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Let $\varphi$ be a nonnegative function defined on a domain $\omega\subset R^n$ such that $\varphi$ is $C^2$ and convex on $\omega$. Does $\varphi$ admit a $C^2-$ extension on $\Omega\supset \overline{\omega}$ which is nonnegative and convex on $\Omega?$

For $n=1$ and $\omega =(a,b)$, a possible extension on $(b,+) $ is $\overline{\varphi}(x)=\varphi'(b)(x-b)+\varphi(b)$. A second idea consists on using odd polynomial, but this does not satisfy the conditions we ask for, unless additional assumtions are imposed. Also I can't see how to extend this construction to $n\ge 2. $

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Even in one dimension you need to assume a bound on the first derivative, otherwise there are simple counterexamples. –  user20266 Aug 22 '12 at 17:02
    
yes, also I tkink that we must also have $\varphi''(b)=0$. –  med Aug 22 '12 at 17:17

1 Answer 1

Every smooth convex function can be written in the form $\varphi(x)=\sup_y \ell_y(x)$ where each $\ell_y$ is an affine function. Specifically, we can take $\ell_y(x)=\varphi(y)+(x-y)^T \nabla \varphi(y)$, the tangent plane at $y$, for each $y\in\omega$. Next, observe that any extension $\overline{\varphi}$ must satisfy $\overline{\varphi}(x)\ge \sup_y \ell_y(x)$ for all $x\in\Omega$ by virtue of convexity. As long as the supremum $\sup_y \ell_y(x)$ is finite for all $x\in\Omega$ (this relates to the 1st derivative bound mentioned by @Thomas) it gives a convex extension of $\varphi$, and the minimal such.

Unfortunately, the minimal extension isn't necessarily $C^2$. You noticed this already in the comments. One has to work harder to get the second derivatives to match (when this is possible at all). This recent paper has a result of this kind, and references to older literature.

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