# Harrison Chapman

## Math 317: Homework 10

Due: Friday, November 22, 2019

Exercises:

1. Consider the function $$f(x) = x^2 \sin (\frac 1x)$$ when $$x \ne 0$$ and $$f(0) = 0$$ which has domain $$\mathbb R$$ and is differentiable.

1. Calculate the derivative $$f’(x)$$ for all $$x \in \mathbb R$$.

2. Use your answer to part (a) to explain why $$f’(x)$$ is discontinuous.

3. We can calculate that $$f’(-\frac 2\pi) = \frac 4\pi$$ and $$f’(\frac 1{2\pi}) = -1$$.

Which theorem guarantees that there exists a number $$c \in (-\frac 2\pi, \frac 1{2\pi})$$ for which $$f’(c) = 0$$?

1. For $$x \ne 0$$ we may use the product and chain rules to find that $$f’(x) = 2x\sin(\frac 1x) - \cos(\frac 1x)$$. When $$x = 0$$ we turn to the definition of the limit and compute

$f’(0) = \lim_{h\to0}\frac{f(h) - f(0)}{h} = \lim_{h\to0}\frac{h^2 \sin(\frac 1h)}{h} = 0.$

2. Using our answer to part (a) we realize that while $$f’(0)$$ exists, $$\lim_{x \to 0}f’(x)$$ does not (it oscillates too rapidly). We thus conclude $$f’(x)$$ is not continuous at $$0$$ and hence is discontinuous.

3. The Intermediate Value Theorem for Derivatives still applies, even though the IVT for continuous functions does not (as $$f’$$ is not continuous on $$[-\frac 2\pi, \frac 1{2\pi}]$$).

2. Recall that a function $$f$$ is Lipschitz continuous on a set $$S$$ provided there exists $$K \ge 0$$ so that for all $$x, y \in S$$ with $$x \ne y$$:

$\left\vert\frac {f(x)-f(y)}{x-y}\right\vert \le K.$

Let $$f$$ be a function which is differentiable on a closed interval $$[a,b]$$ and for which $$f’$$ is continuous on $$[a,b]$$. Prove that $$f$$ is Lipschitz continuous on $$[a,b]$$.

Hint: Use the Mean Value Theorem.

For any $$x, y \in [a,b]$$ with $$x<y$$, by the Mean Value Theorem, there exists $$c \in (x,y) \subseteq [a,b]$$ so that

$\frac{f(x)-f(y)}{x-y} = f’(c).$

As $$f’$$ is continuous on the closed, bounded interval $$[a,b]$$ so too is $$\vert f’\vert$$, so the set $$\left\{ \vert f’(c) \vert \;:\; c \in [a,b] \right\}$$ is bounded above by some non-negative real number $$K \ge 0$$.

Let $$x,y \in [a,b]$$ with $$x < y$$. Then there exists $$c \in [a,b]$$ so that

$\left\vert \frac{f(x)-f(y)}{x-y} \right\vert = \vert f’(c) \vert \le K,$

Notice that if on the other hand $$x,y \in [a,b]$$ and $$x > y$$ then we still have that

$\left\vert \frac{f(x)-f(y)}{x-y} \right\vert = \left\vert \frac{f(y)-f(x)}{y-x} \right\vert \le K.$

So $$f$$ is Lipschitz continuous with constant $$K$$.

3. Show that $$ex \le e^x$$ for all $$x \in \mathbb R$$.

Notice that $$e x \le e^x$$ is true if and only if $$0 \le e^x - ex$$. Let $$g(x) = e^x - ex$$. Then $$g(1) = 0$$. Notice that $$g'(x) = e^x - e$$. When $$x < 1$$, $$g'(x) < 0$$ and when $$x > 1$$, $$g'(x) > 0$$.

Let $$x < 1$$. Then $$g(x) \ge 0$$, as it must be decreasing always until $$x = 1$$, when it is its minimum value of $$0$$. On the other hand, when $$x > 1$$, the function is always increasing from its minimum value of $$0$$, so $$g(x) \ge 0$$ as well. So in all cases $$g(x) \ge 0$$.

4. Let $$f$$ be real-valued and differentiable on $$\mathbb R$$, and let $$g(x) = f(x+1) - f(x)$$. Suppose additionally that $$\lim_{x\to \infty} f'(x) = 0$$.

Prove that $$\lim_{x\to\infty} g(x) = 0$$.

Hint: Use the Mean Value Theorem.

Let $$\epsilon > 0$$, so there exists $$M$$ so that for all $$c > M$$ we have that $$|f'(c)| < \epsilon$$. Let $$x > M$$, and consider the interval $$[x, x+1]$$. By the Mean Value Theorem, there exists an input $$c$$ between $$x$$ and $$x+1$$ for which $$f'(c) = (f(x+1) - f(x))/1 = g(x)$$. But then $$|g(x)| = |f'(c)| < \epsilon$$ as $$c > x > M$$. This holds for all $$x > M$$ and all $$\epsilon > 0$$, so $$\lim_{x \to \infty} g(x) = 0$$.