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Brief Summary of Functional Analysis APPM 5440 Fall 2014 Applied Analysis

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Brief Summary of Functional Analysis APPM 5440 Fall 2014 Applied Analysis Brief summary of functional analysis APPM 5440 Fall 2014 Applied Analysis Stephen Becker, [email protected] Standard theorems. When necessary, I used Royden’s book as a reference. Version 3, 10/20/17 Contents 1 Fundamental theorems 1 2 Hahn-Banach in more detail 2 3 Proof of the Baire category theorem 2 4 Proof of the Uniform Boundedness/Banach-Steinhaus theorem3 5 Open-mapping theorem in more detail4 5.1 Closed graph theorem.............................................6 5.1.1 Royden’s version............................................6 5.1.2 Other versions (with more details filled in).............................7 6 Axiom of choice and Zorn’s lemma 8 1 Fundamental theorems Theorem 1 (Hahn-Banach). See our text for details. Proof requires Zorn’s lemma/Axiom of choice A subset Y ⊂ X is nowhere dense if any open set U ⊂ X contains a ball B ⊂ U ⊂ X such that B ∩ Y = ∅. That is, Y is nowhere dense if the interior of the closure of Y is empty. For example, the Cantor set is a nowhere dense subset of the unit interval. Theorem 2 (Baire Category Theorem). A complete metric space cannot be written as the countable union of nowhere dense sets. Theorem 3 (Uniform Boundedness Theorem, aka Banach Steinhaus Theorem (Thm. 8.39 in our text)). Let Tn : X → Y be a sequence of bounded linear operators from a Banach space X into a normed linear space Y . Assume that for each x ∈ X, there is a real number cx such that kTnxk ≤ cx, ∀ n = 1, 2, ··· , Then there is a real number c such that kTnk ≤ c, ∀ n = 1, 2, ··· ., i.e., sup kTnk ≤ c. n∈N This follows from the Baire category theorem. An open mapping is one that maps open sets to open sets. Theorem 4 (Open Mapping Theorem). Let X and Y be Banach spaces. Then any bounded linear operator T from X onto Y (that is, T is surjective) is an open mapping. Consequently, if T is bijective, then T −1 is continuous and hence bounded (as well as linear). This follows from the uniform boundedness theorem. Theorem 5 (Closed Graph Theorem). Let X and Y be Banach spaces and let D ⊂ X be a subspace. Let T : D → Y be a closed linear operator. If D is closed in X, then T is bounded. This follows from the uniform boundedness theorem. We can also state it like this: 1 Theorem 6 (Closed Graph Theorem, variant). Let X and Y be Banach spaces and let T : X → Y be a linear operator. The graph of T is a subset of X ⊗ Y defined by graph(T ) = {(x, T (x)) | x ∈ X}. Then graph(T ) is a closed subspace iff T is bounded. Theorem 7 (See Thm. 6.29 in our text). Every Hilbert space has an orthonormal basis. Uses Zorn’s lemma in the proof. Theorem 8 (Banach-Alaoglu). See Thm. 5.61, 8.45 in our text. Let X be a normed linear space, then the closed unit ball B∗ of its dual space X∗ is compact with respect to the weak-∗ topology. 2 Hahn-Banach in more detail Our book doesn’t have the most general version, so here is a more general version, but we don’t prove it (proof relies on Zorn’s lemma). Royden and Reed/Simon have proofs, for example. Theorem 9. (Hahn-Banach Theorem) Let X be a linear space over a field F (= R or C). Let p : X → R be a real-valued functional on X satisfying p(x + y) ≤ p(x) + p(y), ∀ x, y ∈ X “sub-linear” p(αx) = |α| p(x), ∀ α ∈ F, x ∈ X “positive homogeneous”. Furthermore, let Z ⊂ X be a subspace of X and let f : Z → F be a linear functional on Z such that |f(x)| ≤ p(x), ∀ x ∈ Z. Then f has a linear extension f˜ : X → F with |f˜(x)| ≤ p(x), ∀ x ∈ X. Note that sub-linearity implies p(x) = 0, and using this with the positive homogeneous property implies p(x) ≥ 0 for all x ∈ X. 3 Proof of the Baire category theorem Let X be a metric space and M ⊂ X be a subspace. A point x ∈ M is called an interior point of M if there is > 0 such that B(x) ⊂ M. We will begin with the following definition regarding metric spaces. Definition 10. Let X be a metric space and M ⊂ X be a subspace. Then M is said to be 1. rare (or nowhere dense) in X if its closure M has no interior point. 2. meager (or of the first category) in X if M is the union of countably many sets each of which is rare in X. 3. nonmeager (or of the second category) in X if M is not meager in X. Royden uses the term hollow for a subset with empty interior. A set is hollow iff its complement is dense. A set is nowhere dense if its closure is hollow. Now we will state and proof the important Baire’s Category Theorem. The proof can be shortened if we used the Cantor Intersection Theorem Theorem 11. (Baire’s Category Theorem) If a metric space X 6= φ is complete, then it is nonmeager in itself. Consequently, if X 6= φ is complete and ∞ [ X = Ak, k=1 then at least one Ak contains a nonempty open subset of X. Royden’s version is that if X is complete, and {On}n∈N is a collection of open dense subsets of X, then the ∞ intersection ∩n=1On is also dense. Or most compactly, an open subset of a complete space is of the second category. 2 Proof. Assume, on the contrary, that X is meager in itself. By definition, ∞ [ X = Mk, k=1 where each Mk is rare in X. Since M1 is rare in X, we have M 1 6= X (otherwise, each point in M 1 is an interior c c 1 point.). So, the complement (M 1) is nonempty and open. Thus, there is a point p1 ∈ (M 1) and 0 < 1 < 2 such that def c B1 = B1 (p1) ⊂ (M 1) . −1 Note that B1 ∩ M1 = φ and 1 < 2 . Since M2 is rare, its closure M 2 does not contain a nonempty open set. In particular, M 2 does not contain c c B 1 (p1). So, the set (M 2) ∩ B 1 (p1) is nonempty and open. Thus, there is a point p2 ∈ (M 2) ∩ B 1 (p1) and 2 1 2 1 2 1 1 0 < 2 < 2 1 such that def c B2 = B2 (p2) ⊂ (M 2) ∩ B 1 (p1). 2 1 −2 Note that B2 ∩ M2 = φ, B2 ⊂ B 1 (p1) ⊂ B1 and 2 < 2 . 2 1 Continuing in this fashion, we obtain a sequence of balls Bk such that def −k Bk = Bk (pk),Bk ∩ Mk = φ, Bk+1 ⊂ B 1 (pk) ⊂ Bk, k < 2 . 2 k −N+1 Now we will show that the sequence {pk} is Cauchy. Given > 0, there is N such that 2 < . For any m, n > N, we have Bm ⊂ BN and Bn ⊂ BN , so −N −N −N+1 d(pm, pn) ≤ d(pm, pN ) + d(pN , pn) < N + N < 2 + 2 = 2 < . This proves {pk} is Cauchy. Since X is complete, there is p ∈ X such that pk → p. Fixed m. For any n > m, using Bn ⊂ B 1 (pm), we have 2 m 1 d(p , p) ≤ d(p , p ) + d(p , p) < + d(p , p). m m n n 2 m n Taking n → ∞, we have 1 d(p , p) ≤ < . m 2 m m ∞ Therefore, p ∈ Bm for all m. Since Bk ∩ Mk = φ, we have p∈ / Mk for all k. Hence p∈ / ∪k=1Mk = X. This is a contradiction. For reference, here is the Theorem 12 (Cantor intersection theorem). Let X be a metric space. Then X is complete if and only if whenever ∞ {Fn}n=1 is a contracting (nested) sequence of non-empty closed subsets of X, there is a point x ∈ X for which ∞ ∩n=1Fn = {x}. The sequence is nested in the sense that Fn+1 ⊂ Fn. We say a sequence is contracting if the diameter of the sets goes to zero. From Royden, “a very rough geometric interpretation of the Cantor Intersection Theorem is that a metric space fails to be complete because it has ’holes’. ” 4 Proof of the Uniform Boundedness/Banach-Steinhaus theorem Theorem 13 (Uniform Boundedness Theorem, aka Banach Steinhaus Theorem). Let Tn : X → Y be a sequence of bounded linear operators from a Banach space X into a normed linear space Y . Assume that for each x ∈ X, there is a real number cx such that kTnxk ≤ cx, ∀ n = 1, 2, ··· , Then there is a real number c such that kTnk ≤ c, ∀ n = 1, 2, ··· . 3 Proof. For each positive integer k, we define Ak as the set of all x such that kTnxk ≤ k, ∀ n = 1, 2, ··· . We will prove that Ak is closed. For each x ∈ Ak, there is a sequence {xj } ⊂ Ak such that xj → x. Since xj ∈ Ak, we have kTnxj k ≤ k, ∀ n = 1, 2, ··· . Letting j → ∞ and using the fact that Tn is continuous, we have kTnxk ≤ k for all n. So, x ∈ Ak. This shows that Ak is closed. By assumption of the theorem, we have ∞ [ X = Ak. k=1 Since X is complete, by the Baire’s Category Theorem (Theorem 11), there is k0 such that Ak0 contains an open ball, namely def B0 = Br(x0) ⊂ Ak0 . Let x ∈ X be arbitrary with x 6= 0.
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