140B HOMEWORK 5 SOLUTIONS Problem 1 Fix ε > 0 and let fn (x) = f (xn ). Since f and xn are continuous functions, fn (x) is continuous, hence integrable. Moreover, since xn → 0 as n → ∞ for all x ∈ [0, 1/2], we have that fn (x) → f (0) pointwise for all x ∈ [0, 1/2]. We claim more, namely uniform convergence fn ⇉ f (0) over [0, 1/2]. Indeed, by continuity of f at 0 there exists δ > 0 such that x < δ Ô⇒ ∣f (x) − f (0)∣ < ε. 1 Choose N such that 2N < δ. Then for all n > N and x ∈ [0, 1/2] we have ∣fn (x) − f (0)∣ = ∣f (xn ) − f (0)∣ < ε since xn − 0 < 1 2N < δ. Thus fn converges uniformly to f (0) on [0, 1/2]. Finally, by Theorem 7.16, we can integrate once we established uniform convergence, yielding 1/2 lim ∫ n→∞ 0 f (xn )dx = ∫ 1/2 0 f (0) dx = f (0) . 2 Problem 2 (i) We must check that d defines a metric: – First, if f ≠ g there exists x such that f (x) ≠ g(x), so d(f, g) ≥ sup ∣f (x) − g(x)∣ > 0, and it is clear that d(f, f ) = 0. – Second, d(f, g) = sup ∣f (x) − g(x)∣ + sup ∣f ′ (x) − g ′ (x)∣ = sup ∣g(x) − f (x)∣ + sup ∣g ′ (x) − f ′ (x)∣ = d(g, f ). – Finally, for any C 1 function h on [a, b] we have d(f, g) = sup ∣f (x) − g(x)∣ + sup ∣f ′ (x) − g ′ (x)∣ = sup ∣f (x) − h(x) + h(x) − g(x)∣ + sup ∣f ′ (x) − h′ (x) + h′ (x) − g ′ (x)∣ ≤ sup ∣f (x) − h(x)∣ + sup ∣h(x) − g(x)∣ + sup ∣f ′ (x) − h′ (x)∣ + sup ∣h′ (x) − g ′ (x)∣ = d(f, h) + d(h, g). In the above, we have made use of the inequality sup ∣u(x) + v(x)∣ ≤ sup ∣u(x)∣ + sup ∣v(x)∣ which was shown in class. Therefore d defines a metric. (ii) To show C 1 ([a, b]) is complete, we need to show that every Cauchy sequence converges to a function in C 1 ([a, b]). Note: it is not sufficient to show that the sequence of functions converges to some function. You need to show that the limit function lives in C 1 ([a, b]). 1 2 140B HOMEWORK 5 SOLUTIONS Let {fn } be a Cauchy sequence in C 1 ([a, b]). For all ε > 0 we can choose N such that n, m > N implies ′ ε > ∣∣fm − fn ∣∣ = sup ∣fm (x) − fn (x)∣ + sup ∣fm (x) − fn′ (x)∣. In particular ′ sup ∣fm (x) − fn (x)∣ < , and sup ∣fm (x) − fn′ (x)∣ < . From the first inequality on the previous line, fn satisfies the Cauchy criterion, so fn converges uniformly to a limit function f . From the second inequality, fn′ also satisfies the Cauchy criterion, hence converges uniformly on [a, b] to some function g. Since each fn′ is continuous and the convergence is uniform, g is also continuous. Furthermore, by Theorem 7.17, f ′ (x) = g(x). Therefore the limit f ∈ C 1 ([a, b]), showing that the metric space is complete. Problem 3 First we’ll show that ∣ sin(α)∣ ≤ α for α > 0. By the Mean Value Theorem there exists c ∈ (0, α) such that sin(α) − sin(0) ∣ = ∣ sin′ (c)∣ ≤ 1 ∣ α−0 so we have ∣ sin(α)∣ ≤ ∣α∣ when α > 0. On (1, ∞), this implies ∣2n sin( 1 2n 2n )∣ ≤ ≤ . 3n x 3n x 3n n Since ∑ ( 32 ) is a geometric series it converges. By the Weierstrass M-test, ∑n≥1 2n sin( 3n1x ) converges uniformly. Let fn (x) = 2n sin( 3n1x ). Then ∣fn′ (x)∣ = ∣ 1 −2n 2 n cos( )∣ ≤ ( ) . 3n x2 3n x 3 n Since ∑n≥1 ( 23 ) converges, the Weierstrass M-test implies that ∑n≥1 fn′ (x) converges uniformly on n 1 (1, ∞). By Theorem 7.17, f is differentiable and f ′ (x) = ∑n≥1 3−2 n x2 cos( 3n x ). Problem 4 (i) Fix a > 0. Then we have for all x ∈ [a, ∞). Since ∑n≥1 formly on [a, ∞). 1 n3 a n n 1 ≤ 4 = 3 4 1+n x n x n a converges, the Weierstrass M-test implies f (x) converges uni- (ii) Suppose that f (x) converges uniformly over (0, ∞). By Cauchy criterion with = 31 , there exists N such that m, n > N implies m k 1 < . 4 3 k=n+1 1 + k x ∑ Make m = n + 1, so that the above sum has only one term, namely m 1 < . 4 1+m x 3 This inequality fails if we set for instance x = m14 since the left hand side is m/2 > 1/3. Therefore the series does not converge uniformly over (0, ∞). 140B HOMEWORK 5 SOLUTIONS 3 (iii) Fix a > 0. We work over the interval [a, ∞) first. Consider the partial sums n k −k 5 ′ Ô⇒ s (x) = . ∑ n 4 4 2 k=1 1 + k x k=1 (1 + k x) n sn (x) = ∑ Since k5 1 1 ≤ 3 2 ≤ 3 2, 4 2 (1 + k x) k x k a and ∑k≥1 k31a2 converges, the Weierstrass M-test implies that {s′n } converges uniformly on [a, ∞). We also know that sn ⇉ f by part (i). By Theorem 7.17, f is differentiable and s′n converges uniformly to f ′ (x) over [a, ∞). Since for any x > 0 we can find an a > 0 such that x > a, it follows that f is differentiable over the entire interval (0, ∞). (iv) We’ll prove f is infinitely many times differentiable by induction. Fix a > 0. Precisely, we show (`) that f is `-times differentiable and that the derivatives of the partial sums sn converge to f (`) uniformly over [a, ∞) for all `. We showed in part (iii) that f can be differentiated once over [a, ∞), which is the inductive base case. (`) We now carry out the inductive step. Suppose that f is `-times differentiable, and that sn converges to f (`) uniformly over [a, ∞). We look at the ` + 1-derivatives of the partial sums sn . The absolute value of the (` + 1)st derivative of 1+kk 4 x is ∣ (` + 1)! (` + 1)! (−1)`+1 (` + 1)!k 4`+5 ∣ ≤ 3 `+2 ≤ 3 `+2 . 4 `+2 (1 + k x) k x k a Since ∑k≥1 k3 a`+2 = a`+2 ∑k≥1 k13 converges, the Weierstrass M-test implies that the derivatives (`+1) (`) sn converge uniformly. By the induction hypothesis, we already know that sn converges to (`) (`) f . By Theorem 7.17, it follows that f is differentiable, so f is (` + 1) times differentiable, and (`+1) that sn converges (uniformly, as we showed) to f (`+1) . This completes the induction step. (`+1)! (`+1)! Therefore, f is infinitely differentiable on [a, ∞), hence also on the union of these intervals which is (0, ∞). Problem 5 See online solutions. Problem 6 Fix ε > 0 and let δ = ε/M . Then for any f ∈ F and any x, y ∈ R such that ∣x − y∣ < δ there exists c ∈ (x, y) such that ε ∣f (x) − f (y)∣ = ∣f ′ (c)∣ ⋅ ∣x − y∣ < M = ε. M Therefore F is equicontinuous. For the particular sequence of functions given, ∣fn′ (x)∣ = ∣ cos(nx) − sin(x + n)∣ ≤ 2 for any choice of n and x. Since the derivatives are uniformly bounded, {fn } is an equicontinuous family. 4 140B HOMEWORK 5 SOLUTIONS Problem 7 Note that the function f is bounded by 1. By the fundamental theorem of calculus, we compute 1 ∣Fn′ (x)∣ = ∣ f (an x)an ∣ = ∣f (an x)∣ ≤ 1. an Thus, the family {Fn } has uniformly bounded derivatives. By Problem 6, {Fn } is an equicontinuous family.
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