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I remember being assigned the following homework problem a few years back.

Let $f:[0,1] \to \mathbb{R}$ be continuously differentiable. Prove that, for every $\epsilon > 0$, there exists $\delta > 0$ such that $0 < |h| < \delta$ implies $\left| \frac{f(x+h) - f(x)}{h} - f'(x) \right| < \epsilon$ for all appropriate $x$.

I also remember how I solved it.

Fix $\epsilon > 0$. Since $f'$ is a continuous function on a compact interval, it is uniformly continuous and we can find a $\delta > 0$ such that $|x-y| < \delta$ implies $|f'(x) - f'(y)| < \epsilon$ for $x ,y \in [0,1]$. Suppose $0 < |h| < \delta$ and that $x$ is such that $x,x+h \in [0,1]$. By the Mean Value Theorem, there is an $a$ between $x$ and $x+h$ such that $f'(a) = \frac{f(x+h) - f(x)}{h}$. Note $|x - a| < \delta$ clearly holds. So, $\left|\frac{f(x+h) - f(x)}{h} - f'(x) \right| = |f'(a) - f'(x)| < \epsilon$ and we are finished.

What had me baffled was that this: we had not covered the MVT at the time the problem was assigned! This suggests there should be a way to prove it without using the MVT. Can anybody think of a way? I don't think I ever did. Thanks.

Keshav
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Mike F
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  • What's definition of uniformly convergence you are using? The definition that I know uses sequences, but here there is no sequence. – Pedro Sep 24 '13 at 11:54

2 Answers2

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Let $\epsilon>0$ be given. For $\delta>0$ let $$U_\delta = \left\{a\in [0,1]\colon 0<|h|<\delta\Rightarrow \left|\frac{f(a+h)-f(a)}h-f'(a)\right|<\epsilon\right\}$$ Clearly, $\delta<\delta'$ implies $U_{\delta'}\subseteq U_\delta$. By continuity of $f$ and $f'$, $U_\delta$ is open and by definition of $f'$, $$[0,1]=\bigcup _{\delta>0}U_\delta.$$ Since $[0,1]$ is compact, there is a finite subcover, i.e. there is a single $\delta>0$ such that $[0,1]=U_\delta$.

Hagen von Eitzen
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    I'm looking at it for 10 minutes and cannot get why $U_\delta$ is open. It would be so for any fixed $h$, but how it follows when $U_\delta$ is effectively an infinite intersection of open sets (one for each $h$)? – Yakov Galka Dec 09 '12 at 18:15
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    @ybungalobill: You're right, more needs to be said. – Mike F Jul 18 '13 at 16:53
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    In general, $U_{\delta}$ is not open, so this approach breaks down at that point. See this post as to why this is the case. – Dean Miller Jan 03 '24 at 01:12
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Some activity on the post reminded me of this 8 year old question. I guess one can add in that the desired statement follows pretty easily from the fundamental theorem of calculus. Choose $\epsilon>0$. Since $f'$ is uniformly continuous, you can choose $\delta>0$ such that $x \in [0,1]$, $|t|<\delta$ implies $|f'(x)-f'(x+t)|<\epsilon$. Then, if $x \in [0,1]$ and $0<|h|<\delta$ we get \begin{align*} \left|\frac{f(x+h)-f(x)}{h}-f'(x)\right| &= \left|\frac{1}{h} \int_0^h (f'(x+t) - f'(x)) \ dt \right| \\ &\leq \frac{1}{|h|} \int_0^h |f'(x+t) - f'(x)| \ dt \\ &\leq \epsilon. \end{align*} This gives another argument, but isn't really a great answer to the question since I'm pretty sure the fundamental theorem of calculus would not have been covered yet either.

Mike F
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