3
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.