I know the mean value theorem hypotheses are:

If $f: [a,b] \mapsto R$ is continuous on $[a,b]$ and differentiable on $(a,b)$.

But can I still invoke MVT under the hypothesis that:

$f: (a,b) \mapsto R$ is differentiable

Does my domain necessarily need to be closed and continuous at those points? I am proving that $f: (a,b) \mapsto R$ is differentiable and $|f'(x)| \leq M$ for all $x \in (a,b)$ then $f$ is uniformly continuous on $(a,b)$. I am pretty certain I need to use MVT for the proof but does it need any special treatment since my hypothesis is a bit different?

Edit: this is my proof

Suppose $f:(a,b) \mapsto \mathbb{R}$ is differentiable and $|f'(x)| \leq M$ for all $x \in (a,b)$. By the Mean-Value Theorem $\frac{f(b) - f(a)}{b-a} \leq M$. Fix $\epsilon >0$ and let $\delta = \frac{\epsilon}{M}$. Now if we have $|b-a|<\delta$ then $f(b)-f(a) \leq M|b-a| < M\delta = \epsilon$.

  • $\begingroup$ With some editing and rearranging your proof can be correct. Start here: "Fix $\epsilon>0$ and let $\delta=\epsilon/M$. Take any points $x$, $y$ in $(a,b)$ for which $|x-y|<\delta$. Then, by the mean-value theorem there is a point $\xi$ between $x$ and $y$ so that $$|f(x)-f(y)|=|f'(\xi)||x-y|\leq M|x-y|< M\delta= \epsilon$$ and so, by definition $f$ is uniformly continuous on $(a,b)$." Your ideas, but logically arranged to end up with the actual definition of uniform continuity. $\endgroup$ – B. S. Thomson Mar 9 '16 at 23:32

Yes you can still apply the MVT, just not for the points $a$ and $b$. If $f$ is differentiable on $(a,b)$, then it is continuous on $[c,d]$ and differentiable on $(c,d)$ for all $c,d \in (a,b)$. Thus you can you the MVT for any points $c<d$ in the interval $(a,b)$.

  • $\begingroup$ Ah thank you, I will still need to check that the endpoints a and b satisfy the uniform continuity definition right? $\endgroup$ – Justin Mar 9 '16 at 23:08
  • $\begingroup$ If $f$ is not defined at $a,b$ you certainly can't check continuity at those points. (Although, bonus exercise: if $f$ is uniformly continuous on $(a,b)$ then $f$ has a uniformly continuous extension to $[a,b]$.) $\endgroup$ – nullUser Mar 9 '16 at 23:11
  • $\begingroup$ Oh of course, appreciate it! $\endgroup$ – Justin Mar 9 '16 at 23:12

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