show that $ \sum \frac{1}{n}e^{-n^2 t}\sin{(nx)}$ converges in $L^2([0,\pi])$ to $ \sum \frac{1}{n}\sin{(nx)}$

Question

I would like to show that

$$ \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty \frac{1}{n}e^{-n^2 t}\sin{(nx)}$$

converges in $L^2([0,\pi])$ to

$$ \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty \frac{1}{n}\sin{(nx)}$$

as $t\to 0$.

Now I have

\begin{align} &\lim_{t\to 0} \int_0^\pi \left| \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty \frac{1}{n}e^{-n^2 t}\sin{(nx)} - \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty \frac{1}{n}\sin{(nx)} \right|^2\\ = &\lim_{t\to 0} \int_0^\pi \left| \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty \frac{1}{n}\sin{(nx)} (e^{-n^2 t} - 1) \right|^2. \end{align}

I am comfortable moving the $\lim_{t\to 0}$ under the sum, since certainly the sum converges uniformly for $t$ near 0. But what would justify my moving it under the integral?

Answer 1

Using Parseval, the problem reduces to showing $$\lim_{t\to 0}\sum_{n=1}^\infty \frac{(1-e^{-n^2t})^2}{n^2}=0$$ which is not hard: the series is dominated by $1/n^2$, so it converges uniformly in $t$.