convolutions and mollification of functions in $L^1_{\text{loc}}(\Omega)$

Question

The following is from the appendix in Leoni's First Course in Sobolev Spaces:

Here is my question:

Could anybody explain what is the meaning of "$u_\varepsilon$ is well-defined"? What is wrong if $\varepsilon\geq \hbox{dist} (x,\partial\Omega)$?

Answer 1

We have the definition

$$u_{\varepsilon}(x) = \int_{\Omega} \varphi_{\varepsilon}(x-y)\, u(y)\,dy.$$

For $u_{\varepsilon}(x)$ to be well-defined, that integral needs to exist. Since measurability is given, the problem is whether

$$\int_{\Omega} \varphi_{\varepsilon}(x-y)\,\lvert u(y)\rvert\,dy < +\infty.$$

If we required $u \in L^1(\Omega)$, or just $\tilde{u} \in L^1_{\text{loc}}(\mathbb{R}^N)$, where $\tilde{u}$ is the trivial extension of $u$ to $\mathbb{R}^N$, then there would be no problem. But as we only require $u \in L^1_{\text{loc}}(\Omega)$, it can be that $u$ grows too fast near $\partial \Omega$.

If $\varepsilon < \operatorname{dist}(x,\partial \Omega)$, then the integrand is nonzero only on a compact subset of $\Omega$ - namely $\overline{B(x,\varepsilon)}$ - and the boundedness of $\varphi$ together with the definition of $L^1_{\text{loc}}(\Omega)$ ensure the existence of the integral.

If $\varepsilon \geqslant \operatorname{dist}(x,\partial \Omega)$, then $\Omega \cap \overline{B(x,\varepsilon)}$ reaches $\partial \Omega$, and the growth of $\lvert u\rvert$ near $\partial \Omega$ may cause the integral to not exist.