Uniform convergence of difference quotients to the derivative

Solution 1:

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 $[0,1]=U_\delta$.

Solution 2:

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.