# Proving uniform continuity via Lagrange's theorem (mean value theorem)

Let $f \colon \mathbb{R} \to \mathbb{R}$ be differentiable on an interval $I$. Prove: $$f \prime \, \text{is bounded on I} \Rightarrow f \, \text{is uniformly continuous on I}$$

I'm supposed to solve this using Lagrange's theorem.

My attempt:

Recall the definition of uniform continuity: $$\forall \varepsilon>0 \exists \delta>0 \forall x,y \in I \colon(|x-y|< \delta \Rightarrow |f(x)-f(y)|<\varepsilon)$$ Now suppose $f \prime$ is bounded on $I$. Then: $$\exists x_1 \in I \colon \forall x_2 \in I \colon |f \prime(x_1)| \geq |f \prime (x_2)|$$ Let $x,y \in I$ be arbitrary. By Lagrange's theorem: $$\exists \xi \in (x,y):|f(x)-f(y)|=|f \prime(\xi)||x-y| \leq |f \prime(x_1)||x-y|< \varepsilon$$ So let's recap: $$|x-y|< \frac{\varepsilon}{|f \prime (x_1)|} \Rightarrow |f(x)-f(y)|< \varepsilon$$ So we have found $\delta$ and we use $\delta:= \frac{\varepsilon}{|f \prime (x_1)|}$

This concludes the proof. Is this valid?

Yes, your proof is entirely correct (except you write "uniform convergence" where you mean "uniform continuity"). Well, to be extremely picky I suppose you should explain what happens when $f'(x_1) = 0$, since then you can't perform the division, but that is a trivial case.
Note that you are actually showing that $f$ is Lipschitz: we say $f: I \rightarrow \mathbb{R}$ is Lipschitz if there is a constant $C$ such that for all $x_1,x_2 \in I$, $|f(x_1)-f(x_2)| \leq C|x_1-x_2|$.