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

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    $\begingroup$ Welcome to math.SE! +1 for including background and your work on the problem. I'm afraid I can't help with your question, however. $\endgroup$ – Alex Becker Jan 5 '13 at 23:47
  • $\begingroup$ Differentiating the $k$th term of the sum however many times in whatever variables can only bring out some polynomial in $k$. The Gaussian factor dominates any polynomial with room to spare. Hence, you have uniform convergence of all derivatives. $\endgroup$ – user53153 Jan 5 '13 at 23:48
  • $\begingroup$ @PavelM Perhaps you could turn that into an answer? $\endgroup$ – Alex Becker Jan 6 '13 at 0:14

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 $$|f_k(t,y)|\le 2\sqrt{2\pi}\exp\left(-\pi^2 a^{-3}k^2\right)$$

  • $\begingroup$ Excellent, exactly what I was looking for. Just some technical detail: I have seen a proof that the series defining the $\vartheta$ function is uniformly convergent, and there the same was shown for $K$ compact (your $K$ is apparently not compact). Does this matter? I mean your proof would work just as well for $|y|\leq c$ instead of $|y|<c$ isn't it? I have edited my question and added the rest of the proof of regularity, is it formally correct? (I'm uncertain about the use of $z, w$ instead of $y, t$) If you elaborate a bit about this, I cannot resist accepting your answer. $\endgroup$ – user55315 Jan 6 '13 at 1:06
  • $\begingroup$ @Karl You were right to be uncertain about the passage to complex numbers. It does not matter how the variables are named, of course, but to use Morera's theorem we need uniform convergence in the complex domain. Uniform convergence for real $t,y$ is not enough. For example, the real-analytic functions $g_n(x)=\sqrt{x^2+1/n}$ converge uniformly on $\mathbb R$ to the non-differentiable function $g(x)=|x|$. I rewrote the answer for complex $t,y$. $\endgroup$ – user53153 Jan 6 '13 at 2:06
  • $\begingroup$ Very nice! Thanks also for your example. I'm sorry if I didn't state this clearly, that I need uniform convergence in $\mathbf{C}$. $\endgroup$ – user55315 Jan 6 '13 at 12:34

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 $\large|\frac{1}{\sqrt{2\pi t}}| ≤ \large\frac{1}{a\sqrt{2\pi}} ≤ 2\sqrt{2\pi}$


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