Chapter 5 21 a) Prove that if A ⊂ n is compact, B ⊂ C(A,m) is compact ⇔ B is closed, bounded and equicontinuous. “⇒” Assume A ⊂ n is compact, B ⊂ C(A,m) is compact. Since if A ⊂ n is compact, B ⊂ C(A,m) is compact, by the Arzela-Ascoli Theorem, B is closed, equicontinuous and pointwise compact. Then since B ⊂ m is compact, by the Heine-Borel theorem, B is both closed and compact. Hence, B is closed, bounded and equicontinuous. “›” For every fn → f ∈ B, ||fn|| ≤ M since B is bounded. Since B ⊂ n, Bx = { f(x) : f ∈ B } is pointwise bounded. Given b ∈ Bx there exists fn with fn(x) → b. Then there exists a subsequence fnk(x) → a converging uniformly. Hence, fnk → f ∈ B. ⇒ B is closed and bounded ⇒ B is compact. b) Let D = { f ∈ C([0,1],) | ||f|| ≤ 1 }. Show that D is closed and bounded but is not compact. Let fn(x) = xn, ε = ½ and x = 1. 22 Let B ⊂ C(A,m) and A ⊂ m be compact. Suppose that for each x0 ∈ A and ε > 0, there is a δ > 0 such that d(x,x0) < δ implies d(f(x),f(x0)) < ε for all f ∈ B. Prove that B is equicontinuous. A ⊂ m is closed and bounded since it is compact. Therefore there exists a max(d(x,x0)) for some x0 ∈ A. Let δ0 > max(d(x,x0)) then for any ε > 0 there exists δ0 such that for any x,x0 ∈ A, d(x,x0) < δ implies that d(f(x),f(x0)) < ε for all f ∈ B. Hence, B is equicontinuous. 25 Let f: [0,1] →  be continuous and one-to-one. Show that f is either increasing or decreasing. Assume that x < c < y or x > c > y. Since the function is continuous and one-to-one, by the intermediate value theorem f(x) < f(c) < f(y) or f(x) > f(c) > f(y). If the function would not be increasing or decreasing then, also by the intermediate value theorem, there would exist x ∫ y such that f(x) = f(y) which contradicts our assumption that f is one-to-one. Therefore, f is either increasing or decreasing.26 Let k(x,y) be a continuous real-valued function on the square U = {(x,y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 } and assume that |k(x,y)| < 1 for each (x,y) ∈ U. Let A:[0,1] →  be continuous. Prove that there is a unique continuous real valued function f(x) on [0,1] such that ∫+=10)(),()()( dyyfyxkxAxf. f is a fixed point for the map φ: f → ∫+10)(),()( dyyfyxkxA. Hence, is φ is a contraction, then there is a unique solution for f(x). The function f is continuous since it satisfies the Lipschitz condition. δL thusand ||)(),()(),( ≤−≤≤∫∫stLdyyfyxkdyyfyxktsts Let k, 0 ≤ k ≤ 1, be such that d(φ(f1),φ(f2)) ≤ kd(f1,f2). Then []),()()(sup)(),()(),(sup))(),((2121102121ffKdyfyfKdyyfyxkyfyxkffdδδφφ=−≤−=∫and δK < 1. Hence, k = δK. Therefore, φ is a contraction and has a unique fixed point which implies that f(x) has a unique solution. 37 Suppose that f:  →  is continuous and f(1) = 7. Suppose that f(x) is rational for all x. Prove that f is constant. Assume that f is not constant. Then there exist a,b ∈  such that f(a) < f(b) or f(b) > f(a). Since f is continuous, there exists c ∈  such that f(a) < f(c) < f(b) which by the density of real numbers need not be rational. Hence, there is a contradiction and f is constant. 41 Show that there is a polynomial p(x) such that |p(x) – |x|3| < 1/10 for -1000 ≤ x ≤ 1000. Let A = [a,b] and B = {q ∈ C([a,b],) where q is a polynomial }. Then by the Stone-Weierstrass theorem if x ∫ y, f(t) = t so that f(x) ∫ f(y). Therefore, B is dense. Then the polynomials p(x) are also dense in C([-1000,1000],). Since |x|3 is continuous, let ε = 1/10 to get p(x).45 a) Let fn: K ⊂ n → m be a sequence of equicontinuous functions on a compact set K converging pointwise. Prove that the convergence is uniform. Since fn is equicontinuous and since K is compact, fn has a uniformly convergent subsequnce fnk. Therefore, fnk → f uniformly on K and f is continuous on K. Since f is continuous and K is compact, f is uniformly continuous on K. Therefore, if f is uniformly continuous, ∀ε > 0 ∃δ > 0 such that x,y ∈ K and d(x,y> < δ implies p(f(x),f(y)) < ε ⇒ f is equicontinuous. Since f is equicontinuous and K is compact, f has a uniformly convergent subsequence fn. Hence, fn converges uniformly. b) Let 222)1()(nxxxxfn−+=, 0 ≤ x ≤ 1.Show that fn converges pointwise but not uniformly. fn(1/n) = 1 does not tend to 0. Hence, <fn>1∞ is not