$\langle u,\phi\rangle=0$ when ${\rm supp}(u)\cap{\rm supp}(\phi)^\circ=\emptyset$?

Question

I know that $\langle u,\phi\rangle=0$ if ${\rm supp}(u)\cap{\rm supp}(\phi)=\emptyset$. For some while I wondered whether it's enough that $\phi$ vanishes on ${\rm supp}(u)$ but that's not true, as one can see with $u=\delta'$ and $\phi=x$ around $0$. But now I asked myself, is it maybe enough to assume ${\rm supp}(u)\cap{\rm supp}(\phi)^\circ=\emptyset$?

Answer 1

Let $\Omega\subseteq{\bf R}^n$ and $\phi\in{\cal D}(\Omega)$ given such that ${\rm supp}(u)\cap{\rm supp}(\phi)^\circ=\emptyset$, I shall construct a sequence $\phi_k$ of test functions with ${\rm supp}(u)\cap{\rm supp}(\phi_k)=\emptyset$ converging to $\phi$. Let $X_k:=\{x\in{\bf R}^n:d(x,{\rm supp}(u)\ge 2/k\}$ and $f_k$ the characteristic function of $X_k$. Further let $j\in{\cal D}(B_1(0))$ be a bump function, i.e. $\int j=1$, and $j_k:=x\mapsto k^n j(kx)$ as well as $\phi_k:=\phi\cdot f_k*j_k$. This is in fact a test function on $\Omega$ with ${\rm supp}(\phi_k)\subseteq{\rm supp}(\phi)$ and ${\rm supp}(\phi_k)\cap{\rm supp}(u)=\emptyset$.

Assume now $\partial^\beta\phi_k$ does not converge uniformly to $\partial^\beta\phi$, because around any point $x$ with $d(x,{\rm supp}(u))>4/k$ we have that $\phi_k$ equals $\phi$, there must be a sequence $(x_k)_k$ with $d(x_k,{\rm supp}(u))\le 4/k$ such that $|\partial^\beta(\phi_k-\phi)(x_k)|$ does not converge. Because $x_k$ lives in the compact set ${\rm supp}(\phi)$, there is some subsequence $(x_{l(k)})_k$ converging to a point $x$ on the boundary of ${\rm supp}(\phi)$ with $|\partial^\beta(\phi_{l(k)}-\phi)(x_{l(k)})|\ge\epsilon$, and since $\partial^\beta\phi(x_{l(k)})$ is converging to $0$, it suffices to show that $|\partial^\beta\phi_{l(k)}(x_{l(k)})|$ is converging to $0$ as well to derive a contradiction.

With $C_\alpha:=\|\partial^\alpha j\|_\infty$ we have $\|\partial^\alpha(f_k*j_k)\|_\infty=\|f_k*(\partial^\alpha j_k)\|_\infty\le k^{|\alpha|}C_\alpha$. So $|\partial^\beta\phi_{l(k)}(x_{l(k)})|\le\sum_{\alpha\le\beta}{\beta\choose\alpha}|\partial^\alpha \phi(x_{l(k)})|l(k)^{|\beta-\alpha|}C_{\beta-\alpha}$ and it suffices to show that $l(k)^{\beta-\alpha}\partial^\alpha\phi_{l(k)}(x_{l(k)})$ converges to $0$ for any $\alpha\le\beta$. But expanding $\partial^\alpha\phi_{l(k)}(x_{l(k)})$ in a Taylor series around $x$, we see immedeately that this must be true.