Functional Analysis Irena Swanson, Reed College, Spring 2011 This is work in progress. I am still adding, subtracting, modifying. Any comments, solutions, corrections are welcome. Table of contents: Section 1: Overview, background 1 Section 2: Definition of Banach spaces 2 Section 3: Examples of Banach spaces 2 Section 4: Lp spaces 6 Section 5: L∞ spaces 9 Section 6: Some spaces that are almost Banach, but aren’t 10 Section 7: Normed vector spaces are metric spaces 11 Section 8: How to make new spaces out of existent ones (or not) 13 Section 9: Finite-dimensional vector spaces are special 18 Section 10: Examples of (infinite dimensional) subspaces 20 Section 11: Sublinear functional and the Hahn–Banach Theorem 22 Section 12: A section of big theorems 27 Section 13: Hilbert spaces 35 Section 14: Perpendicularity 37 Section 15: The Riesz Representation Theorem 41 Section 16: Orthonormal bases 41 Section 17: Section I.5 in the book 46 Section 18: Adjoints of linear transformations 47 Section 19: Self-adjoint operators 49 Section 20: Spectrum 51 Section 21: Holomorphic Banach space-valued functions 55 Appendix A: Some topology facts 59 Appendix B: Homework and exam sets 60 1 1 Overview, background Throughout, F will stand for either R or C. Thus F is a complete (normed) field. [Conway’s book assumes that all topological spaces are Hausdorff. Let’s try to not impose that unless needed.] Much of the material and inspiration came from Larry Brown’s lectures on functional analysis at Purdue University in the 1990s, and some came from my Reed thesis 1987. Functional analysis is a wonderful blend of analysis and algebra, of finite-dimensional and infinite-dimensional, so it is interesting, versatile, useful. I will cover Banach spaces first, Hilbert spaces second, as Banach spaces are more general. 2 Definition of Banach spaces Definition 2.1 Let X be a vector space over F. A norm on X is a function : X || || → [0, ) such that ∞ (1) (Positive semidefiniteness) For all x X, x =0 if and only if x =0. ∈ || || (2) rx = r x for all x X and all r F. || || | | || || ∈ ∈ (3) (Triangle inequality) For all x,y X, x + y x + y . ∈ || || ≤ || || || || Is is easy to verify that for all x,y X, x y x y , and that x y ∈ || − || ≤ || || − || || | || || − || ||| ≤ x y . || ± || Note that if X is a normed vector space, then X has a metric: d(x,y) = x y . || − || Definition 2.2 A vector space X with a norm is called a normed vector space or || || a normed linear space. In a normed linear space, a sequence f is Cauchy (in the { n} norm) if for all ǫ > 0 there exists N such that for all m, n N, f f < ǫ. A sequence ≥ || n − m|| f converges (in the norm) to f if for all ǫ > 0 there exists N such that for all n N, { n} ≥ f f < ǫ. A sequence f in X is convergent (in the norm) if there exists f in X || n − || { n} such that f converges in the norm to f. A Banach space is a normed linear space in { n} which every Cauchy sequence is convergent. Silly/important fact of the day: On February 2, 2011, MathSciNet lists 20331 publications with “Banach” in the title (and only 13539 publications with “Hilbert” in the title). 2 3 Examples of Banach spaces Examples of Banach spaces are given in propositions in this section. Proposition 3.1 For any p [1, ), Fn is a Banach space under the Lp-norm x = ∈ ∞ || ||p ( x p + + x p)1/p. Why is this a norm? H¨older’s inequality helps. For now I am | 1| ··· | n| assuming this, and I am not sure that I will get back to this. Presumably you have seen H¨older’s inequality. And you probably have seen that the open ball topology determined by this norm is equivalent to the standard Euclidean topology (with p =2), in which case, you should be able to prove easily that this is a Banach space. We will actually prove more general facts later. Definition 3.2 For any topological space X, let C(X) = f : X F,f continuous . { → } Note that if X is compact, then the uniform norm, or the sup norm, f = sup f(x) : x X || || {| | ∈ } is indeed a norm on C(X). If X is not compact, sup f(x) : x X may take on the value {| | ∈ } , so we cannot have a norm. ∞ Proposition 3.3 If X is a compact topological space, then C(X) is a Banach space. Proof. Let f be a Cauchy sequence in C(X). Then for all ǫ > 0, there exists N such { n} that for all m, n N, f f < ǫ. In particular, for all x X, f (x) f (x) ≥ || m − n|| ∈ | m − n | ≤ f f < ǫ. It follows that f (x) is a Cauchy sequence in F. Since F is complete, || m − n|| { n } we thus have that for all x X, there exists f(x) F such that f (x) f(x). So f ∈ ∈ n → { n} converges pointwise to a function f. We need to verify that f C(X). It suffices to prove that for every open set W ∈ in F, its f-preimage V in X is an open subset of X. For that it suffices to prove that for every x V there exists an open neighborhood U of x such that U V . So let x V . ∈ ⊆ ∈ Since W is open in a metric space, there exists ǫ > 0 such that B(f(x), ǫ) W . By the ⊆ Cauchy sequence assumption there exists N such that for all m, n N, f f < ǫ/3. ≥ || m − n|| In particular, for all y X, f (y) f (y) f f < ǫ/3, and for any m N, ∈ | m − n | ≤ || m − n|| ≥ fm(y) f(y) = fm(y) lim fn(y) = lim fm(y) fn(y) ǫ/3. | − | | − n→∞ | n→∞ | − | ≤ −1 SincefN is continuous, U = fN (B(fN (x), ǫ/3)) is an open neighborhood of x in X. For all y U, ∈ f(x) f(y) f(x) f (x) + f (x) f (y) + f (y) f(y) < ǫ/3 + ǫ/3 + ǫ/3 = ǫ, | − |≤| − N | | N − N | | N − | so that f(y) B(f(x), ǫ) W . Thus x U V , and U is open, as desired. ∈ ⊆ ∈ ⊆ 3 Finally, we need to prove convergence in the norm. For all m N (with N as above), ≥ f f = sup f (x) f(x) : x X || m − || {| m − | ∈ } = sup f (x) lim f (x) : x X {| m − n | ∈ } ǫ. ≤ And we’re done, without needing to use the Hausdorff condition. Definition 3.4 For any topological space X, define C (X) = f : X F continuous and bounded b { → } with the metric f = sup f(x) : x X . || || {| | ∈ } Proposition 3.5 For any topological space X, Cb(X) is a Banach space with the uniform norm. [If X is compact, then Cb(X) is a subset of the already established Banach space C(X).] Proof. It is straightforward to see that C (X) is a normed linear space. If f is a Cauchy b { n} sequence in C (X), then there exists B such that for all n, f B. Thus any pointwise b || n|| ≤ limit of f is also bounded. The rest of this proof is just like for C(X). { n} Definition 3.6 Let X be a topological space. A function f : X F is said to have → compact support if for all ǫ > 0, the set x : f(x) ǫ is compact. Define { | | ≥ } C (X) = f : X F continuous with compact support . 0 { → } Proposition 3.7 For any topological space X, C0(X) is a closed linear subspace of Cb(X), and hence a Banach space (under the uniform norm). Proof. We first show that C (X) C (X). Let f C (x). For all n, define S x X : 0 ⊆ b ∈ 0 n{ ∈ f(x) n . By assumption, S is compact, and since f is continuous, S is closed. Since | | ≥ } n n f is F-valued, S = . Thus by HW 1.3, there exists n such that S = , which means ∩ n ∅ n ∅ that f is bounded. Clearly C (X) is closed under scalar multiplication. Now let f, g C (X). Let ǫ > 0. 0 ∈ 0 Then x : f(x) + g(x) ǫ x : f(x) ǫ/2 x : g(x) ǫ/2 , { | | ≥ }⊆{ | | ≥ }∪{ | | ≥ } and since f + g is continuous, this says that x : f(x) + g(x) ǫ is a closed subset of { | | ≥ } a compact set, whence compact itself by Proposition 1, which proves that C0(X) is closed under addition. Since 0 C (X), it follows that C (X) is a vector subspace of C (X). ∈ 0 0 b We next prove that C (X) is a closed subset of C (X). If f is a Cauchy sequence 0 b { n} in C (X) in the norm, then f is a Cauchy sequence in C (X) in the norm. Since C (X) 0 { n} b b is Banach, there exists f C (X) such that f f in the norm. Let ǫ > 0. Then there ∈ b n → 4 exists N such that for all n N, f f < ǫ/2. If f(x) ǫ, then by the triangle ≥ || n − || | | ≥ inequality, ǫ f(x) f(x) f (x) + f (x) f f + f (x) < ǫ/2 + f (x) . ≤| |≤| − N | | N | ≤ || − N || | N | | N | Thus f (x) ǫ/2. Thus x : f(x) ǫ x : f (x) ǫ/2 .