Why does the mollified function converge uniformly to the original $W^{1,\infty}$ function

783 Views Asked by At

I'm studying Evans' PDE book and on page 294 it states that if we have $u \in W^{1, \infty } (\mathbb R^n)$ and has compact support, and take $\eta _\epsilon$ as the standard mollifier, then $\eta _\epsilon \ast u$ converges to $u$ uniformly as $\epsilon \to 0$.

I have trouble understanding this since I only know this is true if $u$ is continuous. And I tried to apply the proof for $u \in W^{1, \infty } (\mathbb R^n)$ but failed.

1

There are 1 best solutions below

1
On

Continuity of $u$ follows from Morrey's inequality which was proved a few pages earlier in the book: a function in $W^{1,p}$, $p>n$, has a continuous representative, with which it is identified.


For completeness, a proof of uniform convergence. Continuity and compact support imply uniform continuity. So, for any fixed $\epsilon>0$ there is $\delta$ such that if $r<\delta$, all the values of $u$ involved in the integral $\eta_r* u(x)$ are within $\epsilon$ of $u(x)$. Since mollification averages these values, $|\eta_r* u(x)-u(x)|\le \epsilon$ as desired.