# limit of a sequence of distributions

Let $u\in \mathcal{D}'(\mathbb{R}^d)$, $f\in C_c^\infty(\mathbb{R}^d)$ and $f(x)=1$ if $|x|\leq 1$; $f(x)=0$ if $|x|>2$. Can we get the following conclusion: there exists $v\in \mathcal{D'}(\mathbb{R}^d)$, such that, for any $\phi\in \mathcal{D}(\mathbb{R}^d)$,

$$\lim_{n\to\infty} \langle u \ f(n\cdot),\phi\rangle = \langle v,\phi\rangle,$$ i.e. $u\ f(n\cdot) \stackrel{\mathcal{D'}(\mathbb{R}^d)}{\longrightarrow} v$.

Here is a functional analytic argument: We set $f_n(x)=f(nx)$ and fix $\phi \in \mathscr D=\mathscr D(\mathbb R)$ with $\phi(0)=1$. Assuming that $\langle uf_n,\phi\rangle =\langle u,f_n\phi\rangle$ is convergent (hence bounded) for every $u\in\mathscr D'$ we get that $\lbrace f_n\phi:n\in\mathbb N\rbrace$ is weakly bounded and hence bounded in $\mathscr D$. As this space (with its usual topology) is a Montel space whose bounded sets are metrizable we find a convergent subsquence $f_{n(k)}\phi \to \psi$ for some $\psi\in\mathscr D$. This contradicts the observation that $f_n\phi(0)=1$ and $f_n\phi(x)\to 0$ for all $x\neq 0$.

Edited. Sorry I misunderstood the question at the first reading.

The answer is no. Take $$u=\sum_k\frac{1}{k^2}\delta'_{3/(2k)}.$$ this is a distribution in $D'$. Now take $f$ so that $f'(3/2)\neq 0$. Then $$(uf_n,\phi)=(u,f_n\phi)=\sum_n\frac{1}{n^2}(f_n\phi)'(3/(2n))$$

$$=\sum_n\frac{1}{n^2}(nf'(3/2)\phi(3/(2n))+f(3/2)\phi'(3/(2n))).$$ This diverges for generic $\phi$.

• I think my statement is correct. $u$ is given at begining and define $u_n=u f_n$, my question is: Does $u_n$ convergence in distribution sense. Mar 27, 2017 at 13:42
• If $u=\delta_0$, then $v$ is $u$ itself so nonzero, or did I miss something? Mar 27, 2017 at 14:47
• The OP is not trying to multiply distributions. Mar 27, 2017 at 16:51
• @Eremenko ok, it works, still you need an inequality. $\langle uf_n, \phi\rangle = \sum_k {n \over k^2} f'(3n/2k) \phi(2/3k) + CV$ and the sum is over the $k$ between $3n/4$ and $3n/2$. So assuming $f'\geq 0$ and $\phi$ constant on $[0, 3/2]$, you get a diverging lower bound. Let me remove my previous comments. Mar 27, 2017 at 21:45
• @Alexander: version 2.0 of the answer is much better :)- Do you know if the OP's question works if the intersection of the singular support of $u$ and a neighborhood of the origin is just the origin? Mar 28, 2017 at 12:44