Let $u\in L^1_\text{loc}(\mathbb{R})$ be $T$-periodic. Given $u_n(x)=u(nx)$ for each $n\in\mathbb{N}$, show that $u_n$ converges to a constant $c$ in the sense of the convergence in the distribution space $\mathcal{D}'(\mathbb{R})$ and compute such constant.
So I need to show that exists a constant $c\in\mathbb{R}$ such that for any test function $\varphi\in\mathcal{C}^\infty_c(\mathbb{R})$ $$\langle u_n,\varphi\rangle = \int_{\mathbb{R}}u(nx)\varphi(x)\text{d}x \to \int_{\mathbb{R}}c\varphi(x)\text{d}x = \langle c,\varphi\rangle, \quad \text{for }n\to\infty.$$
Any hints on how to tackle this question?
When $u$ is only assumed to be a $T$-periodic distribution it is a bit trickier than just an integration by parts and $\int_0^x u(y)dy=cx+O(1)$ where $c=\frac1T\int_0^T u(y)dy$.
Say $T=1$. From a convolution with $(1-|x|)1_{|x|< 1}$ construct $\psi \in C^\infty_c$ such that $\sum_k \psi(.+k)=1$.
Then $$\langle u(n.),\phi \rangle= \langle u,\frac1n\phi(\frac{.}n) \rangle=\langle u,\sum_k \psi(.+k)\frac1n\phi(\frac{.}n) \rangle= \langle u,\psi \sum_k \frac1n\phi(\frac{.+k}n) \rangle $$
$ \sum_k \frac1n\phi(\frac{.+k}n)$ is smooth and all its derivatives $\to 0$ uniformly. Whence $\psi \sum_k \frac1n\phi(\frac{.+k}n)\to a \psi$ in the $C_c^\infty$ topology, where $a=\langle 1,\phi\rangle$, and hence $$\langle u,\psi \sum_k \frac1n\phi(\frac{.+k}n) \rangle\to \langle u,a\psi \rangle = c\langle 1,\phi \rangle, \qquad c= \langle u,\psi \rangle$$