# if $f$ convex, $f$ has bounded first derivative iff $f$ uniformly continuous

Formal statement: if $f: \mathbb{R} \rightarrow \mathbb{R}$ is convex and differentiable, it is uniformly continuous iff there exists some $a > 0$ such that $|f'(x)| \leq a$ for all $x$.

One direction seems slightly easier to prove: if $|f'(x)| \leq a$ for all $x$ and we know $f$ is differentiable, we know there's $c$ such that $|f(y) - f(x)| = c \cdot |x - y| \leq a \cdot |x - y|$ we apply Mean Value Theorem (edit: previously cited convexity in ADDITION to the MVT for this part and it was deemed unnecessary). Then we can just let $\delta = \epsilon / a$ to show uniform continuity.

The other direction has me stumped. Any suggestion?

• you don't need convexity for the direction you gave. you should look up mean value theorem – mathworker21 Sep 19 '18 at 21:58
• @mathworker21 I used the Mean Value Theorem in the direction I gave... – 0k33 Sep 19 '18 at 21:59
• you said "because f is convex". idk how that applies – mathworker21 Sep 19 '18 at 22:02
• @mathworker21 so am I to interpret your hint to imply that the MVT also applies to the second part? – 0k33 Sep 19 '18 at 22:04
• i wasn't giving a hint. i was saying "you don't need convexity for the direction you gave" – mathworker21 Sep 19 '18 at 22:09

Let $\delta >0$ be such that $|f(x)-f(y)| <1$ if $|x-y| \leq \delta$. Consider $f(n+\delta)-f(n)$. We can write this as $\int_n^{n+\delta} f'(t) \, dt$ and $f'$ is increasing. Hence $f'(n) \int_n^{n+\delta} \, dt \leq \int_n^{n+\delta} f'(t) \, dt=f(n+\delta)-f(n) <1$. It follows (by monotonicity) that $f'$ is bounded above. Similarly it is bounded below. [Integrate $f'$ over $(-n-\delta , -n)$]. Alternatively, just use MVT on $(n,n+\delta)$ and monotonicity of $f'$ instead of writing $f$ as integral of $f'$.