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

Question

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 1

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.

Answer 2

First, let's show that $f(x)$ is bounded. Let $l = \lim_{x\to a^+} f'(x)$. We therefore have some $\delta > 0$ such that, $$a < x < a + \delta \implies |f'(x) - l| < 1 \implies |f'(x)| < l + 1.$$ Suppose $x \in (a, a + \delta)$. Using the mean value theorem, there must be some $c \in (x, a + \delta)$ such that $$f'(c) = \frac{f(a + \delta) - f(x)}{a + \delta - x}.$$ But $c \in (a, a + \delta)$, hence $$\left|\frac{f(a + \delta) - f(x)}{a + \delta - x}\right| < l + 1 \implies |f(a + \delta) - f(x)| < (l + 1)(a + \delta - x) < \delta(l + 1).$$ Thereore, $$|f(x)| < \delta(l + 1) + |f(a + \delta)|,$$ which shows $f$ is bounded on $(a, a + \delta)$. On the rest of the interval, $[a + \delta, b]$, the function is continuous and hence bounded on the compact interval. Therefore, $f$ is bounded on $(a, b]$.

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Now suppose for the sake of contradiction that $\lim_{x \to a^+} f(x)$ does not exist. Then, due to the boundedness of $f(x)$, we can then construct sequences $(x_n), (y_n)$, both converging to $a$ from above, such that $f(x_n)$ and $f(y_n)$ both converge to different limits. Therefore, it follows that $x_n - y_n \to 0$ but $f(x_n) - f(y_n)$ converges to something non-zero, which implies that $(f(x_n) - f(y_n))/(x_n - y_n)$ becomes unbounded as $n \rightarrow \infty$, which contradicts the MVT and $f'$ being bounded locally to the right of $a$.

Hence $\lim_{x \to a^+} f(x)$ does exist, and so $f$ can be extended to a continuous function on $[a, b]$, which is uniformly continuous, and hence $f$ is uniformly continuous too.