Lemma 2.5.13C40E6BD1222

Proof. (1) Since $\int \rho _\varepsilon =1$, we have for $x\in \Omega _\varepsilon $

If $K\subset \Omega $ is compact, choose $\delta >0$ with $K'=\{ x:\operatorname {dist}(x,K)\leq \delta \} \subset \Omega $. For $\varepsilon <\delta $ we have $K\subset \Omega _\varepsilon $, because the closed ball of radius $\delta $ about each point of $K$ lies in $\Omega $, and the bound tends to zero uniformly on $K$ by the uniform continuity of $u$ on the compact set $K'$.

(2) We first prove the continuity of translation. For $v\in C_ c(\mathbb R^ d)$ it follows from uniform continuity, since the functions $v(\cdot -h)$, $|h|\leq 1$, are supported in a fixed compact set; the general case follows by approximating $v$ in $L^ p$ by such functions (proposition 2.3.5 for Lebesgue measure), since translation preserves the $L^ p$ norm. Since $\int \rho _\varepsilon =1$, we have $v*\rho _\varepsilon (x)-v(x)=\int (v(x-h)-v(x))\rho _\varepsilon (h)\, \mathrm dh$. Hence, by the triangle inequality and Jensen's inequality (lemma 2.2.4) for the convex function $t\mapsto |t|^ p$ on $\mathbb R$, applied to the function $h\mapsto |v(x-h)-v(x)|$, which is integrable for the probability measure $\rho _\varepsilon (h)\, \mathrm dh$ because $|v|\leq 1+|v|^ p$, we obtain

We may take $v$ to be a Borel function, so that the integrand is a Borel function of $(x,h)$. Integrating in $x$ and using Tonelli's theorem, we obtain

Now let $u\in L^ p_{\mathrm{loc}}(\Omega )$ and $K\subset \Omega $ compact, choose $\delta >0$ with $\{ x:\operatorname {dist}(x,K)\leq 2\delta \} \subset \Omega $, and let $v\in L^ p(\mathbb R^ d)$ equal $u$ on $W=\{ x:\operatorname {dist}(x,K)<2\delta \} $ and $0$ elsewhere. For $\varepsilon <\delta $, the values of $u_\varepsilon $ on $K$ only involve $u$ on $W$, so $\| u_\varepsilon -u\| _{L^ p(K)}\leq \| v*\rho _\varepsilon -v\| _ p\to 0$.

(3) The function $\chi *\rho _\varepsilon $ vanishes outside the set of points at distance at most $\varepsilon $ from $\operatorname{Supp}\chi $, which is a compact subset of $\Omega $ for small $\varepsilon $. Differentiation under the integral sign gives $\partial ^\alpha (\chi *\rho _\varepsilon )=(\partial ^\alpha \chi )*\rho _\varepsilon $, and the bound displayed in (1), applied on $\mathbb R^ d$ to the uniformly continuous function $\partial ^\alpha \chi $ in place of $u$, proves uniform convergence.

(4) Choose $\delta >0$ with $3\delta <\operatorname {dist}(K,\mathbb R^ d\setminus U)$, and let $\eta $ be a cutoff function for $\{ x:\operatorname {dist}(x,K)\leq 2\delta \} $ in $\{ x:\operatorname {dist}(x,K)<3\delta \} $. If $\varepsilon <\delta $ is smaller than the distance from $\operatorname{Supp}\eta $ to $\mathbb R^ d\setminus U$, then $\eta *\rho _\varepsilon $ is smooth with values in $[0,1]$ and compact support in $U$, and it equals $1$ on $\{ x:\operatorname {dist}(x,K)<\delta \} $, where the average only involves points at which $\eta =1$.

(5) The support statement is proved as in (3), the functions $f*\rho _\varepsilon $ are smooth by differentiation under the integral sign, and uniform convergence follows from (1) applied on $\mathbb R^ d$, since all these functions vanish outside a fixed compact set. If $g\in L^ p(\mu )$, choose $f\in C_ c(\Omega )$ close to $g$ in $L^ p(\mu )$ by proposition 2.3.5. If all $f*\rho _\varepsilon $ are supported in the compact set $L$, then so is their uniform limit $f$, and $\| f*\rho _\varepsilon -f\| _{L^ p(\mu )}\leq \sup |f*\rho _\varepsilon -f|\, \mu (L)^{1/p}\to 0$. $\square$