In trying to solve with Fourier series the heat equation on the 1-torus, I stumbled into this family of functions: $$\forall t>0, K_t :\mathbb{T}\to\mathbb{C}, s\mapsto\sum_{n∈\mathbb{Z}}e^{-n^2 t}e^{ins},$$ i.e. a family of functions indexed by $t\in(0,+\infty)$ where the $t$-member is a function with Fourier coefficients given by $(e^{-n^2 t})_{n\in\mathbb{Z}}$. It is clear that $\forall t>0, K_t\in C^\infty(\mathbb{T})$ and so, if $f$ is a distribution on $\mathbb{T}$, the distribution $K_t*f$ is actually represented by an element of $C^\infty(\mathbb{T})$. Also, I've proved that the function: $$u_f :\mathbb{T}\times(0,+\infty)\to\mathbb{C}, (x,t)\mapsto (K_t*f)(x)$$ is an element of $C^\infty\big(\mathbb{T\times(0,+\infty)}\big)$ and that: $$\partial_t u_f = \partial_{xx}u_f.$$ So, I really want the family $(K_t)_{t>0}$ to be a summability kernel for $t\to0^+$, because in this case $u_f$ would be a solution for the heat equation on the 1-torus with initial temperature $f\in L^p(\mathbb{T})$ with $1\le p<\infty$, in the sense that $$\|u_f(\cdot,t)-f\|_p, t\to 0^+.$$ So:
Is $(K_t)_{t>0}$ a summability kernel for $t\to0^+$?
I noticed that the coefficients $e^{-t n^2}\to 1, t\to 0^+$ boundedly (by $1$), so $K_t\to\delta_0, t\to 0^+$ in distribution, and then at least there's some hope. I tried with Poisson summation formula obtaining only a messier expression. Any ideas?
By Poisson summation formula, we have $$ \frac1 {2\pi}K_t(s)=\frac{1}{2\pi}\sum_{n\in \Bbb Z} e^{-n^2 t}e^{ins} = \frac 1 {2\sqrt {\pi t}}\sum_{j\in\Bbb Z}e^{-\frac {(s-2\pi j)^2}{4t}}. $$ Extend $f$ $2\pi$-periodically to $\Bbb R$. We obtain for $x\in\Bbb T=[-\pi,\pi]$ $$\begin{eqnarray} (K_t*f)(x) &=&\frac1{2\pi}\int_{-\pi}^\pi K_t(s)f(x-s)\mathrm{d}s\\ &=&\frac 1 {2\sqrt {\pi t}}\sum_{j\in\Bbb Z}\int_{-\pi}^\pi e^{-\frac {(s-2\pi j)^2}{4t}}f(x-s)\mathrm ds\\&=&\frac 1 {2\sqrt {\pi t}}\sum_{j\in\Bbb Z}\int_{-\pi-2j\pi}^{\pi-2j\pi} e^{-\frac {s^2}{4t}}f(x-s-2j\pi)\mathrm ds\\ &=&\frac 1 {2\sqrt {\pi t}}\sum_{j\in\Bbb Z}\int_{-\pi-2j\pi}^{\pi-2j\pi} e^{-\frac {s^2}{4t}}f(x-s)\mathrm ds\\&=& \frac 1 {2\sqrt {\pi t}}\int_{-\infty}^{\infty} e^{-\frac {s^2}{4t}}f(x-s)\mathrm ds. \end{eqnarray}$$ Note that the heat kernel $p_t(x) = \frac 1 {2\sqrt {\pi t}} e^{-\frac {x^2}{4t}}$ is a summability kernel as $t\rightarrow 0^+$. It follows $$\begin{eqnarray} \|K_t*f -f\|_{L^p(\Bbb T)}&\le& \frac 1 {2\sqrt {\pi t}}\int_{-\infty}^{\infty} e^{-\frac {s^2}{4t}}\|f(\cdot-s)-f(\cdot)\|_{L^p(\Bbb T)}\mathrm ds\\&=& \int_{-\infty}^{\infty} p_t(s)\|\tau_s f-f\|_{L^p(\Bbb T)}\mathrm d s\xrightarrow{t\to 0^+} 0. \end{eqnarray}$$