GT MATH 6338 - FUNCTIONAL ANALYSIS LECTURE NOTES

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FUNCTIONAL ANALYSIS LECTURE NOTES:COMPACT SETS AND FINITE-DIMENSIONAL SPACESCHRISTOPHER HEIL1. Compact SetsDefinition 1.1 (Compact and Totally Bounded Sets). Let X be a metric space, and letE ⊆ X be given.(a) We say that E is compact if every open cover of E contains a finite subcover. That is,E is compact if whenever {Uα}α∈Iis a collection of open sets whose union containsE, then there exist finitely many α1, . . . , αNsuch that E ⊆ Uα1∪ · · · ∪ UαN.(b) We say that E is sequentially compact if every sequence {fn}n∈Nof points of Econtains a convergent subsequence {fnk}k∈Nwhose limit belongs to E.(c) We say that E is totally bounded if for every ε > 0 there exist finitely many pointsf1, . . . , fN∈ E such thatE ⊆NSk=1Bε(fk),where Bε(fk) is the open ball of radius ε centered at fk. That is, E is totally boundedif and only there exist finitely many points f1, . . . , fN∈ E such that every elementof E is within ε of some fk.(d) We say that E is complete if every Cauchy sequence of points of E converges to apoint in E.Exercise 1.2. Suppose that X is a complete metric space. Show that, in this case,E is complete ⇐⇒ E is closed.In a finite dimensionsional normed space, a set is compact if and only if it is closed andbounded. In infinite dimensional normed spaces, it is true all compact sets are closed andbounded, but the converse fails in general.We have the following equivalent formulations of compactness for sets in metric spaces.Theorem 1.3. Let E be a subset of a Banach space X. Then the following statements areequivalent.(a) E is compact.(b) E is sequentially compact.12 COMPACT SETS AND FINITE-DIMENSIONAL SPACES(c) E is complete and totally bounded.Proof. (a) ⇒ (c). Exercise.(c) ⇒ (b). Assume that E is complete and totally bounded, and let {fn}n∈Nbe anysequence of points in E. Since E is covered by finitely many balls of radius12, one of thoseballs must contain infinitely many fn, say {f(1)n}n∈N. Then we have∀ m, n ∈ N, d(f(1)m, f(1)n) < 1.Since E is covered by finitely many balls of radius14, we can find a subsequence {f(2)n}n∈Nof{f(1)n}n∈Nsuch that∀ m, n ∈ N, d(f(1)m, f(1)n) <12.By induction we keep constructing subsequences {f(k)n}n∈Nsuch that d(f(k)m, f(k)n) <1kforall m, n ∈ N.Now consider the “diagonal subsequence” {f(n)n}n∈N. Given ε > 0, let N be large enoughthat1N< ε. If m ≥ n > N, then f(m)mis one element of the sequence {f(n)k}k∈N, sayf(m)m= f(n)k. Thend(f(m)m, f(n)n) = d(f(n)k, f(n)n) <1n< ε.Thus {f(n)n}n∈Nis Cauchy, and therefore, since E is complete, it converges to some elementof E.(b) ⇒ (c). Suppose that E is sequentially compact.Exercise: Show that E is complete.Suppose that E was not totally bounded. Then there is an ε > 0 such that E cannotbe covered by finitely many ε-balls centered at points of E. Choose any f1∈ E. Since Ecannot be covered by a single ε-ball, E cannot be a subset of Bε(f1). Hence there existsf2∈ E \ Bε(f1), i.e., f2∈ E and d(f2, f1) ≥ ε. But E cannot be covered by two ε-balls, sothere must exist an f3∈ E \Bε(f1)∪Bε(f2). In particular, we have d(f3, f1), d(f3, f2) ≥ ε.Continuing in this way we obtain a sequence of points {fn}n∈Nin E which has no convergentsubsequence, which is a contradiction.(b) ⇒ (a). Assume that E is sequentially compact. Then by the implication (b) ⇒ (c)proved above, we know that E is also complete and totally bounded.Choose any open cover {Uα}α∈Iof E. We must show that it contains a finite subcover.The main point towards proving this is the following claim.Claim. There exists a number δ > 0 such that if B is any ball of radius δ that intersects E,then there is an α ∈ I such that B ⊆ Uα.To prove the claim, suppose that {Uα}α∈Iwas an open cover of E such that no δ withthe required property existed. Then for each n ∈ N, we could find a ball Bnwith radius1nthat intersects E but is not contained in any Uα. Choose any fn∈ Bn∩ E. Since E issequentially compact, there must be a subsequence {fnk}k∈Nthat converges to an elementCOMPACT SETS AND FINITE-DIMENSIONAL SPACES 3of E, say fnk→ f ∈ E. But we must have f ∈ Uαfor some α, and since Uαis open theremust exist some r > 0 such that Br(f) ⊆ Uα. Now choose k large enough that we have both1nk<r3and d(f, fnk) <r3.Then it follows that Bnk⊆ Br(f) ⊆ Uα, which is a contradiction. Hence the claim holds.To finish the proof, we use the fact that E is totally bounded, and therefore can be coveredby finitely many balls of radius δ. But by the claim, each of these balls is contained in someUα, so E is covered by finitely many of the Uα. Exercise 1.4. Show that if E is a totally bounded subset of a Banach space X, then itsclosureE is compact. A set whose closure is compact is said to be precompact.Exercise 1.5. Prove that if H is an infinite-dimensional Hilbert space, then the closed unitsphere {f ∈ H : kfk ≤ 1} is not compact.Exercise 1.6. Suppose that E is a compact subset of a Banach space X. Show that anycontinuous function f : E → C must be bounded, i.e., supx∈E|f(x)| < ∞.2. Finite-Dimensional Normed SpacesIn this section we will prove some basic facts about finite-dimensional spaces.First, recall that a finite-dimensional vector space has a finite basis, which gives us anatural notion of coordinates of a vector, which in turn yields a linear bijection of X ontoCdfor some d.Example 2.1 (Coordinates). Let X be a finite-dimensional vector space over C. Then Xhas a finite basis, say B = {e1, . . . , ed}. Every element of X can be written uniquely in thisbasis, say,x = c1(x) e1+ · · · + cd(x) ed, x ∈ X.Define the coordinates of x with respect to the basis B to be[x]B=c1(x)...cd(x).Then the mapping T : X → Cdgiven by x 7→ [x]Bis, by definition of basis, a linear bijectionof X onto Cd.4 COMPACT SETS AND FINITE-DIMENSIONAL SPACESWe already know how to construct many norms on Cd. In particular, if w1, . . . , wd> 0are fixed weights and 1 ≤ p ≤ ∞, thenkxkp,w=dXk=1|xk|pw(k)p1/p, 0 < p < ∞,supk|xk| w(k), p = ∞,defines a norm on Cd(and Cdis complete in this norm). If X is an n-dimensional vectorspace, then by transferring these norms on Cdto X we obtain a multitude of norms for X.Exercise 2.2 (`pwNorms on X). Let X be a finite-dimensional vector space over C and letB = {e1, . . . , ed} be any basis. Fix any 1 ≤ p ≤ ∞ and any weight w : {1, . . . , d} → (0, ∞).Using the


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