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This question already has an answer here:

Here are some definitions that was taken of PDE Evans book:

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Here is a proof of a property of mollifiers:

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My (elementary) question is: Why is the convergence uniform on $V$?

Thanks.

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marked as duplicate by Jack, Tom Cooney, user91500, Brevan Ellefsen, Claude Leibovici May 17 '16 at 6:54

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Look at this - https://mathoverflow.net/questions/30664/uniform-convergence-of-difference-quotient . It proves that the fact the function is $C^\infty$ (just $C^2$ would work) and has compact support implies the difference quotient converges uniformly to the derivative.

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  • $\begingroup$ Because $r_p$ as you've labelled it depends on $x$, so that limit as $t\rightarrow 0$ may not be uniform in $x$. $C^2$ comes in as it allows the difference quotient to be bounded by $ |t| * || \frac{ \partial ^2 f } { \partial x_i ^2}||_\infty$ . $\endgroup$ – Matt Rigby Sep 22 '13 at 22:12
  • $\begingroup$ My original comment (I deleted it before see your answer): Why just $C^1$ is not enough? By Taylor's theorem, if $f\in C^1$ then we can write $$f(x+te_i)=f(x)+t\frac{\partial f}{\partial x_i}(x)+r_p(te_i),$$ where $$\lim_{t\to0}\frac{r_p(te_i)}{t}=0.$$ Thus $$\frac{f(x+te_i)-f(x)}{t}=\frac{\partial f}{\partial x_i}(x)+\frac{r_p(te_i)}{t}$$ so that $$\left| \frac{f(x+te_i)-f(x)}{t}-\frac{\partial f}{\partial x_i}(x)\right|=\left|\frac{r_p(te_i)}{t}\right|$$ What's wrong in this argumment? $\endgroup$ – Pedro Sep 22 '13 at 22:13
  • $\begingroup$ Matt, in your opinion is this proof correct? $\endgroup$ – Pedro Sep 25 '13 at 15:13

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