2.5 Convolution and mollifiersE5DAA0D27299

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 $\| f*\rho _\varepsilon -f\| _{L^ p(\mu )}\leq \sup |f*\rho _\varepsilon -f|\, \mu (L)^{1/p}\to 0$. $\square$

Proof. For every $x\in K$ choose an index $j(x)$ with $x\in U_{j(x)}$ and a radius $r_ x>0$ with $\overline{B(x,r_ x)}\subset U_{j(x)}$. Finitely many of the balls $B(x,r_ x)$ cover $K$; let $L_ j$ be the union of the closures of those among them with $j(x)=j$. Then $L_ j$ is a compact, possibly empty, subset of $U_ j$, and $K\subset L_1\cup \cdots \cup L_ n$. By lemma 2.5.1(4), there are smooth cutoff functions $\psi _ j$ for $L_ j$ in $U_ j$ which equal $1$ on neighborhoods $V_ j$ of $L_ j$. Put $\eta _1=\psi _1$ and $\eta _ j=\psi _ j(1-\psi _1)\cdots (1-\psi _{j-1})$ for $2\leq j\leq n$. These functions are smooth with values in $[0,1]$, and $\operatorname{Supp}\eta _ j\subset \operatorname{Supp}\psi _ j$ is a compact subset of $U_ j$. By induction on $n$,

which takes values in $[0,1]$ and equals $1$ on the neighborhood $V_1\cup \cdots \cup V_ n$ of $K$. If $\chi $ is a test function with $\operatorname{Supp}\chi \subset U_1\cup \cdots \cup U_ n$, we apply this to $K=\operatorname{Supp}\chi $; then $\chi =\sum _ j\chi \eta _ j$, and $\operatorname{Supp}(\chi \eta _ j)\subset \operatorname{Supp}\eta _ j$ is a compact subset of $U_ j$. Finally, let a distribution $T$ vanish on each member of a family of open subsets of $\Omega $, and let $\chi $ be a test function supported in their union. The compact set $\operatorname{Supp}\chi $ is covered by finitely many members $U_1,\ldots ,U_ n$ of the family, and the decomposition $\chi =\sum _ j\chi \eta _ j$ just obtained gives $\langle T,\chi \rangle =\sum _ j\langle T,\chi \eta _ j\rangle =0$. $\square$

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 $. $\square$

Proof. Let $B$ be an open ball with $\overline B\subset \Omega $, and fix $\theta \in C_ c^\infty (B)$ with $\int \theta \, \mathrm dx=1$, for instance a translate of $\rho _ s$ with $s>0$ small. Since $\overline B$ is compact, $\overline B\subset \Omega _\varepsilon $ for all small $\varepsilon $. By proposition 2.5.3(1), the smooth function $T_\varepsilon $ has gradient $((\partial _ jT)_\varepsilon )_ j=0$ on $\Omega _\varepsilon $, so it is equal to a constant $c_\varepsilon $ on the convex set $B$. By proposition 2.5.3(2), $c_\varepsilon =\int T_\varepsilon \theta \, \mathrm dx\to \langle T,\theta \rangle =:c_ B$, and therefore, for every $\chi \in C_ c^\infty (B)$,

If two such balls $B$ and $B'$ intersect, testing with a nonnegative function $\chi \neq 0$ in $C_ c^\infty (B\cap B')$ gives $c_ B=c_{B'}$. Any two such balls containing a point $x\in \Omega $ intersect, so $c(x)=c_ B$, for any such ball $B$ containing $x$, is a well-defined function on $\Omega $; it is locally constant, since $c=c_ B$ on $B$. Every test function $\chi $ has compact support covered by finitely many such balls $B_ l$, and by lemma 2.5.2 it is a sum of test functions $\chi _ l$ supported in $B_ l$. Hence $\langle T,\chi \rangle =\sum _ lc_{B_ l}\int \chi _ l\, \mathrm dx=\int c\chi \, \mathrm dx$. Finally, if $\Omega $ is connected, the locally constant function $c$ is constant, because for each value $a$ the set $\{ c=a\} $ is open and so is its complement. $\square$

Proof. Let $K\subset \Omega $ be compact and let $\eta $ be a smooth cutoff function for $K$ in $\Omega $ (lemma 2.5.1(4)). For a real test function $\chi $ with $\operatorname{Supp}\chi \subset K$, the test functions $\| \chi \| _\infty \eta \pm \chi $ are nonnegative, so positivity gives, as in (2.3.1),

We use this bound to extend $T$ to $C_ c(\Omega ,\mathbb R)$. If $f\in C_ c(\Omega ,\mathbb R)$, the real test functions $f*\rho _\varepsilon $ are supported in a fixed compact set for small $\varepsilon $ and converge uniformly to $f$ (lemma 2.5.1(5)). By the displayed estimate, $\langle T,f*\rho _\varepsilon \rangle $ converges as $\varepsilon \to 0$, and its limit $\Lambda (f)$ is also the limit of $\langle T,\chi _ k\rangle $ for any real test functions $\chi _ k$ supported in a common compact set and converging uniformly to $f$. Approximating two functions of $C_ c(\Omega ,\mathbb R)$ by test functions with a common compact support, we see that $\Lambda $ is linear. If $f\geq 0$, then $f*\rho _\varepsilon \geq 0$, so $\Lambda (f)\geq 0$.

By theorem 2.3.3, there is a Radon measure $\mu $ with $\Lambda (f)=\int f\, \mathrm d\mu $. For a real test function $\chi $, the constant sequence $\chi _ k=\chi $ shows that $\Lambda (\chi )=\langle T,\chi \rangle $; applying this to $\operatorname{Re}\chi $ and $\operatorname{Im}\chi $ gives $\langle T,\chi \rangle =\int \chi \, \mathrm d\mu $ for all test functions. The measure $\mu $ is unique by proposition 2.5.3(4). Conversely, if $\mu $ is a Radon measure, then $|\int \chi \, \mathrm d\mu |\leq (1+\mu (K))p_{K,0}(\chi )$ for $\chi \in \mathcal D_ K(\Omega )$, so integration against $\mu $ is a distribution, and it is positive. $\square$