real analysis - Proving uniform continuity of function on a half-open interval whose derivative has a limit at the boundary

I'm given a continuous function $f : (a,b] \rightarrow \mathbb{R}$ for which

• $f'(x)$ exists on $(a,b)$ and


• $\lim_{x \to^+ a} f'(x)$ exists

and asked to prove that $f$ is uniformly continuous.

I am having a bit of trouble with how to show this formally, though I understand the essence of the answer:

• First off, the proof would be immediate if $f$ was defined over the closed interval $[a,b]$ since continuous functions are uniformly continuous on closed domains.




• Second, because the limit of $f'(x)$ exists as $x \to^+ a$, $f$ must be Lipschitz on its domain ($f'$ must be bounded since it is bounded near $a$ and bounded everywhere to the right of $a$ because the domain is closed in that direction).

This is all very well and good for an intuitive answer, but it seems a bit hand wavy. How do I do give a real $\epsilon, \delta$ style argument here? Or is that overkill for this problem?

Thanks.

Answer

Since $f'$ has side limit at $a$, it is bounded on $(a,c)$ for some $c>a$. So $f$ is Lipschitz (and hence uniformly continuous) on $(a,c]$. On the other hand, $f$ is also uniformly continuous on $[c,b]$ by compactness.

Now you can show the following general fact: if a function is uniformly continuous on $(a,c]$ and also on $[c,b]$, then it is uniformly continuous on $(a,b]$. Using this fact and the above observations the proof is finished.

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PS: This is assuming the side limit of $f'$ exists and is finite, otherwise there are counter-examples.

PPS: $f'$ may still be unbounded on $(a,b]$, you can make counter-examples by adding countably-many disjoint tiny blobs with higher and higher derivatives.