Prove $f(x) = x \sin(\frac{1}{x^{2}})$ is uniformly continuous on $(0,\infty)$

Question

I am having trouble with the following problem: Prove $f(x) = x \sin(\frac{1}{x^{2}})$ is uniformly continuous on $(0,\infty)$.

I have tried using the theorem (stated generally):

if $f(x)$ is a continuous function on the closed interval $[a,b]$, then $f(x)$ is uniformly continuous on $(a,b).$ However, I am having difficulty extending the function such that I can use this theorem. (Mainly because of the range of $f(x)$.)

I have also tried using the $\epsilon-\delta$ (definition of uniformly continuous) but I found the algebra to be very messy. I think what is also confusing me is the fact that we have to prove uniform continuity on $(0,\infty)$. I haven't encountered many problems asking uniform continuity of functions containing trigonometric functions, on such an interval.

Thanks.

Answer 1

Hints:

(1) $f$ extends to a continuous function $[0,\infty)\to\mathbb{R}$.

(2) $\lim_{x\to\infty}f(x)=0$.

Answer 2

Notice that $$-x<x\sin(\frac{1}{x^2}<x$$ so by sandwich theorem $$\lim_{x\rightarrow 0} x\sin \left(\frac{1}{x^2}\right)=0.$$

Answer 3

$\lim_{x\to 0^+}f(x)=0$ so we can extend $f$ to $[0,\infty)$ by letting $f(0)=0$. Then show that $f$ is uniformly continuous on $[0,\infty)$ and hence uniformly continuous on the subset $(0,\infty).$

(i). $f$ is continuous on $[0,\infty)$ so f is uniformly continuous on the closed bounded interval $[0,1].$

(ii). If $x\ge 1$ then $|f'(x)|=|\sin (x^{-2})+2x^{-2}\cos(x^{-2})|\le |\sin x^{-2}|+|2x^{-2}|\cdot |\cos (x^{-2})|\le |1|+|2\cdot 1^{-2}|\cdot |1|=3.$

If $1\le y<z$ there exists $x\in (y,z)$ such that $|f(y)-f(z)|=|y-z|\cdot |f'(x)|\le |y-z|\cdot 3.$

So $f$ is uniformly continuous on $[0,\infty).$

(iii). Parts (i),(ii) imply that $f$ is uniformly continuous on $[0,\infty)$: Given $\epsilon >0,$ take $\delta\in (0,\epsilon /6)$ such that $0\le y< z<y+\delta\le 1\implies |f(y)-f(z)|<\epsilon /2.$

Then $1\le y<z\le y+\delta \implies |f(y)-f(z)|\le 3|y-z|<\epsilon/ 2.$

And if $0<y<1<z<y+\delta$ then $|y-1|<\delta$ and $|1-z|<\delta$ so $|f(y)-f(z)|\le |f(y)-f(1)|+|f(1)-f(z)|<\epsilon /2+\epsilon/2=\epsilon.$

Answer 4

You may split it up into two cases and merge them as follows:

• Since $\lim_{x\to 0}f(x) = 0$ you can extend $f$ to a continuous function by setting $f(0) = 0$.

Now, consider $f$ on the closed interval $[0,\color{blue}{2}]$. There it is uniformly continuous.

• On the other hand, consider $f$ on $[\color{blue}{1},\infty)$. There, $f$ is uniformly continuous because of the mean value theorem:

$$|f(x) - f(y)| =|f'(\xi)||x-y|= \left|\sin \frac{1}{\xi^2} - \frac{2\cos \frac{1}{\xi^2}}{\xi^2} \right||x-y|\stackrel{\xi\geq 1}{\leq} 3|x-y|$$

Now, you know that $f$ is uniformly continuous on the two overlapping intervals $I_{\color{blue}{1}}=[\color{blue}{1},\infty)$ and $I_{\color{blue}{2}}=[0,\color{blue}{2}]$.

So, you know

• $\forall \epsilon > 0 \exists \delta_{\color{blue}{1}}(\epsilon)> 0 \forall x,y \in I_{\color{blue}{1}}: |x-y| <\delta_{\color{blue}{1}}(\epsilon) \Rightarrow |f(x) - f(y)|<\epsilon$
• $\forall \epsilon > 0 \exists \delta_{\color{blue}{2}}(\epsilon)> 0 \forall x,y \in I_{\color{blue}{2}}: |x-y| <\delta_{\color{blue}{2}}(\epsilon)\Rightarrow |f(x) - f(y)|<\epsilon$

Since the overlap of the intervals has length $\color{green}{1}=\color{blue}{2} - \color{blue}{1}$, you get uniform continuity on $(0,\infty)$ by choosing $\delta(\epsilon) = \min (\delta_{\color{blue}{1}}(\epsilon),\delta_{\color{blue}{2}}(\epsilon), \color{green}{1})$ for any $\epsilon>0$.