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I am asked to show that the following function is positive for each $(x,s)\in[0,1]^2$ where $t>0$ is a parameter:

$k(x, s) := \displaystyle 1+\sum_{n=1}^\infty 2e^{-t(n\pi)^2}\cos(\pi n s)\cos(\pi nx)$

The background is the following: I am looking at the semigroup $T$ on $L^2[0,1]$ with

$$T(t)f = \sum_{n=0}^\infty e^{-t(n\pi)^2}\langle f,e_n\rangle_{L^2(0,1)} e_n$$ where $e_n(x) = \sqrt{2}\cos(\pi n x)$ if $n>0$ and $e_0 \equiv 1$. Now, for each $t>0$ there is supposed to be a positive kernel $k\in C([0,1]^2)$ with $T(t)f(x) = \int_0^1 k(x,s)f(s)ds$.

I showed that $k$ from the start is indeed a kernel for $T(t)$ but I have no idea how to show positivity. One would need to see that $T(t)$ is a positive operator but I don't see how this would work.

Lukas Betz
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  • Related: https://math.stackexchange.com/questions/1847213/fourier-cosine-transforms-of-schwartz-functions-and-the-fejer-riesz-theorem – Jack D'Aurizio Jun 10 '17 at 17:16

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$$I(t,x,s)=\sum_{n\in\mathbb{Z}}e^{-t(\pi n)^2}\cos(\pi n s)\cos(\pi n x) $$ is a series whose terms are the values over the integers of a Schwartz function.
By Poisson summation formula, the given series equals $\sum_{n\in\mathbb{Z}}\widehat{f}(n)$, where $f(n) = e^{-t(\pi n)^2}\cos(\pi n s)\cos(\pi n x)$. The Fourier transform of $e^{-t\pi^2 n^2}$ is a function of the same gaussian form, the Fourier transform of $\cos(\pi ns)\cos(\pi n x)$ is a combination of Dirac deltas with positive coefficients, hence $\widehat{f}(n)$, as the convolution of two non-negative functions, is non-negative.
This proves $I(t,x,s)\geq 0$ even without the restriction $(x,s)\in[0,1]^2$.

Jack D'Aurizio
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  • Thanks alot. I think i misinterpreted your comment. I was trying to apply the technique to show that this function is not positive everywhere. – Lukas Betz Jun 10 '17 at 20:20
  • @TimB.: it was me changing opinion, not you misunderstanding it. I initially thought that Hadamard's 3-4-1 trick should have been the way to go, then I realized that Poisson's summation formula was the actual way to go. You're welcome. – Jack D'Aurizio Jun 10 '17 at 20:23