Convexity of $\theta(q)$
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 http://dlmf.nist.gov/20.4#E11) $$\theta \,''(q) = 8 \,\theta (q) \sum_{n=1}^\infty \frac{q^{2n-1}}{(1-q^{2n-1})^2}\cdot$$Since $\theta (q)$ and the sum are positive for $q \in [0,1)$ this even implies that $\theta (q)$ is strictly convex on $[0,1)$.