This question may be elementary, I have asked it on math.stackexchange.com but have not received any answer yet. Note that I am not an expert on theta/elliptic functions.

Define Jacobi's theta function with argument zero and nome $q$ (often also denoted by one of the following: $\theta_0$, $\theta_4$, $\vartheta$, $\vartheta_4$ or $\vartheta_0$):

$$\theta(q) = 1+2\sum_{n=1}^\infty (-1)^n q^{n^2}$$

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!