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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 $\theta$ ?

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  • $\begingroup$ no, if f is smooth there's only one subgradient and there's no reason your formula is independent of x $\endgroup$
    – alesia
    Jan 11, 2022 at 19:06
  • $\begingroup$ Indeed, i was mistaken... $\endgroup$
    – lrnv
    Jan 12, 2022 at 10:30

1 Answer 1

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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 $x$.

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