# The uniform convergence of the difference quotient of a smooth compactly supported function

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 $V$?

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?
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 $$\Bigl|f'(x)-\frac{f(x+h)-f(x)}{h}\Bigr|\le\frac{1}{|h|}\int_x^{x+h}|f'(x)-f'(t)|\,dt\le\epsilon.$$
• Thank you! I see that works perfectly. I have a few questions: 1) I'm sure it's easy to extend to the multi variable function case? It seems like the analogous argument would work? 2) in the Evans book, he goes to the effort of changing the set of integration to be a set compactly contained in the support of the integrand. Your proof seems to work anywhere on the entire set. If the uniform convergence holds everywhere - why does he go to the effort to say it converges uniformly on $V$? Could you enlighten me on why he changes to this V and exactly how he gets to choose that V? Thanks again! – beedge89 Jan 14 '15 at 20:30
• 1) The proof can be extended to $\mathbb{R}^n$. 2) I guess to ensure that $f'$ is uniformly continuous. – Julián Aguirre Jan 14 '15 at 22:11