Define Jacobi's (fourth) theta function with argument zero and nome $q$:
$\theta(q) = 1+2\sum_{n=1}^\infty (-1)^n q^{n^2}$
plot of the function via Wolfram|Alpha
I am looking for a simple/standard/illuminating proof of the fact that $\theta(q)$ is convex for $q\in[0,1]$. The proof I found goes like this: We have
\theta'(q) = 2\sum_{n=1}^\infty (-1)^n n^2 q^{n^2-1}
and one can show that for some $q_0\in(0,1)$, $n^2q^{n^2-1} - (n+1)^2q^{(n+1)^2-1}$ is increasing in $[0,q_0]$ for any $n\ge 2$. This gives convexity of $\theta(q)$ in $[0,q_0]$. For the remaining values of $q$, one uses the representation of $\theta$ as a sum over Gaussian kernels:
$\theta(e^{-\pi^2t/2}) = 2 \sqrt{\frac{2}{\pi t}}\sum_{n=1}^\infty \exp\left(-\frac{(2n-1)^2}{2t}\right)$
With this representation, one can show that the second derivative (wrt $q$) of each summand is positive for $q \ge q_1$, with $q_1 < q_0$. This yields convexity of theta.
I don't like this proof, because it requires calculating $q_1$ and $q_0$ explicitly and it is not very illuminating. I tried playing around with the representation of $\theta(q)$ as the infinite product
$\theta(q) = \prod_{n=1}^\infty (1-q^{2n-1})^2(1-q^{2n}),$
but didn't manage to find anything, except that the partial products
$\prod_{n=1}^N (1-q^{2n-1})^2(1-q^{2n})$
all seem to be convex in $[0,1]$, which would prove the statement.
All suggestions are very welcome!