Suppose we have a Schwartz distribution $\phi$ on $\mathbb{R}^d$ such that $$ \forall x, \ \lim\_{\lambda \to 0}| \langle\phi, \psi^{\lambda}\_x \rangle| =0$$ 


where $\psi^{\lambda}\_{x}=\lambda^{-d}{\psi\left(\frac{\cdot - x}{\lambda}\right)}$ approximates a delta in $x$. Here we assume that $\psi$ is in $C\_c^{\infty}$ supported in the unit ball.


Can we conclude that $\phi$ is zero?


This statement is obvious if $\phi$ is a function, but I can't prove it if $\phi$ is a generalized function.


**EDIT:**


I forgot a detail of great importance as shown below: the limit is zero not just for one $\psi,$ but for all $\psi \in C\_c^{\infty}$ with compact support in the unit ball.


**EDIT #2:**


The proof of this fact becomes actually easy in which the limit is zero locally uniformly in $x$, i.e. if for any compact $K \subset \mathbb{R}^d$ we get: $$\lim\_{\lambda \to 0} \sup\_{x \in K}| \langle\phi, \psi^{\lambda}\_x \rangle| =0$$


In this case we just observe that we actually are looking at a convolution: $\langle\phi, \psi^{\lambda}\_x \rangle = \phi \* \psi^{\lambda}(- \cdot) \ (x)$. Now observing that $ h\_{\lambda} = \psi^{\lambda}(- \cdot)$ still approximates a delta, we can use a famous result stating that $\phi \* h\_{\lambda} \to \phi$ in the sense of distributions.


Since our uniform limit estimate actually tells us that we are converging to zero uniformly on all compacts, we get that $\phi = 0.$



Distribution that vanishes against approximated delta is zero
I think that the answer is yes. Assume by contradiction that the support $S$ of $\phi$ is nonempty. Since $S$ is closed, it is in particular a complete metric space.


Let $X:=\{\psi\in C^\infty(\mathbb{R}^d):\psi\equiv 0\text{ on }\mathbb{R}^d\setminus B\_1\}$, which is a Fréchet space.
Apply now, for any fixed $x\in S$, the uniform boundedness principle to the family of functionals
$$X\to\mathbb{R},\qquad\{\psi\mapsto\langle\phi, \psi^{\lambda}\_x \rangle\mid 0<\lambda<1\},$$ obtaining that for some minimal $k(x)\in\{1,2,\dots\}$ it holds
$$ |\langle\phi,\psi\_x^\lambda\rangle|\le k(x)\|\psi\|\_{C^{k(x)}} $$
(for any $0<\lambda<1$). The sets $S\_N:=\{x\in S:k(x)\le N\}$ are closed and cover $S$, so one of them has nonempty interior: say that $B\_r(x)\cap S\subseteq S\_N$ for some $x\in S$.


Now, for any positive integer $M$ large enough, we can easily find a collection $(\rho\_{M,j})\_j\subseteq C^\infty\_c(B\_r(x))$, with cardinality $O(2^{Md})$, of functions whose supports have diameter less than $2^{-M}$ and satisfying $\|\rho\_{M,j}(2^{-M}\cdot)\|\_{C^N}=O(1)$, as well as $\sum\_j\rho\_{M,j}\equiv 1$ on $B\_{r/2}(x)$ (say).


Thus, for any $\psi\in C^{\infty}\_c(B\_{r/2}(x))$,
$$ \langle\phi,\psi\rangle=\sum\_j\langle\phi,\rho\_{M,j}\psi\rangle $$
and we can discard all the terms where $\rho\_{M,j}$ has support disjoint from $S$. As for the other terms,
let $y\_j\in S\cap\text{supp }\rho\_{M,j}$; notice that
$$\langle\phi,\rho\_{M,j}\psi\rangle=2^{-Md}\langle\phi,\eta\_{y\_j}^{2^{-M}}\rangle$$
where $\eta:=\rho\_{M,j}(y\_j+2^{-M}\cdot)\psi(y\_j+2^{-M}\cdot)$. Thus,
$$|\langle\phi,\rho\_{M,j}\psi\rangle|\le 2^{-Md}N\|\rho\_{M,j}(y\_j+2^{-M}\cdot)\psi(y\_j+2^{-M}\cdot)\|\_{C^N}$$
$$\le CN 2^{-Md}\|\psi(y\_j+2^{-M}\cdot)\|\_{C^N}. $$
Summing in $j$ and letting $M\to\infty$, this easily leads to the bound
$$ |\langle\phi,\psi\rangle|=O(\|\psi\|\_\infty), $$
showing that $\phi$ restricts to a measure on $B\_{r/2}(x)$. The hypothesis, plus well-known differentiation theorems, imply that $\phi\equiv 0$ here (see below), contradicting the fact that $x\in S$.


**Addendum.** Proof of the fact that $\phi$ vanishes on $B\_r(x)$:
Let us restrict our attention to $B\_r(x)$. We split the measure $\phi=\phi^+-\phi^-$ into positive and negative part and we set $\mu:=\phi^-+\mathcal{L}^d$, $\nu:=\phi^+$. Assume that some $y\in B\_r(x)$ satisfies
$$\lim\_{\rho\to 0}\frac{\nu(B\_\rho(y))}{\mu(B\_\rho(y))}=+\infty.$$
Pick any radial $\psi\in C^\infty\_c$
with $\psi\ge 0$ and $\int\_{\mathbb{R}^d}\psi\,d\mathcal{L}^n=1$. Notice that
$$\frac{\int\psi\_y^\lambda\,d\nu}{\int\psi\_y^\lambda\,d\mu}=\frac{\int\_0^\infty\nu(\{\psi\_y^\lambda>t\})\,dt}{\int\_0^\infty\mu(\{\psi\_y^\lambda>t\})\,dt}\to+\infty$$ as $\lambda\to 0$ (since the superlevels are smaller and smaller balls). Thus,
$$\frac{\int\psi\_y^\lambda\,d\phi}{\int\psi\_y^\lambda\,d\mu}\ge\frac{\int\psi\_y^\lambda\,d\nu}{\int\psi\_y^\lambda\,d\mu}-1\to+\infty,$$
but $\int\psi\_y^\lambda\,d\mu\ge\int\psi\_y^\lambda\,d\mathcal{L}^d=1$, contradicting the hypothesis. So such $y$ does not exist and by Theorem 2.22 (Besicovitch derivation theorem) in *Ambrosio, Fusco, Pallara, Functions of Bounded Variation and Free Discontinuity Problems* we have $\nu\ll\mu$.


Since $\phi^+$ and $\phi^-$ are mutually singular, we deduce $\phi^+\ll\mathcal{L}^d$.
Repeating all the arguments with $-\phi$ instead of $\phi$, we get $\phi^-\ll\mathcal{L}^d$ as well.
So $\phi$ has a density with respect to $\mathcal{L}^d$. For all $y$ such that $\lim\_{\rho\to 0}\frac{\phi(B\_\rho(y))}{\mathcal{L}^d(B\_\rho(y))}$ exists finite, the ratio $\frac{\int\psi\_y^\lambda\,d\phi}{\int\psi\_y^\lambda\,d\mathcal{L}^d}$ converges to the same limit. Applying the aforementioned theorem and the hypothesis, we obtain that the density is zero.


