This question is the same one as the question found here:
Uniform convergence in a proof of a property of mollifiers in Evans's Partial Differential Equations
However, the answers given just confuse me more.
The question asked in that thread is why is the convergence uniform on $V$?
The top answer refers the reader to this article: https://mathoverflow.net/questions/30664/uniform-convergence-of-difference-quotient
However – there is a big difference between the question at hand and the article used as an example. Namely that Evans goes to the effort of introducing a new set $V$ so he can get uniform convergence on that set. In the proof from the mathoverflow, the uniform convergence is on the whole space, if this result were true in the mollifier case, then wouldn't it make the change of region of integration redundant?
So the question is: does a smooth compactly supported function's difference quotient converge uniformly to its derivative on a compact set? If so, why?
Best Answer
The answer is yes. Let $f\colon\mathbb{R}\to\mathbb{R}$ be a $C^1$ function with $f'$ uniformly continuous (in particular, $f$ could be any smooth function with compact support.) For any $x,h\in\mathbb{R}$ we have $$ f'(x)-\frac{f(x+h)-f(x)}{h}=\frac{1}{h}\int_x^{x+h}(f'(x)-f'(t))\,dt. $$ Since $f'$ is uniformly continuous, given $\epsilon>0$ there exists $\delta>0$ such that $$ |x-t|<\delta\implies |f'(x)-f'(t)|<\epsilon. $$ If $|h|<\delta$, then $$ \Bigl|f'(x)-\frac{f(x+h)-f(x)}{h}\Bigr|\le\frac{1}{|h|}\int_x^{x+h}|f'(x)-f'(t)|\,dt\le\epsilon. $$