Hilbert spaces, Basic Projections Riesz Representation Theorem Application of Riesz representation theorem Chapter 4: Hilbert Spaces I-Liang Chern Department of Applied Mathematics National Chiao Tung University and Department of Mathematics National Taiwan University Fall, 2013 1 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Outline 1 Hilbert spaces, Basic Inner product structure Sobolev spaces 2 Projections Projections in Banach spaces Orthogonal projections 3 Riesz Representation Theorem 4 Application of Riesz representation theorem Solving Poisson equations Error estimates for finite element method 2 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Inner product structure Definition Let X be a complex linear space. An inner product (·; ·) is a bilinear form: X × X ! C which satisfies (a) (x; x) ≥ 0 and (x; x) = 0 if and only if x = 0, (b) (x; y) = (y; x), (c) (x; αy + βz) = α(x; y) + β(x; z). The linear space X equipped with the inner product (·; ·) is called an inner product space. 3 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Examples. The space Cn with X (x; y) := xiyi i is an inner product space. Let A be a symmetric positive definite matrix in Rn. Define hx; yiA := (x; Ay) n Then h·; ·iA is an inner product in R . The space C[0; 1] with the inner product Z 1 (f; g) := f(t)g(t) dt 0 is an inner product space. 4 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem The space L2(0; 1) is the completion of C[0; 1] with the above inner product. In fact, it is the space of all functions whose squares are Lebesgue integrable. Let T be the unit circle and Z 2 2 L (T) := ff : T ! C j jf(t)j dt < 1g T It is the space of all square summable and periodic functions. The space `2(N) is defined to be 1 2 X 2 ` (N) := fx j x = (x1; x2; ··· ); jxij < 1g i=1 equipped with the inner product 1 X (x; y) := xiyi: i=1 5 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Let wn > 0 be a positive sequence. Define 1 2 X 2 `w := fxj x : N ! C; wijxij < 1g i=1 with the inner product: 1 X (x; y) := wixiyi i=1 Let w :(a; b) ! R+ be a positive continuous function. Consider the space Z b 2 2 Lw(a; b) := ff :(a; b) ! C j jf(x)j w(x) dx < 1g a equipped with the inner product Z b (f; g) := f(x)g(x) w(x) dx: a 6 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Cauchy-Schwarz inequality We can define kxk = p(x; x). Theorem Let X be an inner product space. For any x; y 2 X, we have j(x; y)j ≤ kxkkyk: Proof: 1 From non-negativity of (·; ·), we get 2 2 2 0 ≤ (x+ty; x+ty) = kxk +2Re(x; y)t+kyk t for all t 2 R: From this, we obtain one form of Cauchy-Schwarz: jRe(x; y)j2 ≤ kxk2kyk2: 7 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem 2 We claim that jRe(x; y)j ≤ kxk kyk for any x; y 2 X if and only if j(x; y)j ≤ kxk kyk for any x; y 2 X: 3 Suppose (x; y) is not real, we choose a phase φ such that eiφ(x; y) is real. Now we replace x by eiφx. Then jRe(eiφx; y)j ≤ kxk kyk But the left-hand side is j(x; y)j. This proves one direction. The other direction is trivial. 8 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Cauchy-Schwarz and Triangle inequality From kx + yk2 = kxk2 + 2Re(x; y) + kyk2 and (kxk + kyk)2 = kxk2 + 2kxk kyk + kyk2; by comparing the two equations, we get that the triangle inequality is equivalent to the Cauchy-Schwarz inequality. In fact, the following statements are equivalent: (a) For any x; y 2 H, Re(x; y) ≤ kxk kyk; (b) For any x; y 2 H, j(x; y)j ≤ kxk kyk; (c) For any x; y 2 H, kx + yk ≤ kxk + kyk. 9 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Remark If we are care about the cosine law, that is kx − yk2 = kxk2 + kyk2 − 2kxk kyk cos θ; then we should define the angle between x and y by Re(x; y) cos θ := : kxk kyk However, this creates a problem, the orthogonality in this sense may not have (x; y) = 0. This is not what we want. So, we define the acuate angle between two vectors x and y by j(x; y)j cos θ := ; kxk kyk and we give up the traditional cosine law. 10 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Proposition (Parallelogram law) A normed linear space is an inner product space if and only if kx + yk2 + kx − yk2 = 2kxk2 + 2kyk2; for all x; y 2 X: Proof. Suppose a norm satisfies the parallelogram law, we define 1 (x; y) := kx + yk2 − kx − yk2 − ikx + iyk2 + ikx − iyk2 : 4 I leave you to check the parallelogram law implies the bilinearity of the inner product. 11 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Sobolev spaces 1 1 The H space 2 1 The H0 space 3 Poincar´einequality 4 Optimal constant in Poincar´einequality 12 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem The H1 space. Similarly, we define Z b 1 2 0 2 H (a; b) = fu :(a; b) ! Cj ju(x)j + ju (x)j dx < 1g a with the inner product Z b (u; v) = uv¯ + u0v0 dx: a Why is H1(a; b) complete?1 Indeed, it is the completion of C1[a; b], or C1[a; b] under the above inner product. 1 2 0 2 0 If un ! u in L (a; b) and un ! v in L (a; b), then prove u = v. 13 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem 1 The H0 spaces 1 1 Define H0 (a; b) = fu 2 H (a; b)ju(a) = u(b) = 0g: u(a) and u(b) are well-defined for u 2 H1(a; b). In fact, for any two points x1 and x2 near a, we can express Z x2 0 ju(x2) − u(x1)j = j u (x) dxj x1 Z x2 1=2 Z x2 1=2 ≤ 12dx ju0(x)j2dx x1 x1 1=2 0 ≤ (x2 − x1) ku k ! 0 as x1; x2 ! a: 1 1 Alternatively, H0 (a; b) is the completion of C0 [a; b] 1 under the above inner product. Here, C0 [a; b] are those C1 function on [a; b] satisfying zero boundary condition. 14 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Poincar´einequality 1 In H0 (a; b), we can define another inner product Z b hu; vi := u0(x)v0(x) dx: a We check that hu; ui = 0 implies u ≡ 0. R b 0 2 0 From a ju (x)j dx = 0, we get that u (x) ≡ 0 on (a; b). This together with u(a) = u(b) = 0 lead to u ≡ 0. 1 Now, in H0 , we have two norms: 2 2 0 2 2 0 2 kuk1 ≡ kuk + ku k ; kuk2 ≡ ku k : 1 We claim that they are equivalent in H0 . 15 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem We recall that two norms k · k1, k · k2 are equivalent in a normed space X if there exist two positive constants C1;C2 such that for any u 2 X, we have C1kuk2 ≤ kuk1 ≤ C2kuk2: Clearly we have ku0k2 ≤ kuk2 + ku0k2: Theorem (Poincar´einequality) 1 There exists a constant C > 0 such that for any u 2 H0 (a; b), we have kuk2 ≤ Cku0k2: (1.1) 16 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Proof of Poincar´einequality x x 1 R 0 R 0 u(x) = u(a) + a u (y) dy = a u (y) dy: Z x 2 2 0 ju(x)j = u (y) dy a Z x Z x ≤ 12 dy ju0(y)j2 dy a a Z b ≤ (x − a) ju0(y)j2 dy a 2 We integrate x over (a; b) to get Z b (b − a)2 Z b ju(x)j2 dx ≤ ju0(y)j2 dy: a 2 a 17 / 60 Hilbert spaces, Basic Projections Riesz Representation TheoremInner Application product of structure Riesz representation Sobolev spaces theorem Remarks 1 The Poincar´einequality is valid by just assuming u(a) = 0. b 2 R It is also valid by assuming a u(x) dx = 0: 3 Dimension analysis for the Poincar´einequality: Denote the dimensions by [x] = L and [u] = U.