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. $u\ f(n\cdot) \stackrel{\mathcal{D'}(\mathbb{R}^d)}{\longrightarrow} v$. 
 A: 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 $x\neq 0$. 
A: 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 $\phi$.
