Proposition 2.5.30CB24AD25DC3

Proof. (1) Fix $x\in \Omega _ s$ and $\delta >0$ with $\overline{B(x,\delta )}\subset \Omega _ s$. For $|x'-x|\leq \delta $, the test functions $\phi (x'-\cdot )$ are supported in the compact set $K=\overline{B(x,\delta +s)}$, which lies in $\Omega $ because $\overline{B(x',s)}\subset \Omega $ for every such $x'$. Taylor's formula, applied to all derivatives of $\phi $, shows that the difference quotients $t^{-1}(\phi (x+te_ j-\cdot )-\phi (x-\cdot ))$, where $e_ j$ is a coordinate unit vector, converge to $(\partial _ j\phi )(x-\cdot )$ in every seminorm $p_{K,m}$ as $t\to 0$; similarly, $\phi (x'-\cdot )\to \phi (x-\cdot )$ in every $p_{K,m}$ as $x'\to x$. By the estimate of definition 2.4.1 on $\mathcal D_ K(\Omega )$, $T*\phi $ is continuous at $x$ and $\partial _ j(T*\phi )(x)=\langle T,(\partial _ j\phi )(x-\cdot )\rangle =(T*\partial _ j\phi )(x)$. Applying the same argument to the derivatives of $\phi $, which are also supported in $\overline{B(0,s)}$, we find by induction that $T*\phi $ is smooth and $\partial ^\alpha (T*\phi )=T*\partial ^\alpha \phi $.

Since $\partial _ y^\alpha [\phi (x-y)]=(-1)^{|\alpha |}(\partial ^\alpha \phi )(x-y)$, we have $((\partial ^\alpha T)*\phi )(x)=(-1)^{|\alpha |}\langle T,\partial _ y^\alpha [\phi (x-y)]\rangle =(T*\partial ^\alpha \phi )(x)$. The case $\phi =\rho _\varepsilon $, $s=\varepsilon $, gives the assertions about $T_\varepsilon $.

(2) Let $\chi $ be a test function with $\operatorname{Supp}\chi \subset \Omega _\varepsilon $. We first prove that

Since $\rho $ is even, $(\chi *\rho _\varepsilon )(y)=\int \chi (x)\rho _\varepsilon (x-y)\, \mathrm dx$, and $\chi *\rho _\varepsilon $ is supported in the compact set $K\subset \Omega $ of points at distance at most $\varepsilon $ from $\operatorname{Supp}\chi $. Divide a closed cube containing $\operatorname{Supp}\chi $ into $N^ d$ congruent subcubes $Q_ l$ with centers $c_ l$, and put $S_ N(y)=\sum _ l\chi (c_ l)\rho _\varepsilon (c_ l-y)|Q_ l|$. Only centers $c_ l\in \operatorname{Supp}\chi $ contribute, so $S_ N\in \mathcal D_ K(\Omega )$ and $\langle T,S_ N\rangle =\sum _ l\chi (c_ l)T_\varepsilon (c_ l)|Q_ l|$, a Riemann sum for the continuous function $T_\varepsilon \chi $, extended by zero; it tends to $\int T_\varepsilon \chi \, \mathrm dx$.

On the other hand, each derivative $\partial ^\beta S_ N(y)$ is a Riemann sum for the integral over the cube of $x\mapsto \chi (x)\partial ^\beta _ y[\rho _\varepsilon (x-y)]$, and this integral is $\partial ^\beta (\chi *\rho _\varepsilon )(y)$ by differentiation under the integral sign. The integrand is uniformly continuous in $(x,y)$, so the difference between the Riemann sum and the integral is at most the volume of the cube times the largest oscillation of the integrand in $x$ over a subcube, which tends to zero uniformly in $y$. Hence $S_ N\to \chi *\rho _\varepsilon $ in every seminorm $p_{K,m}$, and the estimate for $T$ on $\mathcal D_ K(\Omega )$ gives $\langle T,S_ N\rangle \to \langle T,\chi *\rho _\varepsilon \rangle $. This proves (2.5.4). By lemma 2.5.1(3), the functions $\chi *\rho _\varepsilon $ are supported in a fixed compact set for small $\varepsilon $ and converge to $\chi $ with all derivatives uniformly, so $\langle T,\chi *\rho _\varepsilon \rangle \to \langle T,\chi \rangle $.

(3) If $T$ is positive, then $T_\varepsilon (x)\geq 0$ because $\rho _\varepsilon (x-\cdot )$ is a nonnegative test function.

(4) If $u\in L^1_{\mathrm{loc}}(\Omega )$ defines the zero distribution, then $u_\varepsilon =T_\varepsilon =0$ for every $\varepsilon $, and $u_\varepsilon \to u$ in $L^1(K)$ for every compact $K$ by lemma 2.5.1(2); hence $u=0$ almost everywhere. If two Radon measures $\mu $ and $\nu $ define the same distribution, then $\int f\, \mathrm d\mu =\int f\, \mathrm d\nu $ for every $f\in C_ c(\Omega )$, because $f$ is the uniform limit of test functions supported in a fixed compact set (lemma 2.5.1(5)). The uniqueness in theorem 2.3.3 gives $\mu =\nu $. $\square$