I am still trying to understand Evans' proof on the properties of mollifier. enter image description here

In the proof of (iii), enter image description here

I understand that the crux of the proof is that uniform continuous function on compact set is uniform convergence. However, I do not understand why we need a middle space $W$, isn't that working with $V$ enough? What would happen if do not add such a $W$ in between $V$ and $U$? I was told that, the idea of having $W$ is that, if we can show convergence in $W$, then automatically we can show convergence in a smaller space.

This is intuitive. But I am not convinced mathematically and I wish to see what would happen without $W$ in the proof.


You are using that $f$ is uniform continuous when restricted to each open $W$ so that $W\subset\subset U$.

Then for all $\delta >0$, there is $\epsilon_0 >0$ so that $$|f(x) - f(y)| <\delta$$ whenever $x, y\in W$ and $|x-y| < \epsilon_0$. Now if $x\in V$. Let $\epsilon_0$ be small so that $|y-x|<\epsilon_0 \Rightarrow y\in W$ (it is in general not true that $y\in V$). Then for all $\epsilon <\epsilon_0$,

$$\begin{split} |f(x) - f^\epsilon(x)| &= \left| f(x) - \int_U f(y) \eta_\epsilon (y-x) dy\right| \\ &\le \int_U |f(x) - f(y)| \eta_\epsilon (y-x) dy \\ &\le \delta \end{split}$$

Note that in the second inequality we are using the uniform continuity of $f$ on $W$, not on $V$. As $x\in V$ is arbitrary, $f^\epsilon \to f$ uniformly on $V$.

Remark: The compactness of $V$ is used to guarantee the existence of such a $W$. Indeed, let $d>0$ be such that $d(v, \partial U) > d$ for all $v\in V$ (This $d$ can be found as $V$ is compact. Then let $W = V_{d/2} = \{ x\in U : d(x, V) < d/2\}$. Then $W$ is open and $\overline W$ is compact and is in $U$.

  • $\begingroup$ Im reading it bit by bit. First, "Let $\epsilon_0$ be small so that $|y-x|<\epsilon_0 \Rightarrow y\in W$ (it is in general not true that $y\in V$)." Why is not true $y\in V$? $\endgroup$ – math101 Nov 1 '15 at 4:52
  • $\begingroup$ $x$ might be very close to the boundary of $V$, so that $y\notin V$. @math101 $\endgroup$ – user99914 Nov 1 '15 at 4:53
  • $\begingroup$ ok. my understanding is that, if both $x,y$ are in $V$, then $f$ may not be uniform continuous in some part of $W$ $\endgroup$ – math101 Nov 1 '15 at 5:08
  • $\begingroup$ and in your last remark, what specific theorem is this? Sounds very familiar but I cannot recall it precisely $\endgroup$ – math101 Nov 1 '15 at 5:23
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    $\begingroup$ Just to clarify, $f$ is uniform continuous on $W$. But the uniform limit is found on $V$, i.e. $f^{\epsilon}\to f$ uniformly on $V$, $V$ is that compact subset of $U$ Evans' meant. Right? $\endgroup$ – math101 Nov 1 '15 at 8:05

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