How to prove uniform continuity problem!

Question

A) $f(x)=x^3$ , give an example of an interval where $f$ is uniformly continuous and another where it is not. explain your choose of examples

B) decide if $f(x)= \dfrac{1}{\sin x} - \dfrac{1}{x}$ is uniformly continuous on $(0,1)$

C) explain why $f(x) = \dfrac{1}{1+x^2}$ is uniformly continuous on $(-\infty,\infty)$

D) decide if $g(x) = \sin x$ and $h(x)=\sin(x^2)$ is uniformly continuous on $(-\infty, \infty)$

and to solve these we have;

Theorem 2.1: If $f$ is continuous on a closed, bounded interval $[a, b]$, then $f$ is also uniformly continuous on $[a,b]$.

Satz 2.2: Suppose that the function $f$ is differentiable and that the derivative is bounded on the interval $I$. Then $f$ is uniformly continuous on $I$.

Theorem 2.9: If $f$ is uniformly continuous on a bounded interval $I$, then $f$ is also bounded to $I$. $I$ is of the form; $[a,b]$, $(a,b]$, $[a,b)$ or $(a,b)$.

Satz 2.12: If, for each $h> 0$ is such that $|f(x + h) - f(x)|$ is unlimited in $I$, then $f$ is not uniformly continuous on $I$.

The problem for me is that I don't know how to use any of this to actually do the problems, that's why I put up so many questions since I really wanna learn this once for all! :) so if anyone could give me a hand that would be great. and also on C, I'm pretty sure its uniformly continuous by satz 2.2 since the derivative is bounded? but then can I just say ok that's uniformly continuous done?, I don't think so, I need to some how show it mathematically with some sort of proof I think, if yes then I don't know how to do that either :(

Answer 1

A) By theorem 2.1. $f(x) = x^3$ is uniformly continuous on $[0,1]$. On $(0,\infty)$ on the other hand, by Satz 2.12., it is not uniformly continuous. To see this, pick some $h > 0$. Then $|x^3 + 3x^2 h + 3xh^2 + h^3 -x^3|=|3x^2 h + 3xh^2 + h^3| \ge 3x^2h$ which is unlimited in $x$.

Geometrically this is so because the slope of $x^3$ gets arbitrarily large as $x$ gets large.

B) Let $f(x)= \dfrac{1}{\sin x} - \dfrac{1}{x} = { x - \sin x\over x \sin x}$. Then $f$ is continuous for $x \in (0,1]$ because $x, \sin x$ are continuous and sums, differences and quotients of continuous functions are continuous (if the denominator is non-zero which is the case here). Note that $\lim_{x \to 0}f(x) = 0$. (to see this apply de l'Hôpital's rule) Hence we may continuously extend $f$ to the closed and bounded interval $[0,1]$ (by defining $f(0):= 0$). It then follows from theorem 2.1. that $f$ is uniformly continuous on $[0,1]$ and since $(0,1) \subseteq [0,1]$ also on $(0,1)$.

C) Let $f(x) = \dfrac{1}{1+x^2}$. Then $f'(x) = -\dfrac{2x}{(1+x^2)^2}$. This function is uniformly continuous on $[-1,1]$ (by theorem 2.1) and therefore bounded on $[-1,1]$. Outside $[-1,1]$ we have $$\left | -\dfrac{2x}{(1+x^2)^2} \right | = \left | -\dfrac{2x}{(1+x^2)^2} \cdot {{1\over x} \over {1 \over x}}\right | = \left | -\dfrac{2}{((1+x^2){1\over \sqrt{x}})^2} \right | = {2 \over (1+x^{3/2}))^2}\le {1\over 2}$$

Here you want to apply Satz 2.2. and to this end you want to show that $f'$ is bounded. How you do this depends on the function in question but in this case you can achieve this by dividing the domain into two sets and showing that $f'$ is bounded on each and hence also on the union.

Why did I "multiply by ${1 \over x}$"? Well, I wanted to get a bound on the expression and thought about how to eliminate $x$ from the numerator. So I tried multiplying by $1={{1\over x} \over {1 \over x}}$ to see if it did the trick and it did. Once you have a constant in the numerator it's easy to see that it's bounded for $|x|$ greater some non-zero constant.

D) Apply Satz 2.2. to $g(x) = \sin x$ to deduce that it is uniformly continuous.

For the last one you can apply Satz 2.12. This is a bit tricky, you can find a full solution here.