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Let $f$ be a $C^2-$function on an open set $\omega\subset R^n$ 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 $\Omega \supset \omega$ such that $f\times \Delta f \ge 0$ on $\Omega$.

NB. $C^2$ can be replaced with the Sobolev space $H^2$.

Thanks

flag	asked Aug 20 at 9:25 hardy 23●5

Cf. math.stackexchange.com/questions/184456/… and math.stackexchange.com/questions/184533/… over at MSE – Yemon Choi Aug 20 at 9:31

Alexandre Eremenko · Answer 1 · 2012-08-20 10:17:31Z UTC

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No extension is possible unless some additional conditions are imposed. A positive subharmonic function in the unit disc in the plane can tend to infinity at the boundary and certainly has no extension to a twice larger disc.

answered Aug 20 at 10:17

Alexandre Eremenko
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	conditions such as $\omega$ is bounded and $f$ is continuous in $\overline{\omega}?$ – hardy Aug 20 at 11:46
	At the very least you need to assume $C^2$ on the boundary as well, since there are subharmonic, indeed harmonic, functions with continuous nondifferentiable boundaries. – Will Sawin Aug 20 at 13:55
	For my question, I can take $\omega$ smooth enough. – hardy Aug 20 at 15:51

extension of “subharmonic” function

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