Math 5616H Midterm 1 with solutions

Spring 2013 March 8, 2013

Total 80 points

1. (15 points) Let f (x) and g(x) be real continuous functions on an interval [a, b], such that b b

f 2 (x) dx = a a b

g 2 (x) dx = 1.

Prove that a f (x)g(x) dx ≥ −1, and that a b

f (x)g(x) dx = −1 if and only if f ≡ −g on [a, b]. Answer: Since f and g are continuous, so is (f + g)2 , which is therefore integrable. We compute: b b b b b

0≤ a b

[f (x)+g(x)]2 dx = a f (x)2 dx+2 a f (x)g(x) dx+ a g(x)2 dx = 1+2 a f (x)g(x) dx+1,

so a f (x)g(x) dx ≥ −1. If it is = −1, then the first “≤” must be “=”, so the continuous function [f (x) + g(x)]2 ≡ 0, and f ≡ −g on [a, b]. 2. (25 points) Let α(x) be a strictly increasing function on the interval [0, 1], such that α(0) = 0 and α(1) = 1. Show that the Riemann-Stieltjes integral
1

α(x) dα(x),
0

exists if and only if α is continuous on [0, 1], and evaluate this integral if it is continuous. Answer: Consider any partition P of [0, 1] : P = {0 = x0 , . . . , 1 = xn }. Since α is increasing, Mi := supx∈[xi−1 ,xi ] α(x) = α(xi ) and mi = α(xi−1 ). Then n n

U (P, α, α) − L(P, α, α) = i=1 (Mi − mi )∆αi = i=1 (∆αi )2 .

Suppose α is continuous; then since [0, 1] is compact, α is uniformly continuous. Thus, for any given ε > 0 there is δ > 0 so that if |x − y| < δ then |α(x) − α(y)| < ε. Hence if P ∗ is a refinement of P which satisfies xi − xi−1 < δ for all i = 1, . . . , n, we have n U (P ∗ , α, α) − L(P ∗ , α, α) ≤ ε i=1 ∆αi = ε(α(1) − α(0)) = ε.

Math 5616H Spring 2013 This shows that α ∈ R(α).

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For the converse, suppose that α ∈ R(α). Then for all ε > 0 there is a partition P so that n U (P, α, α) − L(P, α, α) = i=1 (∆αi )2 < ε

√ But the terms in this sum are ≥ 0, so ech term ∆αi < ε. Let 2δ = min{∆x1 , . . . , ∆xn }. If |x − y| < δ, then x and y are either (1) in the same subinterval [xi−1 , xi ] or (2) in successive √ subintervals [xi−1 , xi ] and [xi , xi+1 ]. In case (1), since α is increasing, |α(x)−α(y)| < ∆αi < ε; √ in case (2), |α(x) − α(y)| < |α(x) − α(xi )| + |α(xi ) − α(y)| < 2 ε, which is arbitrarily small. This shows that α must be continuous. 3. (20 points) Suppose {fn : n = 1, 2, 3, . . . } is an equicontinuous sequence of functions fn : [a, b] → C, and that fn converges pointwise on [a, b]. Show that fn converges uniformly on [a, b]. Answer: Let ε > 0 be given. Since the family is equicontinuous, we know that there is δ > 0 so that if x, y ∈ [a, b] and |x − y| < δ, then |fn (x) − fn (y)| < ε/3 for all n = 1, 2, . . . . Let f : [a, b] → C be the pointwise limit of fn as n → ∞. Then for any x, y ∈ [a, b] with |x − y| < δ, we have |f (x) − f (y)| = limn→∞ |fn (x) − fn (y)| ≤ ε/3. Let P = {x0 , x1 , . . . , xm } be a partition of [a, b] with xi − xi−1 < δ for i = 1, 2, . . . , m. Since fn → f pointwise on [a, b], with the same ε > 0 there is an N so that if n ≥ N , then |fn (xi ) − f (xi )| < ε/3 for all i = 1, 2, . . . , m. Then for any x ∈ [a, b], there is an interval [xi−1 , xi ] which contains x. This implies that |x − xi | < δ, and hence |fn (x) − fn (xi )| < ε/3 for all n = 1, 2, . . . as well as |f (x) − f (xi )| < ε/3. So if n ≥ N , we conclude |fn (x) − f (x)| ≤ |fn (x) − fn (xi )| + |fn (xi ) − f (xi )| + |f (xi ) − f (x)| ≤ ε/3 + ε/3 + ε/3 = ε. This shows that fn converges to f uniformly on [a, b]. 4. (25 points) We have a theorem which says that if (a) fn is a sequence of continuous functions fn : [a, b] → R; (b) fn → f pointwise on [a, b], where f : [a, b] → R is continuous; and (c) fn+1 (x) ≤ fn (x) for all x ∈ [a, b]; then fn → f uniformly. Is this still true without hypothesis (c)? If it is still true, give a proof, by reducing the general case to the case in the theorem. If it is not true, give a counterexample by sketching the graph of the functions fn . Answer: This theorem is not true without hypothesis (c). For a counterexample, consider [a, b] = [0, 1] and for n ≥ 2, let fn be piecewise linear: fn (x) = nx if 0 ≤ x ≤ fn (x) = 2 − nx if and fn (x) = 0 if 1 ; n

2 1 ≤x≤ ; n n

2 ≤ x ≤ 1. n

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2 Then fn (0) = 0 for all n. If x > 0, then fn (x) = 0 for all n ≥ x . Thus fn → f pointwise on 1 [0, 1], where f ≡ 0, which is certainly continuous. But for each n, fn ( n ) = 1, so fn does not converge uniformly to 0.

A rational function counterexample is given on p. 146 of Rudin.