Problem Set 5: Solutions Math 201A: Fall 2016

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1 Problem Set 5: s Math 21A: Fall 216 Problem 1. Define f : [1, ) [1, ) by f(x) = x + 1/x. Show that f(x) f(y) < x y for all x, y [1, ) with x y, but f has no fixed point. Why doesn t this example contradict the contraction mapping theorem? For every x, y 1 with x y, we have f(x) f(y) = x y 1 1 xy < x y. If f(x) = x, then 1/x =, which is impossible. Since 1 1/xy 1 as xy, there is no constant θ < 1 such that f(x) f(y) θ x y for all x, y 1, so the contraction mapping theorem does not apply. 1

2 Problem 2. Suppose that X is a compact metric space and T : X X satisfies d (T (x), T (y)) < d(x, y) for all x, y X with x y. Prove that T has a unique fixed point in X. Why doesn t the example in Problem 1 contradict this theorem? Define f : X R by f(x) = d(t (x), x). The function f is continuous, since if x n x and y n = T (x n ), then y n T (x) by the continuity of T, and by the continuity of d. f(x n ) = d(y n, x n ) d(t (x), x) = f(x) Since f : X R is a continuous function on a compact set, it attains its minimum value at some a X. If T (a) a, then f (T (a)) = d (T (T (a)), T (a)) < d (T (a), a) = f(a), contradicting the fact that f attains its minimum at a. Hence, a is a fixed point of T. If a, b X are distinct fixed points of T, then d(a, b) = d (T (a), T (b)) < d(a, b), which is impossible, so T has a unique fixed point. The example in Problem 1 doesn t contradict this theorem because X = [, ) is complete but not compact. 2

3 Problem 3. Consider the nonlinear integral equation u(x) 1 4 (x + y) u 2 (y) dy = 1 2. Show that there is a unique continuous solution u : [, 1] R of this equation with the property that u(x) 1 for all x 1. Define Φ : C([, 1]) C([, 1]) by Φ(f) = (x + y) f 2 (y) dy. Then u C([, 1]) is a solution of the of the integral equation if and only if it is a fixed point of Φ. If k : [, 1] [, 1] R is continuous, then k is uniformly continuous. Given ɛ >, there exists δ > such that k(x 1, y 1 ) k(x 2, y 2 ) < ɛ if x 1 x 2 + y 1 y 2 < δ, so x 1 x 2 < δ implies that k(x 1, y) dy k(x 2, y) dy k(x 1, y) k(x 2, y) dy < ɛ, meaning that k(, y) dy is continuous on [, 1]. In particular, Φ(f) is continuous if f is continuous, so Φ maps C([, 1]) into itself as claimed. Alternatively, since k(x, y) M and M dy <, the same result follows from the Lebesgue dominated convergence theorem. Let If f X, then 1 4 X = {f C([, 1]) : f(x) 1 for all x [, 1]}. (x + y) f 2 (y) dy 1 4 so 1/2 Φ(f) 7/8, and Φ : X X. (x + y) dy = x , 3

4 Similarly, if f, g X, then f 2 g 2 2 f g, and Φ(f) Φ(g) = sup Φ(f)(x) Φ(g)(x) x [,1] 1 4 sup x [,1] 1 2 f g 3 4 f g, (x + y) f 2 (y) g 2 (y) dy sup x [,1] (x + y) dy which shows that Φ : X X is a strict contraction. The space X is a closed subset of the complete space C([, 1]), so X is complete, and the contraction mapping theorem implies that Φ has a unique fixed point u X. Remark. The integral equation in this problem is degenerate, meaning that the image of Φ is finite-dimensional, and we can find the explicit form of its solutions. For any u C([, 1]), we have Φ(u)(x) = A + Bx, where A = yu 2 (y) dy, B = 1 4 u 2 (y) dy. Thus, Φ maps C([, 1]) into the two-dimensional space of linear functions, and any fixed point must lie in this space. Moreover, u(x) = A + Bx is a solution if and only if the constants A, B satisfy the algebraic equations A = y (A + By) 2 dy = A AB B2, B = 1 4 (A + By) 2 dy = 1 4 A AB B2. 4

5 Problem 4. (a) Let A be a dense subset of C([, 1]) with respect to the sup-norm. Prove that f = if f C([, 1]) and f(x)g(x) dx = for every g A. (b) Prove that f = if f C([, 1]) and x n f(x) dx = for every n N. (a) Since A is dense in C([, 1]), there is a sequence (g n ) in A such that g n f uniformly as n. Then f 2 dx = 1 2 lim n = 1 2 lim n = 1 2 lim n =, which implies that f =. ( ) f 2 + gn 2 dx ( ) f 2 2fg n + gn 2 dx (f g n ) 2 dx (b) Let h(x) = xf(x). Then the condition in the problem implies that xn h(x) dx = for every n N = {, 1, 2,... }, so p(x)h(x) dx = for every polynomial p : [, 1] R. By the Weierstrass approximation theorem, the polynomials form a dense set in C([, 1]), so h = from (a). It follows that f(x) = for < x 1, so f() = by continuity, and f =. 5

6 Remark. The proof in (a) shows, more generally, that if f L 2 (, 1) and 1 fg dx = for all g A where A is dense in L2 (, 1), then f =. The set of polynomials is dense in L 2 (, 1), but we don t need all of them. According to the Müntz theorem for L 2 (, 1), if (λ n ) n= is a sequence of distinct real numbers with λ n > 1/2 (so that the corresponding power is square-integrable), then the linear span of the functions {x λ, x λ 1, x λ 2,... } is dense in L 2 (, 1) if and only if 2λ n + 1 (2λ n + 1) =. n= The original Müntz theorem (1914) gave a criterion for density in C([, 1]) with respect to the sup-norm. In that case, we do need to include the constant functions, since all positive powers x λ vanish at x =. If (λ n ) n=1 is a sequence of distinct positive real numbers with inf n N λ n >, then the linear span of the functions {1, x λ 1, x λ 2,... } is dense in C([, 1]) if and only if n=1 1 λ n =. In particular, to get a uniformly dense set of polynomials, we only need to include enough integer powers λ n to ensure that 1/λ n diverges. 6

7 Problem 5. Let R = [, 1] [, 1] and suppose that F C(R) consists of all functions f C(R) of the form f(x, y) = N g n (x)h n (y) n=1 for some g n, h n C([, 1]) and some N N. Show that F is dense in C(R). (You can use any appropriate theorems.) Linear combinations and products of functions in F clearly belong to F, so F is an algebra. Taking g n = h n = 1, we see that the constant functions belong to F. If P 1 = (x 1, y 1 ) and P 2 = (x 2, y 2 ) are distinct points in R then at least one of their coordinates are distinct, say x 1 x 2. Then f(x, y) = x 1 F has different values on P 1 and P 2, so F separates points. Since R is compact, the Stone-Weierstrass theorem implies that F is dense in C(R). Alternatively, note that F includes all polynomials in (x, y), and the polynomials are dense in C(R), so F is dense in C(R). 7

8 Problem 6. Let C (R) denote the Banach space of continuous functions f : R R such that f(x) as x, equipped with the sup-norm. (a) For n N, define f n C (R) by { 1 if x n, f n (x) = n/ x if x > n. Show that F = {f n : n N} is a bounded, equicontinuous subset of C (R), but (f n ) has no uniformly convergent subsequence. Why doesn t this example contradict the Arzelà-Ascoli theorem? (b) A family of functions F C (R) is said to be tight if for every ɛ > there exists R > such that f(x) < ɛ for all x R with x R and all f F. Prove that F C (R) is precompact in C (R) if it is bounded, equicontinuous, and tight. (a) Clearly, f = 1 for every n N, so F is uniformly bounded. (i) If x, y n, then f n (x) f n (y) =. (ii) If x, y n, then f n (x) f n (y) = n x y xy x y. n (iii) If y n x (and similarly if x n y ), then f n (x) f n (y) = x n x x y x x y. n It follows that f n (x) f n (y) x y for every n N and x, y R, which implies that F is uniformly equicontinuous (not just pointwise equicontinuous). We have f n (x) 1 as n for every x R. If a subsequence (f nk ) converged uniformly, it would converge to the pointwise limit 1, but f n 1 = 1 for every n N, so (f n ) has no uniformly convergent subsequence, and F is not precompact. This example does not contradict the Arzelà-Ascoli theorem because R is not compact. 8

9 (b) We are told that C (R) is a Banach space and therefore complete, but we give a proof anyway. Let (f n ) be a Cauchy sequence in C (R). The completeness of C([ R, R]) for every R > implies that (f n ) converges pointwise and uniformly on compact sets to a continuous function f : R R. Let ɛ >. Since (f n ) is Cauchy, there exists N N such that f m f n < ɛ/2 for all m, n N. Taking the limit of this inequality as m, we get that f f n ɛ/2 for all n N, so (f n ) converges uniformly to f on R. Since f N C (R), there exists R > such that x > R implies that f N (x) < ɛ/2, and then x > R implies that f(x) f f N + f N (x) < ɛ. Thus, f C (R), meaning that C (R) is complete. Since C (R) is complete, a family F C (R) is precompact if and only if it is totally bounded. We will prove that a bounded, equicontinuous, tight family of functions is totally bounded. (The converse is also true.) Let ɛ >. By the tightness condition, there exists R > such that f(x) < ɛ/2 for x R and all f F. Let f R = f [ R,R] denote the restriction of f F to [ R, R]. Then the family F R = { f R : f F } is bounded and equicontinuous in C([ R, R]), so by the Arzelá-Ascoli theorem it is precompact and therefore totally bounded. Hence, F R has a finite ɛ-net {f1 R, f2 R,..., fn R } for some f i F. If f F and x R, then f(x) f i (x) f(x) + f i (x) < ɛ. Hence, f R B ɛ (fi R ) implies that f B ɛ (f i ), and it follows that {f 1, f 2,..., f n } is a finite ɛ-net for F, so F is totally bounded. Remark. 1. An alternative proof is to extract from a sequence (f n ) in F successive subsequences (fk N) k=1 that converge uniformly to f C(R) on [ N, N] for each N N, and then use the tightness condition to show that the diagonal sequence (fk k) k=1 converges uniformly to f on R. 2. The tightness condition excludes a loss of compactness due to functions escaping to infinity, as in (a), or in the sequence (f n ) where f n (x) = f(x n) for some f C (R). 9

10 Problem 7. Let X be a compact metric space and (f n ) an increasing sequence of continuous functions f n : X R. If f n f pointwise and f : X R is continuous, prove that f n f uniformly. Let ɛ >. The pointwise convergence of (f n ) implies that for each x X, there exists N x N such that f(x) f Nx (x) < ɛ 3. The continuity of f, f Nx at x implies that there exists δ x > such that f(x) f(y) < ɛ 3, f N x (x) f Nx (y) < ɛ 3 for every y B δx (x). Since {B δx (x) : x X} is an open cover of the compact space X, it has a finite subcover {B δxi (x i ) : i = 1,..., m}. Let N = max{n xi : i = 1,..., m}. If x X, then x B δxi (x i ) for some 1 i m. In that case, f(x) fnxi (x) f(x) f(xi ) + f(xi ) f Nxi (x i ) < ɛ, + fnxi (x i ) f Nxi (x) so the monotonicity of (f n ) implies that if n > N, then f(x) f n (x) f(x) fnxi (x) < ɛ. It follows that (f n ) converges uniformly to f. Remark. This result is called Dini s theorem (1878). It describes one of the few situations in which we can deduce uniform convergence from pointwise convergence. 1