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Question: I have a sequence $(T_n)$, where $T_n$ is given by the locally integrable function $ne^{-\dfrac{n^{2}x^{2}}{2}}$, converges in $C^{-\infty}(\mathbb{R})$ and compute its limit. I suspect that the limit tends towards zero since the exponential tending towards infinity will become zero. Is this enough to prove this, in conjunction with the definition? I've already written the definition of a locally integrable function and I already understand the definition (sometimes called the 'weak-dual convergence') of the convergence of a sequence of distributions. I've not considered any topologies in these cases, as I don't understand Frechet spaces and the like.

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up vote 2 down vote accepted

Let $\phi\in\mathcal{D}(\mathbb{R})$ (i.e., $\phi$ is a $C^\infty$ function with compact suport). Then $$ \langle T_n,\phi\rangle=\int_\mathbb{R}n\,e^{-n^2x^2}\phi(x)\,dx=\int_\mathbb{R}\,e^{-x^2}\phi\bigl(\frac xn\bigr)\,dx. $$ We have $$ \lim_{n\to\infty}e^{-x^2}\phi\bigl(\frac xn\bigr)=e^{-x^2}\phi(0)\quad\forall x\in\mathbb{R} $$ and $$ \Bigl|e^{-x^2}\phi\bigl(\frac xn\bigr)\Bigr|\le \|\phi\|_\infty e^{-x^2}. $$ The dominated convergence theorem implies that $$ \lim_{n\to\infty}\langle T_n,\phi\rangle=\phi(0)\int_\mathbb{R}\,e^{-x^2}dx=\sqrt\pi\,\phi(0), $$ that is, $T_n$ converges in the distribution sense to Dirac's $\sqrt\pi\,\delta_0$.

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