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.


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 $\phi = 0.$

  • $\begingroup$ You are aware that the with the edit the $ψ^λ_x$ form a generating set of $C^∞_c$? $\endgroup$ Feb 19, 2017 at 19:42
  • $\begingroup$ Yes (if by generating you mean that the span is dense), but I only know that the limit is zero, not that $\langle \phi, \psi^{\lambda}_x \rangle = 0.$ So I don't really know how to use this fact. Of course if the span where not dense it would directly follow that my statement is false. $\endgroup$
    – Kore-N
    Feb 19, 2017 at 22:17

2 Answers 2


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.

  • $\begingroup$ Thank you very much for the answer. First the obvious question: what differentiation theorem (all the rest I could follow)? Second: am I wrong in understanding that what you prove is that for every distribution we can find a set such that the distribution is actually a measure if restricted to that set, just for the sake of my understanding? Third: the first part of the proof is very pretty, so, no offense, I wonder whether you found it in a book: I'd like to know more about the subject. $\endgroup$
    – Kore-N
    Feb 19, 2017 at 22:15
  • 1
    $\begingroup$ (1) I will add some explanation on the last part. (2) I prove this fact assuming that the limit in your question exists and is finite (but yes, we do not need that it is zero to reach this intermediate conclusion), otherwise the uniform boundedness principle cannot be applied. (3) Sorry, I found the problem interesting and made this up by myself (but I remember having seen proofs in topology where Baire's theorem is applied to a carefully chosen set, as here $S$ instead of the naive $\mathbb{R}^d$). $\endgroup$
    – Mizar
    Feb 19, 2017 at 23:16
  • $\begingroup$ Thank you for the addendum. This proof is really pretty and unexpectedly deep :) $\endgroup$
    – Kore-N
    Feb 20, 2017 at 6:42

No. Take $\phi=\delta'$ and $\psi$ even. Then your condition holds for $x=0$ because $\psi'(0)=0$, and for $x\neq 0$ due to the compact support of $\psi$.

  • $\begingroup$ Thank you for the answer. This example is very nice :) So I am all the more sorry that I forgot to write the (with hindsight crucial) detail that the condition holds for all $\psi$ satisfying some assumptions. $\endgroup$
    – Kore-N
    Feb 19, 2017 at 11:35

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