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I have a couple of inequalities that I want to prove. The proof is easy using fourier analysis but I am wondering whether there is a proof that does not use fourier analysis.

1) For any $c, s > 0$, $$ \sum_{x \in \mathbb{Z}} e^{-\pi x^2 s^2} \ge \sum_{x \in \mathbb{Z}} e^{-\pi (x - c)^2 s^2}\;. $$

2) For any $c,s >0$, $$ \sum_{x \in \mathbb{Z}} e^{-\pi x^2 s^2} \cos(2\pi xc) > 0\;. $$

Can someone suggest some other simple proof? In particular, for the second inequality, is it possible to partition the sum into parts each containing finitely many terms and each is individually greater than $0$.

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    $\begingroup$ For part 2 note that the term $x=0$ is $1$, and the rest of the terms are at least $-2 \sum_{n=1}^{\infty} e^{-\pi n^2} = -0.086\ldots$. $\endgroup$ – Lucia Oct 18 '14 at 8:17
  • $\begingroup$ Lucia, Seva: Thanks for replying. Sorry, but when I posted the question, I was trying to simplify what I had in mind, but I ended up oversimplifying it. I have edited the question, and now it does not seem as trivial as what you pointed out. $\endgroup$ – user47772 Oct 18 '14 at 17:59
  • $\begingroup$ Are you sure about termwise differentiation? Some of the terms are ${}$ increasing functions of $c$; equality holds for all integers $c$, not just $c=0$. $\endgroup$ – Noam D. Elkies Oct 18 '14 at 20:20
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These are both tantamount to known properties of the heat kernel $K_t(c)$ for $c$ in the circle ${\bf R} / {\bf Z}$. (In (1) the heat kernel is obtained by starting at $t=0$ with a row of delta functions $K_0(c) = \sum_{x \in \bf Z} \delta_x(c)$, and in (2) it's obtained by separation of variables.) Each of the properties can be proved in various ways, not all of which involve Fourier analysis. Here's one way.

(1) Let $K_t(c) = t^{-1/2} \sum_{x \in \bf Z} \exp (-\pi t^{-1} (x-c)^2)$, so the desired inequality is (after multiplying by $s$) $K_t(0) \geq K_t(c)$ for $t = s^{-1/2}$. Clearly $K_t$ is an even function of $c$; it is also $\bf Z$-periodic: $K_t(c)=K_t(c')$ when $c' \equiv c \bmod \bf Z$. Thus $K_t(0) = K_t(c)$ for $c \in \bf Z$, and we may assume $0 < c < 1$. We shall show that then $K_t(0) > K_t(c)$.

Now $K_t$ satisfies the partial differential equation (heat equation) $$ \frac{\partial^2 K_t(c)}{\partial^2 c} = 4 \pi \frac{\partial K_t(c)}{\partial t} $$ (which holds for for each term $t^{-1/2} \exp (-\pi t^{-1} (x-c)^2)$ separately). Hence the function $D_t(b) := K_t(b) - K_t(c-b)$ satisfies the same equation $$ \frac{\partial^2 D_t(b)}{\partial^2 b} = 4 \pi \frac{\partial D_t(b)} {\partial t}. $$ For all $t$ we have $D_t(b) = -D_t(c-1-b) = -D_t(c-b)$, whence $D_t((c-1)/2) = D_t(c/2) = 0$ for all $t$. For small $t$ it is clear that $D_t(b) > 0$ for $b \in ((c-1)/2, c/2)$, because the $x=0$ term in the sum for $K_t(b)$ dominates everything in $K_t(c-b)$. Therefore $D_t(b)$ remains positive in $b \in ((c-1)/2, c/2)$ for all $t$. In particular $0 < D_t(0) = K_t(0) - K_t(c)$, QED.

(2) We could also show that $\sum_{x \in \bf Z} e^{-\pi x^2 s^2} \cos(2\pi x c)$ is proportional to a solution of the heat equation, but the fact that it approaches a delta function as $s \rightarrow 0$ seems to be a matter of Fourier analysis. Instead we write the sum as a theta function $$ \sum_{x \in \bf Z} e^{-\pi x^2 s^2} e^{2\pi i x c} = \sum_{x \in \bf Z} q^{x^2} z^{2x} $$ for $q = e^{-\pi s^2}$ and $z = e^{\pi i c}$ (average the $x$ and $-x$ terms on the RHS side to recover the LHS). We can now use the Jacobi triple product formula $$ \sum_{x \in \bf Z} q^{x^2} z^{2x} = \prod_{m=1}^\infty (1-q^{2m}) (1+q^{2m-1} z^2) (1+q^{2m-1} z^{-2}) $$ and observe that in our setting each of the factors is positive (note that $(1+q^{2m-1} z^2)$ and $(1+q^{2m-1} z^{-2})$ are complex conjugates because $q \in \bf R$ and $|z|=1$), whence the product is positive as desired.

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