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The function $f(x,y)=x^3+y-1$ in $\omega = (1,2)^2$ is such that $f\times \Delta f \ge 0$ on $\omega.$ I am wondering about the existence of a $C^2-$extension $F$ of $f$ in $\Omega = (0,2)^2$ such that $f\times \Delta f \ge 0$ on $\Omega$.

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    6 questions, 0 accepted answers, last seen August 20. Typical ask-and-run.2012-09-20

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This particular function can be written as $f(x,y)=(x^3-1/2)+(y-1/2)$ where both summands are 1-variable functions that are convex and strictly positive on $(1,2)$. It is not hard to extend such functions to $(0,2)$ (or the entire line $\mathbb R$, if you wish) while keeping both convexity and positivity. (And of course, convexity implies subharmonicity).

Here is a concrete extension to $\Omega$, using the Iverson bracket: $f(x,y)=\left((x^3-1/2)+10(1-x)^3[x<1]\right)+\left((y-1/2)+(1-y)^3[y<1]\right)$ You can check directly that both expressions in big parentheses are positive on $(0,1)$. Convexity is clear, as is $C^2$ smoothness.