MATH 242 – Principles of Analysis Selected Solutions for Problem Set 7, “B portion” April 8, 2011

§3.5/15. A comment first: A fairly common idea that people tried to use in solutions for (a) was this. Let 0 < c < 1. Then 0 < c^{2} < 1 as well. Since f(0) < 0 and f(1) > 1, this means c^{2} is between f(0) and f(1). So by the Intermediate Value Theorem there is some x_{0} ∈ [0, 1] where f(x_{0}) = c^{2}. The problem with this is that the question asked you to show that there is some c where f(c) = c^{2}. The x_{0} above will not usually equal c. (Indeed that statement shows ∀c ∈ [0, 1], ∃x_{0} ∈ [0, 1] s.t. f(x_{0}) = c^{2}.) The correct way to apply the IVT here is to apply it to a different function. We want to find c where f(c) = c^{2}, so f(c) − c^{2} = 0.

(a) Let g(x) = f(x)−x^{2}. By Theorem 3.4.10, this is continuous on [0, 1] since f is assumed to be continuous and x^{2} is continuous everywhere. Using the assumptions on f, we have

\begin{aligned}g(0) & = f(0) − 0 = f(0) < 0 \\ g(1) & = f(1) − 1 > 0 \end{aligned}

Hence since k = 0 is a value between g(0) and g(1), there is a c ∈ (0, 1) where g(c) = f(c) − c^{2} = 0. This shows f(c) = c^{2}.

(b) Let h(x) = f(x) − g(x). By Theorem 3.4.10, this is continuous on [0, 1] since f, g are assumed to be continuous. Since f(0) < 0, g(0) ≥ 0, f(1) > 1, and g(1) ≤ 1, we have

\begin{aligned}h(0) & = f(0) − g(0) < 0 \\ h(1) & = f(1) − g(1) > 0 \end{aligned}

Hence since k = 0 is a value between h(0) and h(1), there is a c ∈ (0, 1) where h(c) = f(c) − g(c) = 0. This shows f(c) = g(c).

22. This is FALSE. The function

f(x) = \frac{cos(π/x)}{x}

on (0, 1] is one counterexample. Notice that at x_{n} = \frac{1}{n} we have

f(1/n) = n cos(nπ) = (−1)^{n}n.

Since x_{n} → 0 as n → ∞, the interval (0, 1] contains points where f takes both positive and negative values that are arbitrarily large in absolute value:

f(1/n) = \{ \begin{matrix}n & if n is even \\ −n & if n is odd \end{matrix}

It follows that f has no maximum or minimum on (0, 1].

23. (a) This is FALSE. There are many non-continuous functions that have the “intermediate value property” (that is, for which the conclusion of the IVT holds). For example, on [0, 1], the function

f(x) = \{ \begin{matrix}2x & if 0 ≤ x ≤ 1/2 \\ 2x − 1 & if 1/2 < x ≤ 1 \end{matrix}

has a jump discontinuity at x = 1/2. But it takes every value between f(0) = 0 and f(1) = 1 at least once (most of them twice, in fact).

(b) This is FALSE. Here’s a counterexample, and an explanation: Let

f(x) = \{ \begin{matrix}sin(1/x) & if x ≠ 0 \\ 0 & if x = 0 \end{matrix}

and consider the interval [a, b] = [−1, 1]. First, f fails to be continuous at 0 because lim_{x→0} f(x) does not exist. But f does have the “intermediate value property” on every subinterval [c_{1}, c_{2}] ⊂ [−1, 1]. We can see this as follows. First, if 0 ≠ [c_{1}, c_{2}], then f is actually continuous on [c_{1}, c_{2}] and the IVT applies. Next, if 0 ∈ [c_{1}, c_{2}], then since f oscillates more and more quickly as x → 0, f must take every value between −1 and 1 on that interval. This certainly implies that f takes every value between f(c_{1}) and f(c_{2}), since those are also in the range [−1, 1].

Comment: A lot of people tried to construct a counterexample f by somehow using a pair c_{1} and c_{2} in the definition of f. That cannot work here because the property we want is that f is not continuous, but for all a ≤ c_{1} < c_{2} ≤ b, the intermediate value property holds for f on [c_{1}, c_{2}]. For instance if you tried to make a counterexample with jump discontinuities at particular c_{1} and c_{2}, then on closed subintervals straddling one of those discontinuities there can be some y-values that are omitted by f. So the intermediate value property would fail for some intervals of x’s.

§3.6/4 Comment: Many people had the correct idea on this one after talking to me in office hours. However, in order to give a complete proof that

f(x) = sin(2π[x]x)

is not uniformly continuous on R, you have to supply more details than many people did. Here is a complete argument.

Aiming for a contradiction, suppose f is uniformly continuous on R. Then for all ε > 0, there exists δ > 0 such that |f(x) − f(x′)| < ε for all x, x′ with |x − x′| < δ. However, take ε = 1. Assuming there is a δ > 0 that works here, we can find n ∈ N such that \frac{1}{4n} < δ. Consider x = n and x′ = n + \frac{1}{4n}. Then

|x − x′| = \frac{1}{4n} < δ,

but f(n) = sin(2πn^{2}) = 0 and

f (n + \frac{1}{4n}) = sin (2π [n + \frac{1}{4n}] (n + \frac{1}{4n}))

= sin (2πn (n + \frac{1}{4n}))

= sin (2πn^{2} + \frac{π}{2} )

= 1

Hence |f(x) − f(x′)| = 1, which is not less than ε = 1. This contradiction shows that f is not uniformly continuous on R.

§4.1/11. Comment: Several people overlooked the fact that the statement to be proved is an “if and only if” statement, so both implications must be shown.

⇒: Suppose h(x) is differentiable at c. Then the limit of the difference quotient of h:

h′(c) = lim_{x→c} \frac{h(x) − h(c)}{x − c}

must exist. This is equivalent to saying the two one-sided limits

lim_{x→c^{−}} \frac{h(x) − h(c)}{x − c}, lim_{x→c^{+}} \frac{h(x) − h(c)}{x − c}

exist and are equal. By the definition of h, though,

lim_{x→c^{−}} \frac{h(x) − h(c)}{x − c} = lim_{x→c^{−}} \frac{f(x) − f(c)}{x − c},

and that limit exists and equals f′(c) since we assume f is differentiable at c. Similarly,

lim_{x→c^{+}} \frac{h(x) − h(c)}{x − c} = lim_{x→c^{+}} \frac{g(x) − g(c)}{x − c},

and that limit exists and equals g′(c) since we assume g is differentiable at c. It follows that f′(c) = g′(c).

⇐: Conversely, suppose f′(c) = g′(c). Then taking the limits of the difference quotients on only one side, we can say:

lim_{x→c^{−}} \frac{f(x) − f(c)}{x − c} = lim_{x→c^{+}} \frac{g(x) − g(c)}{x − c}.

Because of the way h is defined piecewise, the first limit here is the same as

lim_{x→c^{−}} \frac{h(x) − h(c)}{x − c}

and the second is

lim_{x→c^{+}} \frac{h(x) − h(c)}{x − c}.

This shows that the two one-sided limits

lim_{x→c^{−}} \frac{h(x) − h(c)}{x − c}, lim_{x→c^{+}} \frac{h(x) − h(c)}{x − c}

exist and are equal. Therefore,

h′(c) = lim_{x→c} \frac{h(x) − h(c)}{x − c}

must exist, so h is differentiable at c.