# Erroneous proof that derivative tending to $\infty$ implies that the function is not uniformly continuous.

I have a proof for the following statement (which I know is wrong):

Let $f:(a,b)\rightarrow \mathbb{R}$ be a differntiable function such that $\underset{x\rightarrow a^+}{\lim} \vert f'(x) \vert=\infty$. Then $f$ is not uniformly continuous on $(a,b)$.

This is false because of $\sqrt{x}$ for example, but I have the following proof which I cannot find where it fails:

It suffices to show that there for all $\delta>0$, there exists $\vert x-y\vert< \delta$ such that $\vert f(x)-f(y)\vert\geq 1$. By the given limit, there exists $b>X_0>a$ such that for all $a<x<X_0$ we have:

$\vert f'(x) \vert >\frac{2}{\delta}$

Consider:

$y=X_0-\delta, \quad x=X_0-\frac{\delta}{2}$

By the mean value theorem, there exists $c\in (x,y)$ such that:

$f'(c)=\dfrac {f(y)-f(x)} {y-x}$

Hence:

$\vert f(y)-f(x) \vert =\vert f'(c)\vert \cdot \vert y-x\vert =\frac{\delta}{2}\cdot \vert f'(c)\vert \overset{c<X_0}{>}\frac{\delta}{2}\cdot \frac{2}{\delta}=1$

Which shows that $f$ is not uniformly continuous. I can not seem to pinpoint what is false here.

## 1 Answer

You cannot evaluate $f$ at $X_0-\delta$ or $X_0-\frac\delta2$ because it is not necessarily the case that $X_0>a+\delta$.

• Will this proof work for showing that $f$ is not uniformly continuous on $(a,\infty)$ if $\underset{x\rightarrow \infty}{\lim} \vert f'(x) \vert=\infty$? – Keen-ameteur Jun 19 '18 at 16:15