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$A_n(s)$ is a sequence of convex random functions defined on an open set $S\in \mathbb{R}^p$ which converges in probability to some $A(s)$ for each $s$. I'm trying to show that $\sup_{s\in K} \big| A_n(s)-A(s) \big|\stackrel{P}{\to} 0$ for $K\subset S$ compact.

I think I could use the fact that since $K$ is compact, then $\sup_{s\in K} \big| A_n(s)-A(s) \big| > \epsilon$ is an event contained in $\bigcup_{s\in K} \left\{\big| A_n(s)-A(s) \big| > \epsilon\right\}$ (since continuous functions reach their supremums on compact sets). But after that, I'm stuck because I can't see how to show the probability of $\bigcup_{s\in K} \left\{\big| A_n(s)-A(s) \big| > \epsilon\right\}$ goes to $0$.

Suggestions?

Kashif
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  • Are they continuous AND convex? – Bunder Apr 28 '14 at 10:31
  • Convex functions are continuous. – Kashif Apr 28 '14 at 15:25
  • @Glassjawed No, in general, this is not correct. Only convex functions on finite-dimensional spaces are continuous (more precisely, they are continuous in the interior of the domain). – saz Apr 28 '14 at 18:11
  • Oh okay. Yeah I was just thinking in $\mathbb{R}^p$. I can't think of a counterexample in $\mathbb{R}^{\infty}$ off the top of my head though.

    Also any suggestions for how to go about this problem?

     – Kashif Apr 28 '14 at 18:14
  • This question might be helpful; there it is proved that pointwise convergence implies uniform convergence on compact sets if the functions are convex (and continuous). – saz Apr 28 '14 at 18:43

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