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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?

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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 $x_1,x_2 \in I$, $|f(x_1)-f(x_2)| \leq C|x_1-x_2|$.

It is helpful to think in terms of the stronger conclusion, because Lipschitz functions have many other nice properties. (For instance, any Lipschitz function is absolutely continuous and thus differentiable almost everywhere. These are more advanced concepts and you may not have encountered them yet, but if you take a graduate level course on real analysis, you certainly will.)

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