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Let $\mathcal{R}$ be an open subset of $\mathbb{R}^M$ with $M\geq 2$.

Consider the function $$a\equiv(a_1,...,a_M) \in \mathbb{R}^M\mapsto G(a)\equiv \int_{\{r\equiv (r_1,..., r_M)\in \mathcal{R}\}} \ \Big[\max_{i\in \{1,...,M\}} (a_i+ r_i)\Big] f(r) dr $$ where $f: \mathbb{R}^M\rightarrow \mathbb{R}$ is

  • strictly positive on $\mathcal{R}$ and zero on $\mathbb{R}\setminus \mathcal{R}$

  • $\int_{\{r\equiv (r_1,..., r_M)\in \mathcal{R}\}} f(r) dr =1$

Is $G$ convex? Is it strictly convex? Could you help me to show it? I know that the function within the square brackets is convex but not strictly convex (as kindly explained here)

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    $\begingroup$ The integral is a convex combination of the functions within the square brackets, hence, convex. $\endgroup$ – A.Γ. May 25 '18 at 13:31
  • $\begingroup$ Thank you very much. What about strict convexity? Are there stronger conditions on $f$ that can ensure strict convexity (even though, the function within the square brackets is not strictly convex)? $\endgroup$ – TEX May 25 '18 at 13:33
  • $\begingroup$ Also, would convexity hold for ANY function $f$ (not necessarily with the properties described above)? $\endgroup$ – TEX May 25 '18 at 14:33
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    $\begingroup$ It is easy to prove that the integral is convex for any $f\ge 0$. However, strict convexity is problematic here. For example, if the set $R$ is bounded then the integral cannot be strictly convex. $\endgroup$ – A.Γ. May 25 '18 at 14:57

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