Regularity of an infinite series arising with the heat equation Let $(t,y)\in(0,\infty)\times\mathbf{R}$, and $\displaystyle f(t,y) \equiv \sum_{k=-\infty}^{\infty}\frac{\exp(-(y-2\pi k)^2/2t)}{\sqrt{2\pi t}}$. This infinite series arises if one attempts to solve the one-dimensional heat equation on $S^1$. I wish to establish the following: $f$ is smooth. I guess one way (and probably the easiest) to do this is, is to show that the series converges uniformly for fixed $(t,y)$ and then to apply Morera's theorem. I wish to apply the Weierstrass M-test, but I did not suceed in finding a series to compare with. Any ideas? Thanks in advance
 A: Let's make $t,y$ complex from the beginning. Specifically, let's shoot for the analyticity on the set $\Omega =\{(t,y)\in\mathbb C^2:  \operatorname{Re}t>0, \ y\in\mathbb C\}$. This open set is the union of compact sets of the form $K=\{(t,y)\in\mathbb C^2: a^{-1}\le \operatorname{Re}t \le |t|\le a, \ |y|\le a\}$ where $a>0$ is positive. For large values of $|k|$, we must estimate the supremum of the modulus of 
$$f_k(t,y)=\frac{\exp(-(y-2\pi k)^2/(2t))}{\sqrt{2\pi t}} $$
on $K$. The modulus of the denominator is bounded from below by $a\sqrt{2\pi} $. The modulus of exponential comes from the real part of its argument. We must show that 
$$\operatorname{Re}\frac{(y-2\pi k)^2}{2t} $$
is positive and large. Rewrite it as 
$$\operatorname{Re}\frac{4\pi^2k^2-4\pi ky+y^2}{2t} = \frac{2\pi^2 \operatorname{Re} t }{|t|^2}k^2 +\operatorname{Re}\frac{4\pi ky+y^2}{2t}\tag{1}$$
where the quadratic term is positive:
$$\frac{2\pi^2 \operatorname{Re} t }{|t|^2}k^2\ge 2\pi^2 a^{-3}k^2 \tag{2}$$ 
and the remainder (which is $O(k)$) can be estimated roughly: 
$$\operatorname{Re}\frac{4\pi ky+y^2}{2t}\ge -\frac{4\pi |k||y|+|y|^2}{2|t|} \ge  -\frac{a}{2}(4\pi a |k|+a^2)\tag{3}$$
Putting (2) and (3) into (1), we get
$$\operatorname{Re}\frac{(y-2\pi k)^2}{2t} \ge 2\pi^2 a^{-3}k^2 -\frac{a}{2}(4\pi a |k|+a^2) \tag{4}$$
To finish, let $M$ be large enough so that $\pi^2 a^{-3}k^2 \ge \frac{a}{2}(4\pi a |k|+a^2)$ whenever $|k|\ge M$. Then for $|k|\ge M$ we have 
$$\operatorname{Re}\frac{(y-2\pi k)^2}{2t} \ge \pi^2 a^{-3}k^2$$
and therefore
$$|f_k(t,y)|\le 2\sqrt{2\pi}\exp\left(-\pi^2 a^{-3}k^2\right)$$
A: Is the first approximationin 5PM's post correct for all a?
We wish to find an upper bound for $\large |\frac{1}{\sqrt{2\pi t}}| = \large\frac{1}{\sqrt{2\pi}\sqrt{Re(t)}} ≤ \large \frac{\sqrt{a}}{\sqrt{2\pi}} $
Somehow (I don't see why this is true for every $a>0$) 5PM is able to limit $\large|\frac{1}{\sqrt{2\pi t}}| ≤ \large\frac{1}{a\sqrt{2\pi}} ≤ 2\sqrt{2\pi}$
