MATH 205B NOTES 1. 1/10 the Topic of the Course Is Functional Analysis
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MATH 205B NOTES MOOR XU NOTES FROM A COURSE BY RAFE MAZZEO Abstract. These notes were taken during Math 205B (Real Analysis) taught by Rafe Mazzeo in Winter 2012 at Stanford University. They were live-TEXed during lectures in vim and compiled using latexmk. Each lecture gets its own section. The notes are not edited afterward, so there may be typos; please email corrections to [email protected]. 1. 1/10 The topic of the course is functional analysis and applications. These include harmonic analysis and partial differential equations. These applications were the reason much of the theory was created. Today we will review some facts about topological spaces and metric spaces, and state some fundamental theorems. Definition 1.1. We have metric spaces (M; d). Here M is some set, and d : M × M ! R+ satisfying (1) d(x; y) ≥ 0 (2) d(x; y) = d(y; x) (3) d(x; y) = 0 iff x = y (4) d(x; y) ≤ d(x; z) + d(z; y). The simplest examples are things we know: n pP 2 Example 1.2. Euclidean space R with d(x; y) = (xi − yi) Other examples include the following: Example 1.3. • Finite sets with metrics. Here M is a finite set. • On the interval a ≤ x ≤ b, the real-valued continuous functions are C([a; b]). This has a metric d(f; g) = supa≤x≤b jf(x) − g(x)j. We can define B(f; ") as the set of all g such that d(f; g) < ". These are all continuous functions that never stray too far from f, inside a tube around f. • C1([a; b]) are continuously differentiable functions, so f 2 C1 if f 2 C0 and f 0 2 C0. This means that there are no sharp corners, and the derivative doesn't stray too far 0 0 either. The metric is d(f; g) = supx2[a;b](jf(x) − g(x)j + jf (x) − g (x)j). • L2[a; b] are functions with metric d(f; g) = (R jf(x) − g(x)j2 dx)1=2. Here, d(f; g) = 0 iff f = g almost everywhere; this notion is one reason why the Lebesgue integral exists. In fact, L2 functions are actually equivalence classes of functions, where f ∼ g if they agree almost everywhere. This eliminates some interesting functions. 1 p • L [a; b] for 1 ≤ p ≤ 1. Here kfk1 = sup jfj. The triangle inequality holds by the Holder inequality. All of these are special cases of normed spaces (V; k·k). Here, M is the vector space V , and the norm is k·k : V ! R+. By definition, we can write d(v; w) = kv − wk. A consequence is that d(v + b; w + b) = d(v; w). All of linear functional analysis is the interaction between the vector space and the underlying topology. Slightly more generally, we can talk about topological vector spaces. We want to apply our finite dimensional intuition, and we can for Hilbert spaces, but not for more exotic cases. But we don't have compactness, even for closures of balls! We'll start with the simplest examples and build up to more generality. At the bottom, we have Hilbert spaces (i.e. having inner products, like L2). More generally, there are Banach spaces (having norms such as Lp or Ck), but they are very ubiquitous but harder to deal with. Then there are Frechet spaces (with a countable set of norms like C1), and then locally convex topological vector spaces (e.g. distributions). Let's go back to general metric spaces. The really important property is the notion of completeness. Definition 1.4. A sequence of elements fxng in M is a Cauchy sequence if given any " > 0 there exists N such that m; n > N then d(xm; xn) < ". Definition 1.5. A space (M; d) is called complete if every Cauchy sequence has a limit in M. Example 1.6. Here are examples of complete spaces. • R or Rn are complete. The rational numbers are not complete, but its completion is all of R. But this makes many elements of R somewhat intangible { all we know is that they are limits of Cauchy sequences. • C0(I) or any Ck(I). • L1(I) is complete. (Riesz-Fisher Theorem) Example 1.7. Here are some counterexamples. 1 1 Consider (C ; k·kC0 ). That's certainly a normed metric space because C is a subset of C0, but it is not complete; there are differentiable functions that approach a sharp corner. The sup norm doesn't force the derivative to converge. Theorem 1.8. C0([a; b]) is complete. Proof. Let ffng be Cauchy with this norm. We want to produce a function which is the limit, and we need to show that it is still in C0. If x 2 I then ffn(x)g is a sequence in R, and it is still Cauchy, because jfn(x) − fm(x)j ≤ kfn − fmkC0 ! 0. So define f(x) = limn!1 fn(x). So now we've shown that this sequence ffng converges pointwise. Now, we need to show that f is C0, i.e. if " > 0 is fixed, then there exists δ > 0 such that we want jf(x) − f(y)j < " if jx − yj < δ. (This is actually uniform continuity, but we're working on a compact interval, so that's ok.) To show this, we have jf(x) − f(y)j ≤ jf(x) − fN (x)j + jfN (x) − fN (y)j + jfN (y) − f(y)j. Fix " > 0 and choose N such that kfn − fN k < " if n ≥ N. Then supx jf(x) − fN (x)j = supx limn!1 jfn(x) − fN (x)j ≤ supx supn≥N jfn(x) − fN (x)j = supn≥N kfn − fN k < ". Now the result follows by choosing N appropriately. 2 Now, we will head toward the main theorem that we will do today: the Arzela-Ascoli theorem. The question is: What sets of continuous functions are compact? Definition 1.9. Suppose that F ⊂ C0(I) is some collection of continuous functions. This is called uniformly equicontinuous if given any " > 0 there exists δ such that if jx − yj < δ and any f 2 F we have jf(x) − f(y)j < ". There are a lot of natural examples of equicontinuous functions, and this is the main motivating example. 1 0 Example 1.10. Let F = ff 2 C (I): kfkC1 ≤ Ag. Then F ⊂ C is uniformly equicontin- uous. This is because Z x Z y 0 0 jf(x) − f(y)j = f (t) dt ≤ jf (t)j dt ≤ Ajy − xj: y x Here are a couple of preparatory theorems. 0 Theorem 1.11. Suppose ffn 2 C (X; Y )g which is equicontinuous. If ffng converges point- wise to f, then f is continuous. Proof. Left to your vivid imaginations. Theorem 1.12. Let Y be complete, and let D ⊂ X be a dense set in X. Suppose that fn is equicontinuous on X and fn converges pointwise on D. Then, in fact, fn ! f uniformly. (Uniform needs that X is compact.) Proof. First, if x 2 D, we need to define f(x). Take any sequence xk ! x with xk 2 D. Then the limit limn!1 fn(xk) is well-defined. Define f(x) = lim lim fn(xk): k!1 n!1 There's something to check: any other sequence yk ! x gives the same limit. This comes from equicontinuity. We have d(xk; yl) < δ, so therefore ρ(fn(xk) − fn(yl)) < ", and this is well-defined. We still need to show that f is actually continuous, and the convergence is uniform. Go through the details yourself. We illustrate this with an example: Theorem 1.13. Suppose that ffng is a uniformly equicontinuous family on [0; 1] and D ⊂ [0; 1] is dense with pointwise convergence of fn on D. Then fn ! f uniformly. Proof. Fix ". There exists δ such that jfn(x) − fn(y)j < " if jx − yj < δ. For that δ, choose y1; : : : ; ym 2 [0; 1] such that yj 2 D and B(yj; δ) cover I. Then choose N so large that jfn(yi) − f(yi)j < " if n ≥ N. Now, jf(x)−f(~x)j ≤ jf(x)−fN (x)j+jfN (x)−fN (yj)j+jfN (yj)−fN (~x)j+jfN (~x)−f(~x)j. Theorem 1.14 (Arzela-Ascoli Theorem). Suppose that ffng is a family with fn : [0; 1] ! R uniformly bounded (i.e. there exists A such that jfn(x)j ≤ A for all n, x. Suppose that fn is equicontinuous. Then there exists fnj which converges uniformly. This is extremely useful and is the template for all compactness theorems in infinite dimensional spaces. The proof uses the diagonalization argument. 3 Lemma 1.15. Suppose that ffn(m)g is bounded in m and n uniformly. Then there exists nj such that fnj (m) converges for each m. Proof. For m = 1, fn(1) is a bounded sequence. Choose fn(1)j (1) convergent. Take a further subsequence such that fn(2)j (2) converges. So we get numbers n(k)j. Then define nj = n(j)j. This is the diagonalization. Proof of Theorem 1.14. Let D = Q \ I be dense. Call D = fxkg. For every rational, we have fn(xk) is uniformly bounded in n; k, so we can apply the diagonalization lemma above to choose nj ! 1 so fnj (xk) ! f(xk) for all xk. Now, using Theorem 1.12 gives the desired result. Remark. This is in fact a very general theorem about maps between general metric spaces.