Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\Omega$.

Thanks

share|improve this question
    
6 questions, 0 accepted answers, last seen August 20. Typical ask-and-run. –  user31373 Sep 20 '12 at 0:09
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.