Limit of $f_{n}(x)=\sqrt{\frac{n}{\pi}}e^{-n^2.x^2}$ in $\mathcal{D}'$

Question

Is the limit of $f_{n}(x)=\sqrt{\frac{n}{\pi}}e^{-n^2.x^2}$ in $\mathcal{D}'$ when $n$ tends to infinity equal to zero? This is what I tried: $$\lim_{n\rightarrow \infty }\langle f_n(x),\phi(x)\rangle=\lim_{n\rightarrow \infty }\int f_n(x) \phi(0) dx+\int x f_n(x)\phi'(tx),$$ how to calculate $\int x f_n(x)\phi'(tx)$? Why $t$ is in $]0,1[$?

Answer 1

It more or less follows from an application of the dominated convergence theorem.

Note that $$\lim_{n\rightarrow\infty}\int\limits_{-\infty}^\infty\sqrt{\frac{n}{\pi}}e^{-n^2x^2}\varphi(x)\, dx=\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n\pi}}\int\limits_{-\infty}^{\infty} \varphi\left(\frac{u}{{n}}\right) e^{-u^2}\, du.$$ Since $\varphi\in\mathcal{D},$ it is bounded, say by $M$. If we consider the sequence $$g_n(u)=\frac{1}{\sqrt{n\pi}}\varphi(u/{n})e^{-u^2},$$ then $g_n(u)\rightarrow 0$ pointwise, and $|g_n(u)|\leq \frac{M}{\sqrt{\pi}}e^{-u^2}\in L^1.$ By the dominated convergence theorem, \begin{align*}\lim_{n\rightarrow\infty}\frac{1}{\sqrt{\pi}}\int\limits_{-\infty}^{\infty} \varphi\left(\frac{u}{{n}}\right) e^{-u^2}\, du&=\lim_{n\rightarrow\infty}\int\limits_{-\infty}^\infty g_n(u)\, du=\int\limits_{-\infty}^\infty \lim_{n\rightarrow\infty}g_n(u)\, du\\ &=0. \end{align*} So, it converges in $\mathcal{D}'$ to $0.$