I have an integral to minimize that writes like $$F: \mathbb R^d \to \mathbb R: \theta \mapsto \int\_{[0,1]^d} f(\langle x,\theta\rangle) dx$$.


The function $f$ is a convex function, which makes $F$ a convex function.


Q : Let $x \in [0,1]^d$. Is $\frac{df(\langle x,\theta\rangle)}{d\theta}$a subgradient of $F$ at $\theta$ ?



Subgradient of a convex integral
For $d=1$ and $f(z)=z^2 $, $dF/d\theta = 2\theta /3$, but $df(x\theta )/d\theta = 2\theta x^2 $. Not a subderivative of $F$ for almost all $x$.


