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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

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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.

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  • $\begingroup$ conditions such as $\omega$ is bounded and $f$ is continuous in $\overline{\omega}?$ $\endgroup$
    – hardy
    Aug 20, 2012 at 11:46
  • $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Aug 20, 2012 at 13:55
  • $\begingroup$ For my question, I can take $\omega$ smooth enough. $\endgroup$
    – hardy
    Aug 20, 2012 at 15:51

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