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Let $x=[0,1]^n,\;p: [0,1]^n \rightarrow [0,1]^n$. Define $V(x) = p(x).x$, i.e. the inner product of $p(x)$ and $x$. We are given $V(.)$ is convex. Can we say $p(x)$ is Lipschitz?

My approach: $V(.)$ is convex, therefore locally Lipschitz. We know, locally Lipschitz on a compact set implies Lipschitz. Therefore $V(.)$ is Lipschitz. My question is, does this imply $p(x)$ is Lipschitz or at least almost everywhere differentiable?

I tried to do the following but could not proceed further:

$\lim\limits_{u \rightarrow u_0} |V(u_0) - V(u)| = \lim\limits_{u \rightarrow u_0} |p(u_0).(u - u_0) + u.(p(u) - p(u_0))| = \lim\limits_{u \rightarrow u_0} |u.(p(u) - p(u_0))|$

Canine360
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1 Answers1

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When $n=1$ and $p(x)=\sqrt x$ is is clear that $p$ is not Lipschitz but $xp(x)$ is convex and Lipschitz (it has a bounded derivative).

Kavi Rama Murthy
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  • As I said in the question, almost everywhere differentiability is fine too. $\sqrt x$ is almost everywhere differentiable in $[0,1]$ or $\mathbb{R_{+}}$. Is there a counterexample where the function in question is not almost everywhere differentiable? That would be great. Thank you in advance. – Canine360 Aug 15 '20 at 05:26