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Let $d,n\ge 1$ be fixed integers. Given some compact subset $E\subset \mathbb R^d$, consider the function $f: E^n\ni (x_1,\ldots, x_n) \longrightarrow f(x_1,\ldots, x_n)\in \mathbb R$ defined by

$$f(x_1,\ldots, x_n):= \max_{(c_1,\ldots,c_n)\in\mathbb R^n}\left\{\int_E \left(\min_{1\le i\le n}|y-x_i|^2-c_i\right)p(y)dy + \sum_{i=1}^n \alpha_i c_i\right\},$$

where $p:E\to \mathbb R_+$ is a probability density on $E$ and $\alpha_1,\ldots, \alpha_n>0$ s.t. $\sum_{i=1}^n \alpha_i =1$. Under which conditions (on $E, \rho$) $f$ is differentiable (almost everywhere) on $E^n$?

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1 Answer 1

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Suppose that there is an open subset $U$ of $E$ such that the Lebesgue measure of $E\setminus U$ is $0$. Since $E$ is compact, the function $E^n\ni(x_1,\dots,x_n)\mapsto|y-x_i|^2$ is $L$-Lipschitz for some real $L>0$ and each $i\in\{1,\dots,n\}$ and each $y\in E$. Therefore and because the $\max$, $\min$, and integration (with respect to a probability measure) operations preserve the $L$-Lipschitz condition, $f$ is $L$-Lipschitz.

So, by Rademacher's theorem, $f$ is differentiable almost everywhere (a.e.) on $U^n$ and hence a.e. on $E^n$.

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  • $\begingroup$ Thanks for the answer. Do you think there is some way to compute its gradient? $\endgroup$
    – Fawen90
    Commented Nov 3, 2022 at 16:33
  • $\begingroup$ @Fawen90 : I think there should be a way to compute the gradient where it exists. However, that would be a different, much more difficult question -- in particular, that would require an explicit determination of the points where the gradient exists (which would probably depend on the density $p$). As your posted question has been answered, you may want to post additional questions separately. $\endgroup$ Commented Nov 3, 2022 at 16:47
  • $\begingroup$ Of course. Will post it in an alternative one. Thanks again for your help. $\endgroup$
    – Fawen90
    Commented Nov 4, 2022 at 8:01

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