My problem is
$u_n(x)=ne^{inx}$ converges to zero in $\mathcal{D}'(\mathbb{R})$?
Let $\varphi \in C^{\infty}_{c}(\mathbb{R})$, \begin{align*} \langle u_n, \varphi \rangle&=\int_{\mbox{supp}\, \varphi} u_n (x) \varphi(x)\, dx\\ &=\int e^{iy} \varphi(\frac{y}{n})\, dy\\ & \rightarrow \int e^{iy}\, \varphi(0)\, dy\\ &=\int e^{iy}\, dy \, \langle \delta, \varphi \rangle. \end{align*}
$\int e^{iy}\, dy=0 \quad \mbox{or} \quad \ne 0$? I am confused about it.
Thank you.
$$f_n(x)=\frac{e^{inx}}{-n}$$ It is obvious $f_n\to 0$ in the sense of distributions thus so does $f_n''(x)=n e^{inx}$
(by definition of the distributional derivative as $<f_n'',\varphi>= <f_n,\varphi''>$)
This is indeed the point of the distribution topology : making the linear map $u \mapsto u'$ continuous.