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I've been recently introduced to the concept of uniform continuity and I'm having some trouble deciding whether a function is uniformly continuous or not. The function is: $$f(x)=\begin{cases}\dfrac{\sin(x^5)}{x} & x\neq0 \\ 0 & x=0\end{cases}$$

I mean, I know that if the derivative of the function is bounded, then we can conclude that it is uniformly continuous, but if it isn't, I really don't know how to proceed.

Sergamar
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1 Answers1

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The function $f$ is uniformly continuous on $\mathbb{R}$.

Note that $\frac{\sin x^5}{x} = x^4 \frac{\sin x^5}{x^5} \to 0 \cdot 1 = 0$ as $x \to 0$ (from the left or the right). Hence, $f$ is continuous and uniformly continuous on any compact interval $[-a,a]$.

Since $f(x) \to 0 $ as $|x| \to \infty$ the function is also uniformly continuous on intervals $(-\infty,-a]$ and $[a,\infty)$. Any continuous function with a finite limit as $x \to \pm \infty$ is uniformly continuous -- proved many times on this site.

It is a common misconception that a function with a derivative that is unbounded for large $|x|$ cannot be uniformly continuous. The condition that $|f'(x)| \to +\infty$ as $|x| \to +\infty$ guarantees non-uniform continuity.

RRL
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  • Could you show me some proof of the fact that if $|f'(x)|\to \infty$ as $|x|\to \infty$ then it collides with uniform convergence? – Jakobian Dec 26 '18 at 01:59
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    @Jakobian: Sure. $|f(x) - f(y)| = |f'(\xi)||x - y|$ by the MVT. For any $\delta > 0$ take $y = x + \delta/2$ so that $|x-y| < \delta$ and $|f(x) - f(y)| = |f'(\xi)|\delta/2$ Now this stays bigger than $1$ for some $x$ and $\xi > x$ sufficiently large if $|f'(x)| \to +\infty$. – RRL Dec 26 '18 at 02:07
  • @Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity? – ImNotTheGuy Dec 26 '18 at 02:09
  • @RRL oh, thank you, it was a typo and it got me confused – Jakobian Dec 26 '18 at 02:10
  • @Jakobian It got me confused too, lol. – ImNotTheGuy Dec 26 '18 at 02:10
  • Yes, its about uniform continuity. – RRL Dec 26 '18 at 02:11