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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

plot of the function via Sage

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!

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I was hoping the partial products would be log-concave, but if you do the calculation, they're actually log-convex with every term negative. – J Swanson Jan 19 '14 at 14:13
Whoops! Reverse "concave" and "convex" in the comment above: I was hoping the partial products would be log-convex, but they're log-concave, with every term in the second derivative negative. – J Swanson Jan 20 '14 at 4:43

Do you accept the following proof with a semi-explicit formula as simple, standard, or illuminating?

The second derivative of $\theta (q) = \theta_4 (q)$ is given by (see $$\theta \,''(q) = 8 \,\theta (q) \sum_{n=1}^\infty \frac{q^{2n-1}}{(1-q^{2n-1})^2}\cdot$$Since $\theta (q)$ and the sum are positive for $q \in [0,1)$ this even implies that $\theta (q)$ is strictly convex on $[0,1)$.

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Uhm, no. That $'$ is $\mathrm{d}/\mathrm{d}z$. By the heat equation, the double-prime $''$ can be replaced with something proportional to $q\,\mathrm{d}/\mathrm{d}q$, but that yields only the first derivative with respect to $q$. – ccorn Aug 12 '13 at 10:37
One could use identities such as $\left(q\frac{\mathrm{d}}{\mathrm{d}q}\right)^2 (\theta_4^{-2}) = \frac{1}{4}\theta_2^4\theta_3^4\theta_4^{-2}$, but you will need Halphen's psi functions to prove that. One of Halphen's psi functions is (proportional to) the sum you have given. – ccorn Aug 12 '13 at 10:51

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