Also check that p h i satisfies the conditions of a

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Also, check that p , ·i satisfies the conditions of a norm. Definition 20.3. A Hilbert space is a complete inner product space. Also c.f., a Banach space is a complete normed space. For example, L 2 , H 1 , H 1 0 are Hilbert spaces, where we defined h u, v i H 1 = ´ Ω ( uv + u · ∇ v ) d x . Also, L p is a Banach space. Recall: If Ω is bounded, then k u k L p C k u k L q for p q . This means that L q (Ω) L p (Ω), and they are not the same if p < q . 64
LECTURE 20. MAY 2 65 Problem: Find an example u L 1 [ - 1 , 1] such that u / L 2 [ - 1 , 1]. Solution: u ( x ) = 1 x L 1 , / L 2 . L 1 is not a Hilbert space, but it is still a Banach space with norm k u k L 1 = ´ Ω | u | d x . 20.1.1 Projection Given a vector x and a plane V , we can find a projection of x onto V . Theorem 20.4. Let V be a closed subspace of a Hilbert space H . Then, given x H , there exists a unique P V x V H such that k P V x - x k = inf v V k v - x k . Moreover, P V x = x if and only if x V , and Q V x = x - P V x satisfies h Q V x, P V x i = 0 , and k x k 2 = k Q V x k 2 + k P V x k 2 (Pythagorean identity). Definition 20.5. A separable space is a space containing a countable dense subset, namely, there exists { x m } m =1 such that any open subset U 6 = contains some x i . The spaces C ( Ω ) and C ( Ω ) are separable (e.g., by Stone-Weierstrass Theorem). Also, C ( Ω ) is dense in L 2 (Ω) so L 2 (Ω) is separable, and H 1 0 (Ω) is the closure of C c (Ω) in H 1 so this is separable too. Definition 20.6 (Basis) . An orthogonal basis of a separable Hilbert space H is a sequence of elements { w k } k =1 H such that h w i , w j i = 0 for i 6 = j , k w j k = 1, and x = i =1 h x, w i i w i for all x H (in the k·k H sense). Proposition 20.7. A separable Hilbert space H admits an orthogonal basis. Idea of Proof. Choose { z k } k =1 H so that { z k } k =1 is dense. Set w 1 = z 1 k z 1 k , ˜ w 2 = z 2 - h z 2 , w 1 i w 1 , w 2 = ˜ w 2 k ˜ w 2 k , ˜ w 3 = z 3 - h z 3 , w 1 i w 1 - h z 3 , w 2 i w 2 , . . . Then, { w k } k =1 is a basis. Example 20.8. L 2 [0 , 2 π ] has the basis n 1 2 π , sin θ π , cos θ π , sin(2 θ ) π , . . . o . Check!! ˆ 2 π 0 1 2 π 2 d θ = ˆ 2 π 0 sin( ) π 2 d x = 1 ,
LECTURE 20. MAY 2 66 ˆ 2 π 0 sin( ) π cos( ) π d θ = 0 , if m 6 = n . If f L 2 , then D 1 2 π , f E 1 2 π + k X m =1 D sin( ) π , f E sin( ) π + k X m =1 D cos( ) π , f E cos( ) π f in L 2 as k → ∞ . A linear operator is a linear map L : H 1 H 2 , where H 1 , H 2 are Hilbert spaces (c.f. A : R m R n ). Example 20.9. For Δ = L , H 1 = W 2 , 2 , H 2 = L 2 . Then, u W 2 Δ -→ Δ u L 2 . We could also consider Banach spaces W 1 ,p L p or C 2 → C α . Definition 20.10. The kernel is N ( L ) = { x H 1 : Lx = 0 } and the range is R ( L ) = { y H 2 : Lx = y for some x H 1 } . Definition 20.11. We say L is bounded if k Lx k H 2 C k x k H 1 . Proposition 20.12. A linear operator is bounded if and only if it is continuous. Proof. Suppose L is bounded. Then, k L ( x - x 0 ) k H 2 C k x - x 0 k H 1 , so if k x - x 0 k → 0, then k L ( x - x 0 ) k H 2 0. (C.f., if | f ( x ) - f ( x 0 ) | ≤ c | x - x 0 | , then f is continuous; in fact, it is Lipschitz.) Suppose L is continuous. In particular, L is continuous at x = 0, namely, there exists some δ > 0 such that k Lx k H 2 1 if k x k H 1 δ. (20.1) So, for any x 6 = 0, x H 1 , Lx = k x k δ x k x k , k Lx k H 2 = k x k δ x k x k k x k δ · 1 by ( 20.1 ).
Lecture 21 May 7 21.1 Weak Convergence Let H be a Hilbert space. Suppose that a sequence { x i } i =1 H satisfies h

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