Let $f\in C([0,\infty))$. Define
$$K(x,t)=2\pi\sum_{n=1}^\infty n\sin(n\pi x)e^{-n^2\pi^2t},\qquad x\in[0,1],\ t>0.$$
Show that
$$\lim_{x\to0^+}\int_0^tK(x,t-\tau)f(\tau)\,d\tau=f(t), \qquad t>0.$$
I can show the result by assuming that
$$\lim_{x\to0^+}\int_0^tK(x,t-\tau)\,d\tau=1,\tag{1}$$
and
$$\int_0^1 |K(x,t)|\,dt<\infty \text{ for all } x \text{ close to }0.\tag{2}$$
For any $\varepsilon>0$, we can find $\delta>0$ so that $|f(\tau)-f(t)|<\varepsilon$ for all $\tau\in(t-\delta,\delta)$. In $[\delta,t)$, the series defining $K$ is convergent absolutely and we can change the order of limitation and integration to get
$$\lim_{x\to0^+}\int_0^{t-\delta}K(x,t-\tau)f(\tau)\,d\tau=0.$$
As for the integration in $(t-\delta,t)$, we have
$$\int_{t-\delta}^t |K(x,t-\tau)||f(\tau)-f(t)|\,d\tau\leq \varepsilon\int_{t-\delta}^t |K(x,t-\tau)|\,d\tau\lesssim \varepsilon,$$
by $(2)$. And now the result follows from $(1)$.
I have no idea how to show $(1)$ and $(2)$. It seems that we need to carefully use the definition of $K$. For example, if we simply apply Fubini to $(2)$, we will get a divergent integral.
Any help would be appreciated!
Best Answer
The alternative representation $$K(x,t)=\frac1{t\sqrt{\pi t}}\sum_{n=-\infty}^\infty\left(n+\frac x2\right)\exp\left[-\frac1t\left(n+\frac x2\right)^2\right]\tag{$\ast$}\label{altrep}$$ makes all the claims easy to obtain. Say, for $0<x<2$ we get $$\newcommand{\erfc}{\operatorname{erfc}}\int_0^t K(x,\tau)\,d\tau=\erfc\frac{x}{2\sqrt{t}}+\sum_{n=1}^\infty\left[\erfc\frac1{\sqrt{t}}\left(n+\frac x2\right)-\erfc\frac1{\sqrt{t}}\left(n-\frac x2\right)\right].$$ This tends to $1$ as $x\to 0^+$, which shows $(1)$. If we replace $[\ldots-\ldots]$ with $[\ldots+\ldots]$ above, we get an upper bound for $\int_0^t|K(x,\tau)|\,d\tau$, sufficient to see why $(2)$ holds.
To prove \eqref{altrep}, apply the Poisson summation formula $$\sum_{n=-\infty}^\infty g(n)=\sum_{n=-\infty}^\infty\hat{g}(n),\qquad\hat{g}(\eta)=\int_{-\infty}^\infty g(\xi)e^{-2\pi i\eta\xi}\,d\xi$$ to $g(\xi)=\pi\xi\sin x\pi\xi\ e^{-(\pi\xi)^2t}$, and obtain $\hat{g}(\eta)$ using known integrals $$\int_{-\infty}^\infty e^{-a^2 x^2}\cos 2bx\,dx=\frac{\sqrt\pi}{a}e^{-(b/a)^2}\underset{\partial/\partial b}{\implies}\int_{-\infty}^\infty xe^{-a^2 x^2}\sin 2bx\,dx=\frac{b\sqrt\pi}{a^3}e^{-(b/a)^2}.$$