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I'm confused how to use the following theorem:

19.6 Theorem. Let $f$ be a continuous function on an interval $I$ [$I$ may be bounded or unbounded]. Let $I^◦$ be the interval obtained by removing from $I$ any endpoints that happen to be in $I$. If $f$ is differentiable on $I^◦$ and if $f′$ is bounded on $I^◦$, then $f$ is uniformly continuous on $I.$

So far, I have encountered examples. $f(x)= \sqrt{x}, g(x)= \frac{1}{x}, h(x)= x^2$ They are each on the interval $(0,\infty)$

I know $f$ is uniformly continuous, but $g$ and $h$ are not. However, the derivatives for each of these functions is unbounded on $(0,\infty)$

To show that a continuous function is not uniformly continuous on $(0, \infty)$, do I need to show the derivative is unbounded for every interval $[a, \infty )$ , where $a>0$?

If so, how would I prove the function is unbounded from $[a, \infty )$?

I would appreciate a worked out example with one of the functions above or one of your choosing.

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If $f$ is differentiable on $I^\circ$ and $f'$ is bounded on $I^\circ$, then $f$ is uniformly continuous on $I$

is correct.

If $f$ is uniformly continuous on $I$, then $f$ is differentiable on $I^\circ$ and $f'$ is bounded on $I^\circ$

is not correct. The counterexample is exactly $f(x)=\sqrt x$.

To show that $f$ is not uniformly continous on $I$, it is enough to show that it is not uniformly continuous on $(0,1]$. And uniformly continous functions on a bounded interval are always bounded.

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  • $\begingroup$ Could you explain how I could prove $g(x)$ and $h(x)$ are not uniformly continuous? I believe $g$ cannot be extended to a continuous function at $x=0$. How do you prove h is not uniformly continuous? $\endgroup$ – Zslice Oct 23 '14 at 8:14
  • $\begingroup$ @Zslice Look at the last paragraph I wrote. $\endgroup$ – 5xum Oct 23 '14 at 8:15
  • $\begingroup$ How does the last paragraph apply to $h(x)= x^2$? $\endgroup$ – Zslice Oct 23 '14 at 8:37

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