to a mapping T −1 : Y → X is continuous if and only if for any closed set M ⊂ X, T (M ) is a closed set in Y . (*) Let M be a closed subset in X. Since X is a compact metric space, M is compact (Prob. 2.5-9). Since T is continuous, according to Theorem 2.5-6 we have that T (M ) is compact. By using Lemma 2.5-2, we have that T (M ) is closed and bounded. Since M is arbitrary and T is bijective, by (*) we have that T −1 is continuous. Therefore, T is a homeomorphism. Ex. 2.5.10 Let X and Y be metric spaces, X be compact, T : X → Y bijective and continuous and F = T −1 . • Let V ⊆ X and assume that V is closed and bounded so that it is compact. • Since T is continuous, it follows that T (V ) is compact. • This implies that T (V ) is closed and bounded. • But T (V ) = F −1 (V ) so F −1 (V ) is closed and bounded. Hence F is continuous, and thus T is a homeomorphism. Ex. 2.5.10 We study the continuous and bijective operator T : X → Y where X is compact. By theorem 2.5-6, Y is compact. Lets assume that T −1 is discontinuous, that is there is a y ∈ Y and x ∈ X such that dy (yn , y) → 0 dx (xn , x) > δ > 0 (2.0.2) (2.0.3) where the sequences xn , yn satisfy T xn = yn and T x = y. However, X is compact, so there is a subsequence of xn such that xk → x0 6= x. We have dy (T x0 , T x) ≤ dy (T xk , T x) + dy (T x0 , T xk ) = dy (yk , y) + dy (y0 , yk ). Since T is continuous, we have dy (y0 , yk ) = dy (T x0 , T xk ) → 0 as k → ∞. Therefore, we have dy (T x0 , T x) ≤ dy (yk , y) + dy (y0 , yk ) → 0 100 as k → ∞. However, according to (8) we should have for bijective T dy (T x, T x0 ) > > 0, which is a contradiction. T −1 must therefore be continuous and T is then a homeomorphism. Ex. 2.5.10 A Homeomorphism is a continuous bijective mapping whose inverse is continuous. Since we have a continuous and bijective mapping T : X −→ Y (X and Y metric spaces, X compact), we need to show that the inverse T −1 is continuous. Generally, a mapping T : X −→ Y is continuous if and only if, for any closed set K ⊂ Y , the inverse image T −1 (K) is a closed set in X (cf. Prob. 14, Sec. 1.3). So here, to show that T −1 : Y −→ X is continuous, we want to show that T (M ) is a closed set in Y , for any closed set M ⊂ X. 1. Let M be a closed subset in X-compact, then from Prob. 9, Sec. 2.5, M is also compact. 2. Since T is continuous, and M is compact ⇒ T (M ) is compact (Theorem 2.5-6) 3. Since T (M ) is a compact subspace of the metric space Y , T (M ) is closed and bounded (Lemma 2.5-2) Ex. 2.7.2 • Consider T bounded. If there exists M > 0 such that kxk ≤ M , then kT xk ≤ c kxk = cM and therefore T maps bounded sets in X into bounded sets in Y . • Let us consider T which maps bounded sets in X into bounded sets in Y . If x = 0, then T is trivially bounded since T (0) = T (0R z) = 0R T (z) = 0 ∀z ∈ X ⇒ kT 0k = 0. Let us consider x 6= 0. We know that exists C > 0 such that kT xk ≤ C, ∀x 6= 0 such that kxk = 1. Therefore x T ≤C kxk and for the homogeneity of the norm we have kT xk ≤ C kxk. 101 Ex. 2.7.2 Let X and Y be normed spaces. A linear operator T : X → Y is bounded iff T maps bounded sets in X into bounded sets in Y . If T is bounded: We want to show that for any bounded set M ⊂ D(T ) s.t. for any x, y ∈ M we have that kx − yk ≤ r < ∞ than also kT x − T yk < ∞. Since D(T ) is a vector space then x, y ∈ M ⊂ D(T ) → x − y = z ∈ D(T ). Furthermore kT zk ≤ ckzk since T is bounded. Hence kT x − T yk = kT (x − y)k = kT zk ≤ ckzk < ∞. If T maps bounded sets in X into bounded sets in Y : We know that for any x, y ∈ M bounded than kT x − T yk < ∞. Then we call δ(M ) = sup kx − yk and δ(T (M )) = x,y∈M sup T x,T y∈T (M ) kT x − T yk. For any zD(T ), z 6= 0 we define x=y+ δ(M ) z kzk ⇒ kx − yk = δ(M ). Then kT x − T yk = kT (x − y)k = kT which means that kT zk = kT x − T yk where c = δ(M ) δ(M ) z k= kT zk kzk kzk δ(T (M )) kzk ≤ kzk = ckzk. δ(M ) δ(M ) δ(T (M )) . δ(M ) Ex. 2.7.2 A linear operator T : D(T ) −→ Y , (where D(T ) ⊂ X and X and Y are normed spaces), is bounded if there is a real number c such that for all x ∈ D(T ), kT xk ≤ ckxk. (2.0.4) Further, a subset K in a normed space X is bounded if and only if there is a positive number a such that kxk ≤ a (2.0.5) for every x ∈ K. Let M be a bounded set in X. 102 1. Show that if T is bounded ⇒ T (M ) is a bounded set in Y : Let x ∈ M and T x ∈ T (M ), then from (2.0.4) and (2.0.5) we have kT xk ≤ ckxk ≤ ca (2.0.6) where ca is a positive number. Since x ∈ M is arbitrary this proves T (M ) is a bounded set in Y (according to the definition (2.0.5) of a bounded subset). 2. Show that if T (M ) is a bounded set in Y ⇒ T is bounded: Let x ∈ M and T x ∈ T (M ), then using the definition for the operator norm we have kT kkxk = kT xk kT xk kxk ≥ kxk = kT xk, kxk x∈M,x6=0 kxk sup x 6= 0. (2.0.7) Since kT k is a real number, call it c, we have kT xk ≤ ckxk, x 6= 0. (2.0.8) For the case when x = 0, since for linear operators T 0 = 0, we have kT xk = kT 0k = 0 ≤ ckxk, x = 0. (2.0.9) Since inequality (2.0.4) holds for all x ∈ M , T is bounded. Ex. 2.7.7 Given: T : X → Y is bounded and linear.Furthermore, ∃b > 0 3 ∀x ∈ X, kT xk ≥ bkxk ⇒ kxk ≤ kT xk b (2.0.10) We want to show that T −1 : Y → X exists and is bounded. According to [Theorem 2.6-10(a)], T −1 exists if and only if T x = 0 implies x = 0. In our case, T x = 0 ⇒ 0 ≥ bkxk Since b > 0 the only way for the latter inequality to be valid is that, x = 0. Hence, T −1 exists. Also, based on the definition of mapping we have, ∀x ∈ X ∃y ∈ Y 3 y = T x ∧ x = T −1 y kT xk kyk ⇒ kT −1 yk = kxk ≤(1) = b b 1 ⇒ kT −1 yk ≤ kyk b therefore, according to [Definition 2.7-1], T −1 is bounded. 103 (2.0.11) (2.0.12) (2.0.13) Ex. 2.7.8 If we consider T : l∞ → l∞ , x 7→ y defined by x = {αj } , y = {βj } , s.t. βj = αj j the bounded set y = {1} ⊂ R (T ) is mapped by T −1 into x = {j}j∈N , that is not bounded. Ex. 2.7.8 Let X be a normed space with points x = {ξk } being sequences of real or complex numbers that has only finitely many non-zero terms. Further let kxk = supk |ξk |. Now define the operator T : X → X as T x = (ξ1 , ξ2 /2, . . . ) = {ξk /k}. If y = {ηk } and α and β are numbers then T (αx + βy) = {(αξk + βηk )/k} = α{ξk } + β{ηk } = αT (x) + βT (y) so T is a linear operator. Also note that kT xk = sup |ξk |/k ≤ kxk sup 1/k ≤ kxk k k so T is bounded. Now, note that T −1 x = {kξk }. Consider the sequence x = {δk,n } ∈ X. Then kxk = 1 and kT −1 xk = n. Since n is an arbitrary integer and x and infinite sequence, we may let n → ∞, which shows that T −1 is not bounded. Ex. 2.8.6 The space C 1 [a, b] is the normed space of all continuously differentiable functions on J = [a, b] with norm defined by kxk = max |x(t)| + max |x0 (t)|. t∈J (2.0.14) t∈J (N1) Since |x(t)| ≥ 0 and |x0 (t)| ≥ 0, kxk ≥ 0. (N2) ”⇒”: kxk = 0 ⇔ max |x(t)| = − max |x0 (t)|⇔ x(t) = 0 and x0 (t) = 0. t∈J t∈J ”⇐”: x = 0 ⇒ kxk = max |0| + max |0| = 0 t∈J t∈J (N3) kαxk = max |αx(t)| + max |αx0 (t)| = |α| max |x(t)| + |α| max |x0 (t)| = |αk|xk t∈J t∈J t∈J t∈J (N4) kx + yk = {differentiation is linear} = max |x(t) + y(t)| + max |x0 (t) + y 0 (t)| t∈J t∈J ≤ max |x(t)| + max |y(t)| + max |x0 (t)| + max |y 0 (t)| = kxk + kyk. t∈J t∈J t∈J t∈J 104 Define f (x) = x0 (c), c = (a + b)/2. This functional on C 1 [a, b] is bounded since a + b a + b a + b 0 0 kxk = max |x(t)| + max |x0 (t)| ≥ x + x ≥ x t∈J t∈J 2 2 2 ⇔ |f (x)| ≤ kxk. (2.0.15) Further, f is linear since f (αx + βy) = {differentiation is linear} = αx0 = αf (x) + βf (y). a + b 2 + βx0 a + b 2 (2.0.16) If f is now considered as a functional on the subspace of C[a, b] which consists of all continuously differentiable functions, we want to show that f is not bounded. We consider the sequence of continuous and differentiable functions (xε ) = arctan( εt ), where ε is a real number. If we let ε −→ 0, then for c = 0, lim x0ε (c) = ∞, ε−→0 see Figure 2.1. So, given a constant C, there exists an ε such that |f (xε )| > Ckxε k, which proves that f is not bounded (since kxε k = max | arctan( εt )| ≤ t∈J ε). π 2 independent of Ex. 2.8.6 The space C 1 [a, b] is the normed space of all continuously differentiable functions on J = [a, b] with norm defined by kxk = max |x(t)| + max |x0 (t)|. t∈J t∈J (2.0.17) Show that the axioms of a norm are satisfied. Show that f (x) = x0 (c), c = (a + b)/2, defines a bounded linear functional on C 1 [a, b]. Show that f is not bounded, considered as a functional on the subspace of C[a, b] which consists of all continuously differentiable functions. Solution: First, we show that the four axioms of a norm are satisfied in this problem. (N1). By the definition of maximum and absolute function, kxk ≥ 0. (N2). If kxk = 0, as both max|x(t)| and max|x0 (t)| are nonnegative, |x(t)| = |x0 (t)| = 0. t∈J t∈J If x(t) = 0 is a constant function, then |x0 (t)| = 0. Therefore, kxk = 0. 105 2 1.5 arctan(x/eps) 1 0.5 0 −0.5 −1 −1.5 −2 −100 −80 −60 −40 −20 0 20 40 60 80 100 x Figure 2.1: The function arctan( xε ). Smaller and smaller values of ε leads to a steeper and steeper function. (N3). kαxk = max |αx(t)| + max |αx0 (t)| t∈J t∈J = |α|(max |x(t)| + max |x0 (t)|) t∈J = |α|kxk. 106 t∈J (2.0.18) (N4). The triangle inequality is satisfied kx + yk = max |x(t) + y(t)| + max |x0 (t) + y 0 (t)|. t∈J t∈J ≤ max |x(t)| + max |y(t)| + max |x0 (t)| + max |y 0 (t)|. t∈J t∈J t∈J t∈J = kxk + kyk. (2.0.19) Then, we show that functional f (x) is linear and bounded on C 1 [a, b]. As the space C 1 [a, b] is a normed space, it defines the algebraic operations. For x, y ∈ C 1 [a, b] and any scalar α, the operation f satisfies f (x + y) = (x(c) + y(c))0 = f (x) + f (y), (2.0.20) f (αx) = (x(c) + y(c))0 = f (x) + f (y). (2.0.21) and This shows that f is linear. Consider the norm in C 1 [a, b], |f (x)| = |x0 (c)| ≤ max |x0 (t)| ≤ max |x(t)| + max |x0 (t)| = kxk, t∈J t∈J t∈J (2.0.22) followeding the definition in 2.8-2, f is bounded on C 1 [a, b]. However, on the space C[a, b], where the norm is defined by kxk = max |x(t)|, t∈J (2.0.23) let xn (t) = sin(ntπ/c), where n ∈ N. Then kxn k ≤ 1 and nπ f (xn ) = x0n (c) = cos(nπ). (2.0.24) c So kf (x)k = nπ/c. As n ∈ N is arbitrary, this shows that there is no fixed number α such that kf (x)k ≤ αkxk, so f is unbounded. Ex. 2.10.4 Let X and Y be normed spaces and Tn : X → Y , n = 1, 2, 3, .., be bounded linear operators. Show that convergence Tn → T implies that for every > 0 there is an N such that for all n > N and all x in any given closed ball we have kTn x − T xk < . ˜ 0 ; r) ⊂ X. Then, for all x ∈ B(x ˜ 0 ; r), Solution: Denote the given closed ball by B(x there exists a positive number C such that kxk ≤ C. From the convergence of Tn , for a given > 0, there exists N ∈ N such that for all n > N , (2.0.25) kTn − T k ≤ . C ˜ 0 ; r) Therefore, for all x ∈ B(x kTn x − T xk ≤ kTn − T kkxk ≤ . 107 (2.0.26) Ex. 2.10.8 ∞ Since {ek }∞ k=1 with ek = (δkj )j=1 is a Schauder basis for c0 , for every x ∈ c0 we can write ∞ X x= λk e k , k=1 where x = (λk ) and limn→∞ λk = 0 and we can approximate x by sn = n X λk ek , k=1 then limn→∞ k sn − x k∞ = 0. Let f ∈ c00 , which is the dual space of c0 . This means that f : c0 → R is a bounded linear functional, so f (sn ) → f (x), as n → ∞. So f (x) = ∞ X λk f (ek ). k=1 n) for f (en ) 6= 0 and γn = 0 Now we show that f (ek ) is in l1 . Define the sequence γn = |ff (e (en )| for f (en ) = 0. Consider the sequence xn = (γ1 , γ2 , · · · , γn , 0, 0, · · · ) = γ1 e1 + · · · + γn en . It is obvious that k xn k∞ ≤ 1, so |f (xn )| = So Pn k=1 n X k=1 |f (ek )| ≤k xn k∞ k f k≤k f k . |f (ek )| ≤k f k for all n. Since this inequality holds for all n, we have ∞ X k=1 |f (ek )| ≤k f k . (2.0.27) This means that (f (ek )) ∈ l1 . Now we want to show that for every y = (λk ) ∈ l1 , we can obtain a corresponding bounded linear operator g : c0 → R. Let g(x) = ∞ X λk µk , x = (µk ). k=1 Clearly g is linear and |g(x)| = | ∞ X k=1 λk µk | ≤ sup |µj | j ∞ X k=1 |λk | ≤k x k∞ ∞ X k=1 |λk |. So g is bounded. Taking the supremum over all x with norm 1, we get |g(x)| ≤ ∞ X k=1 108 |λk |. This inequality and the inequality (2.0.27) gives ∞ X k=1 |f (ek )| =k f k . So the mapping f : c00 → l1 with f → (f (ek )) is bijective and preserves norms and hence isomorphism. Ex. 2.10.8 We consider f a bounded linear functional on c0 . Since a Schauder basis for c0 is {ek }k∈N , with ek = δkj , then every x = {αk } ∈ c0 has a unique representation x= ∞ X αk ek k=1 and, from the linearity of f , we know also that f (x) = ∞ X αk f (ek ) . k=1 We can write |f (x)| = ∞ X αk f (ek ) ≤ k=1 ∞ X k=1 |αk | |f (ek )| ≤ kxk∞ ∞ X k=1 |f (ek )| = c kxk∞ , where c = |f (ek )| < ∞ because f (x) is bounded. This implies that {f (ek )} ⊂ l1 and 0 therefore c0 ⊆ l1 . On the other hand, for every b = {βk }k∈N ∈ l1 , we can obtain a bounded linear functional g in c0 . In fact, we may define g on c0 by f (x) = ∞ X αk βk . k=1 Then g is • Linear g (λx + µy) = ∞ X (λαk + µγk ) βk = k=1 ∞ X =λ k=1 αk βk + µ ∞ X k=1 where y = {γk }k∈N ∈ c0 . 109 γk βk = λg (x) + µg (y) , λ, µ ∈ R • Bounded ∞ ∞ ∞ X X X |g (x)| = α k βk ≤ |αk βk | ≤ kxk∞ βk = kxk∞ kbk1 . k=1 0 k=1 k=1 0 Then g ∈ c0 and therefore l1 ⊆ c0 . P We finally prove that kgk = kbk1 . Indeed, since ∞ k=1 |βk | converges, then ∞ X ∀ε > 0, ∃N = N (ε) s.t. k=n+1 |βk | < ε, ∀n > N. Let us consider x s.t. αk = x ∈ c0 , and ( |βk | , βk 0, k < N, k≥N . ∞ ∞ ∞ X X X |g (x) − kbk1 | = αk βk − |βk | = |βk | < ε. k=1 k=1 k=N +1 0 This shows that the bijective linear mapping on l1 onto c0 defined by g is an isomorphism. Ex. 2.10.8 Show that the dual space of the space c0 is l1 . c0 is the space of all sequences of scalars converging to zero. The norm on c0 is k · k∞ . 0 0 Let c0 be the dual space of c0 . We want to show that c0 is isomorphic with l1 . The norm on l1 is k · k1 . We prove it step by step. Step 1: Let x ∈ c0 , then x has a unique representation x= ∞ X ξk ek , k=1 where (ek ) = (δkj ) is a Schauder basis for c0 , (δkj ) has one in the k th place P and zeros otherwise, and ξk → 0 as k → ∞. Since f is linear, we have f (x) = ∞ k=1 ξk f (ek ). Since f is bounded on c0 , we have kf k = sup x∈c0 110 |f (x)| < ∞. kxk∞ We construct s sequence (¯ xk ) such that ( x¯k = |f (ek )| f (ek ) 0 k ≤ n and f (ek ) 6= 0 k > n or f (ek ) = 0 |f (¯ x)| Obviously, (¯ xk ) ∈ c0 . Therefore, k¯ < ∞. Since the elements of (¯ xk ) can only take xk∞ values 0 , 1, or -1, we have k¯ xk∞ = 1. We notice that f (¯ x) = ∞ X x¯k f (ek ) = k=1 n X k=1 |f (ek )|, f (¯ x) ≤ kf kk¯ xk∞ = kf k Since n is arbitrarily large, we have ∞ X k=1 |f (ek )| ≤ kf k. (2.0.28) Step 2: Take any x ∈ c0 , we have |f (x)| = | ∞ X ξk f (ek )| k=1 ≤ kxk∞ ∞ X k=1 |f (ek )|. Taking all x such that kxk∞ = 1 gives kf k ≤ ∞ X k=1 |f (ek )|. (2.0.29) P 1 By (2.0.28) and (2.0.29), we get that kf k = ∞ k=1 |f (ek )|. Therefore, f is in l . This also 1 shows that given P∞ a sequence (ηk ) in l , there exists a linear bounded operator g on c0 such that g(x) = k=1 ξk ηk , where x = ξk ∈ c0 . g is linear by construction. Boundedness can be proved similar to (2.0.29). 0 We conclude that c0 is isomorphic with l1 . 111 Chapter 3 Inner product spaces 112 Ex. 3.3.4 A X is a Hilbert space (complete) and M ⊂ X is closed. A closed subset of a complete one is also complete, so M is complete. M is also a vector space, which is convex by definition. Hence, M has all the properties needed for Theorem 3.3-1. B Appolonius identity is 1 1 ||x − yn ||2 + ||x − ym ||2 = ||yn − ym ||2 + 2||x − (yn + ym )||2 . 2 2 (3.0.1) Since M is convex, we have 1 2||x − (yn + ym )||2 ≥ 4δ 2 2 where δ = ||x − y|| the minimum distance. Putting vn = yn − x and using this in (3.0.1), one obtains 1 ||vn ||2 + ||vm ||2 ≥ ||vn − vm ||2 + 2δ 2 . (3.0.2) 2 Using the parallelogram equality on the left-hand side of (3.0.2), one can conclude that ||vn − vm ||2 ≥ 4δ 2 which is the first part of the proof. The rest of the proof can then proceed as usual. Ex. 3.3.4 (a) Show that the conclusion of Theorem 3.3-1 also holds if X is a Hilbert space and M ⊂ X is a closed subspace. (b) How could we use Appolonius’ identity (Sec. 3.1, Prob. 5) in the proof of Theorem 3.3-1? Solution The problem states that: Let X be a Hilbert space and a closed subspace M ⊂ X. For every given x ∈ X there exists a unique y ∈ M , such that δ = inf kx − y˜k = y˜∈M kx − yk. 1) Existence. From theorem 3.2-4, we know that M is complete. From theorem 3.3-4, the Hilbert space X can be represented as X = M ⊕ M ⊥ . Then for any given x ∈ X there exists y ∈ M and z ∈ M ⊥ such that x = y + z. Then we prove such a y satisfies kx − yk ≤ kx − y˜k for any y˜ ∈ M . Since y − y˜ ∈ M , the norm kx − y˜k2 = = = = ≥ kx − y + y − y˜k2 kz + y − y˜k2 kzk2 + < z, y − y˜ > + < y − y˜, z > +ky − y˜k2 kzk2 + ky − y˜k2 kzk2 . 113 2) Uniqueness. Suppose there are two different elements in M satisfy the minimizing condition, denote as y1 and y2 such that δ = inf kx − y˜k = kx − y1 k = kx − y2 k. By using the Appolonius’ y˜∈M identity, ky1 − y2 k2 = 2kx − y1 k2 + 2kx − y2 k2 − 4kx − = 2δ 2 + 2δ 2 − 4kx − y1 + y2 2 k. 2 y1 + y2 2 k 2 (3.0.3) 2 2 2 As y1 6= y2 , we have kx − y1 +y k < δ 2 . Then, y = y1 +y is a better approximation for 2 2 the norm kx − yk in M ., contradiction. Therefore, there is a unique y ∈ M satisfies the minimizing condition. Ex. 3.3.4 (a) Show that the conclusion of Theorem 3.3-1 (minimizing vector) holds also if X is a Hilbert space and M ⊂ X is a closed subspace. (b) How could we use Appolonius’ identity in the proof of Theorem 3.3-1? Solution: (a) The assumptions in Theorem 3.3-1 are that X is an inner product space and M 6= ∅ is a convex subset which is complete. We now assume that X is a Hilbert space and M is a closed subspace. Since M is closed and X is a Hilbert space, M is complete. Since M is a subspace, M is also convex. Thus, all the assumptions in Theorem 3.3-1 are fulfilled and the conclusions of the Theorem follow. (b) Appolonius’ identity reads kc − ak2 + kc − bk2 = 1 1 ka − bk2 + 2kc − (a + b)k2 2 2 or, equivalently, 1 ka − bk2 = 2 kc − ak2 + 2 kc − bk2 − 4kc − (a + b)k2 . 2 (3.0.4) To prove uniqueness of the minimizing vector in Theorem 3.3-1, assume that both y ∈ M and y0 ∈ M satisfy ky − xk = δ and ky0 − xk = δ. By Appolonius’ identity (3.0.4) with a = y, b = y0 and c = x, we obtain 1 ky − y0 k2 = 2 kx − yk2 + 2 kx − y0 k2 − 4kx − (y + y0 )k2 2 1 = 2δ 2 + 2δ 2 − 4kx − (y + y0 )k2 . 2 Because M is convex, 21 (y + y0 ) ∈ M , so that kx − 12 (y + y0 )k ≥ δ, which implies ky − y0 k2 ≤ 2δ 2 + 2δ 2 − 4δ 2 = 0. 114 But since ky − y0 k2 ≥ 0, we must have ky − y0 k = 0 and thus y = y0 , which proves that the minimizing vector is unique. Ex. 3.3.10 We want to show that if Y ⊂ H is a closed subspace such that M ⊂ Y implies that M ⊥⊥ ⊂ Y. As a first step let y⊥ ∈ Y ⊥ . This implies hy⊥ , mi = 0 ∀m ∈ M since M ⊂ Y , which shows us that y⊥ ∈ M ⊥ . This means that Y ⊥ ⊂ M⊥ (3.0.5) since y⊥ was arbitrary. L ⊥ Let now x ∈ M ⊥⊥ . Since Y is a closed subspace we have that X = Y Y . So x can ⊥ be decomposed as x = y + y⊥ with y ∈ Y and y⊥ ∈ Y . But from (3.0.5) we have that hx, y⊥ i = 0 so we get 0 = x, y ⊥ = hy + y⊥ , y⊥ i = ky⊥ k2 , which means that y⊥ = 0. Thus x = y ∈ Y , and since x was arbitrary we get M ⊥⊥ ⊂ Y. Ex. 3.3.10 If M ⊂ H, show that M ⊥⊥ is contained in any closed subspace Y ⊂ H such that M ⊂ Y . There are two cases: Case I) if M is closed; then, M = M ⊥⊥ by Theorem (3.3-6). Since M ⊂ H, then M ⊥⊥ is the smallest subset of H that contains M . Case II) if M is not closed; we know that M ⊥⊥ contains M ; i.e., ∀x ∈ M, x⊥M ⊥ ⇒ x ∈ M ⊥⊥ ⇒ M ⊂ M ⊥⊥ If there exists a closed subspace Y ⊂ H, then, M ⊂ Y ⇒ Y ⊥ ⊂ M ⊥ ⇒ M ⊥⊥ ⊂ Y ⊥⊥ = Y Y was arbitrary, so M ⊥⊥ is the smallest closed subset of H which contains M . 115 Ex. 3.4.6 (Minimum property of Fourier coefficients) Let {e1 , . . . , en } be an orthonormal set in an inner product space X, where n is fixed. Let x ∈ X be any fixed element and y = β1 e1 + . . . + βn en . Then kx − yk depends on β1 , . . . , βn . Show by direct calculation that kx − yk is minimum if and only if βj = hx, ej i, where j = 1, . . . , n. Solution: kx − yk2 = hx − y, x − yi = hx, xi − hx, yi − hy, xi + hy, yi = kxk2 + kyk2 − hx, yi − hx, yi = kxk2 + kyk2 − 2Re {hx, yi} ≥ kxk2 + kyk2 − 2|hx, yi| n n X X = kxk2 + |βk |2 − 2|hx, βk ek i| k=1 ≥ kxk2 + 2 = kxk + By choosing βk = hx, ek i, n X k=1 n X k=1 k=1 |βk |2 − 2 n X k=1 |βk ||hx, ek i| 2 2 2 (|βk | − |hx, ek i|) − |hx, ek i| ≥ kxk − n X k=1 |hx, ek i|2 k = 1, . . . , n, we obtain kx − yk2 = kxk2 + kyk2 − 2Re {hx, yi} ) ( n n X X βk e k i = kxk2 + |βk |2 − 2Re hx, ( n k=1 ) X |βk |2 − 2Re βk hx, ek i k=1 = kxk2 + n X = kxk2 − n X k=1 k=1 ) ( n n X X |hx, ek i|2 = kxk2 + |hx, ek i|2 − 2Re k=1 k=1 k=1 |hx, ek i|2 which shows that the choice βk = hx, ek i, k = 1, . . . , n minimizes kx − yk 116 Ex. 3.4.6 Since X is an inner product space and x, y ∈ X, x − y is also in X, and we can use the Bessel inequality for x − y: 2 kx − yk ≥ = n X k=1 n X k=1 = = n X k=1 n X k=1 |hx − y, ek i|2 |hx, ek i − hy, ek i|2 |hx, ek i − n X j=1 βj hej , ek i|2 |hx, ek i − βk |2 . Thus, the minimun of kx − yk is obtained if and only if βk = hx, ek i. Ex. 3.4.6 Let {e1 , ..., en } be an orthonormal set in an inner product space X, where n is fixed. Let x ∈ X be any fixed element and y = β1 e1 + ... + βn en . We want to show that kx − yk is minimum if and only if βi = hx, ei i. The set {e1 , ..., en } is a basis of a closed finite vectorial subspace Y ⊂ X, i.e. Y = span(e1 , ..., en ), dim(Y ) = n. which is a convex and complete (see Theorem 2.4-2) subspace of X (see Exercise 3.3.4). Then we can say that y = β1 e1 + ... + βn en ∈ Y for any list of n scalars (β1 , ..., βn ). From Theorem 3.3-1, we know that for every given x ∈ X there exists a unique y ∈ Y such that kx − yk = inf kx − y˜k, y˜∈Y (3.0.6) which in our context we can rephrase as follows: there exists a unique list of n scalars (β1 , ..., βn ) such that y = β1 e1 + ... + βn en minizes kx − yk. From Lemma 3.3-2 we also know that a such y must be the orthogonal projection of x on Y , i.e. z = x−y ∈ Y ⊥ . Then we need to show that if we choose the coeffincients of y to Pbe βi = hx, ei i, then z = x − y is orthogonal to any other element of Y . We take v ∈ Y , nk=1 αk ek and we multiply it by 117 z with respect to the inner product of X: hz, vi = hx − y, vi = hx, vi − hy, vi = hx, = = = = Pn k=1 αk ek i − h Pn αk hx, ek i − Pn αk hx, ek i − k=1 Pn k=1 k=1 Pn k=1 αk hx, ek i − Pn k=1 Pn k=1 Pn k=1 Pn k=1 βk ek , Pn βk hek , βk Pn k=1 Pn j=1 αk ek i k=1 αk ek i αj hek , ej i βk α k αk [hx, ek i − βk ] which prove Pnthat hz, vi = 0 iff βi = hx, ei i . Note that in the second last equality we have used that j=1 αj hek , ej i = αk since hek , ej i = 1 only when k = j and zero otherwise. 3.4.8 From Bessel’s inequality, we have ∞ X |hx, ei i|2 j=1 ||x||2 ≤ 1. Assume we have nm inner products hx, ei i satisfying |hx, ei i| > nm X |hx, ek i|2 k=1 ||x||2 > nm X k=1 (3.0.7) 1 . m We have 1 nm = . m2 ||x||2 m2 ||x||2 We must then (at least) have nm < m2 ||x||2 , since (3.0.7) cannot hold otherwise. This proves the statement. Ex. 3.5.4 If (xj ) is a sequence in an inner product space X such that the series kx1 k + kx2 k + kx3 k + . . . converges, show that Sn is a Cauchy sequence, where Sn = x1 + . . . + xn 118 (3.0.8) Solution: We let S n = x1 + x2 + x3 + . . . + xn (3.0.9) Then kSn − Sm k = kxm + xm+1 + · · · + xn k ≤ kxm k+kxm+1 k+. . .+kxn k → 0 as n, m → ∞ (3.0.10) Since the series (3.0.8) converges. Ex. 3.5.6 Given: x = P∞ i=1 αi ei and y = hx, yi = h = ∞ X αi ei , i=1 ∞ ∞ XX P∞ j=1 ∞ X j=1 βj ej where, (ei ) is an orthonormal sequence. Then, ∞ X ∞ ∞ X ∞ X X βj ej i = hαi ei , βj ej i = αi βj hei , ej i αi βj δij = i=1 j=1 i=1 j=1 ∞ X i=1 j=1 αi βi i=1 It must be added that the series of x and y are absolutely convergent. For example, for x: ∞ ∞ ∞ ∞ X X X X kxk = hx, xi = h αi e i , αj e j i = αi αi = |αi |2 2 i=1 i=1 i=1 i=1 Ex. 3.6.4 We study hx + y, x + yi = hx, xi + hx, yi + hy, xi + hy, yi which can be re-written by using Parsevals identity X X X hx + y, ek ihx + y, ek i = hx, ek ihx, ek i + hy, ek ihy, ek i + hx, yi + hy, xi. (3.0.11) By using the linear properties of the inner product and rearranging, (3.0.11) can be written X X hx, yi + hy, xi = hy, ek ihx, ek i + hx, ek ihy, ek i. k k P P P Since hx, yi+hy, xi = 2Re(hx, yi) and k hy, ek ihx, ek i+ k hx, ek ihy, ek i = 2Re( k hy, ek ihx, ek i) we can conclude that X Re(hx, yi) = Re( hx, ek ihy, ek i). k 119 As for the imaginary part, we have hx + iy, x + iyi = hx, xi − ihx, yi + ihy, xi + hy, yi, where i is now the usual imaginary number. Repeating the same procedure on this, one arrives at X X i(−hx, yi + hy, xi) = i( hy, ek ihx, ek i − hx, ek ihy, ek i). k k P P By using i(−hx, yi + hy, xi) = 2Im(hx, yi) and i( k hy, ek ihx, ek i − k hx, ek ihy, ek i) = P 2Im( k hy, ek ihx, ek i), one can conclude X Im(hx, yi) = Im( hx, ek ihy, ek i). k Since both the real and imaginary parts are equal, we must have X hx, ek ihy, ek i, hx, yi = k which completes the proof. Ex. 3.6.4 Let M be a total orthonormal set in a Hilbert space H. Then ∀ x, y ∈ H we know that X X x= hx, em i em and y= hy, en i en (3.0.12) m n where en and em are the elements of M which generate a countable set of Fourier coeffincients (Lemma 3.5-3) for x and y respectively. In the next step we consider (ek ) = (el ) = (em ) ∪ (en ) which is a countable set since it is the countable union of countable sets. P P hx, yi = h k hx, ek i ek , l hy, el i el i = = = P k hx, ek i hek , P P k P k P l hy, el i el i l hx, ek i hy, el i hek , el i hx, ek i hy, ek i which is another version of the Parseval relation. 120 (3.0.13) Ex. 3.6.4 P P We have Parseval’s relation k |hx, ek i|2 = kxk2 and wish to show that hx, yi = k hx, ek ihy, ek i. Consider kx − yk. We have kx − yk2 = hx − y, x − yi = kxk2 + kyk2 − 2<[hx, yi] but also from Parseval’s relation that X X kx − yk2 = |hx − y, ek i|2 = |hx, ek i − hy, ek i|2 X X X = |hx, ek i|2 + |hy, ek i|2 − 2<[ hx, ek ihy, ek i] X = kxk2 + kyk2 − 2<[ hx, ek ihy, ek i], where we again have used Parseval’s relation in the final equality. Comparing the two expressions tells us that X <[hx, yi] = <[ hx, ek ihy, ek i]. k Repeating the exercise using kx + yk yields the analogous result for the imaginary parts, X =[hx, yi] = =[ hx, ek ihy, ek i]. k Consequently the desired result follows. Ex. 3.6.4 Let x ∈ X, M = {ek } ⊂ X be a total orthonormal set and X x˜ = hx, ej i ej . j Since hx − x˜, ej i = 0 ∀i we have x − x˜ ⊥ M. This implies x = x˜ by Theorem 3.6.2 a) since M is total in X. Thus we can write any element y ∈ X as X y= hy, ek i ek . k This now gives hx, yi = = * X j X k hx, ej i ej , X k hy, ek i ek + hx, ek i hy, ek i . 121 = XX j k hx, ej i hy, ek i hek , ej i = Ex. 3.6.10 Let M be a subset of a Hilbert space H, and let u, v ∈ H. Suppose that hv, xi = hw, xi for all x ∈ M implies v = w. If this holds for all v, w ∈ H, show that M is total in H. Solution: By the linearity of the inner product hv, xi = hw, xi ⇐⇒ hv − w, xi = 0. Thus, hv, xi = hw, xi for all x ∈ M implies v = w, for all v, w ∈ H ⇐⇒ hv − w, xi = 0 for all x ∈ M implies v = w, for all v, w ∈ H ⇐⇒ hu, xi = 0 for all x ∈ M implies u = 0 ⇐⇒ u⊥M implies u = 0 ⇐⇒ M is total in H. In the last step we used Theorem 3.6-2. Here it is important that H is complete. Ex. 3.6.10 Let us consider M subset of H Hilbert space and v, w ∈ H. Moreover we admit that hv, xi = hw, xi , ∀x ∈ M ⇒ v = w holds ∀v, w ∈ H. We want to prove that M is total: in order to do this we consider z ⊥ M , that satisfies for construction hz, xi = 0, ∀x ∈ M . But since also h0, xi = 0, ∀x ∈ M , then we have hz, xi = h0, xi and this implies that z = 0. Thus, by Theorem 3.6.2 the claim follows. Ex. 3.8.10 Sesquilinearity follows by h(x1 + x2 , y) = hx1 + x2 , yi = hx1 , yi + hx2 , yi = h(x1 , y) + h(x2 , y) h(x, y1 + y2 ) = hx, y1 + y2 i = hx, y1 i + hx, y2 i = h(x, y1 ) + h(x, y2 ) h(αx, y) = hαx, yi = α hx, yi = αh(x, y) ¯ h(x, βy) = hx, βyi = β¯ hx, yi = βh(x, y). 122 From the Cauchy-Schwartz inequality we get: khk = sup sup x6=0 y6=0 | hx, yi | kxk kyk ≤ sup sup = 1. kxk kyk x6=0 y6=0 kxk kyk But note that for x = y we have h(x, x) kxk2 = =1 kxk2 kxk2 so we get khk = 1. Ex. 3.8.14 Schwarz inequality. Let X be a vector space and h a Hermitian form on X × X. This form is said to be positive semidefinite if h(x, x) ≥ 0 for all x ∈ X. Show that then h satisfies the Schwarz inequality |h(x, y)|2 ≤ h(x, x)h(y, y). Since h is positive semidefinite, we have h(x − λy, x − λy) ≥ 0. By using the fact that h is a Hermitian sesquilinear form, we obtain ¯ ¯ h(x, x) − λh(x, y) − λh(y, x) + λλh(y, y) ≥ 0. Choose λ= (3.0.14) h(x, y) . h(y, y) Thus, ¯ = h(x, y) = h(y, x) . λ h(y, y) h(y, y) Substituting this into inequality (3.0.14), we obtain h(x, y) h(x, y)h(y, x) h(y, x) h(x, y) − h(y, x) + h(y, y) h(y, y) h(y, y) h(y, x) = h(x, x) − h(x, y) h(y, y) |h(x, y)|2 = h(x, x) − . h(y, y) 0 ≤ h(x, x) − Therefore, |h(x, y)|2 ≤ h(x, x)h(y, y). Remark: It may not be straightforward to get the correct λ value. The idea is that the equality in the Schwarz inequality is taken when x and y are linear dependent. Therefore, we use λ to project y so that λy and x are linear dependent. 123

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