Let $f\geq0$ be a continuous function satisfying $\Delta f=0$ in $\{x\in U:f(x)>0\}$.

I was wondering if one could follow $\Delta f=0$ in $U$, especially in the cases $f\in C^2$ or $\Delta f=0$ in a weak sense.

So I've tried to show it for $\Delta u=0$ in weak sense, e.g. $$ \int_U\nabla f\nabla\phi=0 $$for all $\phi\in C_0^\infty(U)$. I've considered instead of just $\phi$ the function $\phi \eta_\delta(f)$ with a bounded function $\eta_\delta$ satisfying $\eta_\delta(x)=0$ if $x\leq \delta$ and $\eta_\delta(x)=1$ if $x\geq 2\delta$. Then we get for the left side $$ \int \eta_\delta(f)\nabla f\nabla\phi+\int\eta_\delta'(f)\nabla f\nabla f $$Now I've tried to focus on some other conditions for $\eta_\delta'$ but it didn't get me any further.

Any hints how you can show $\Delta f=0$ in $U$?

  • $\begingroup$ Did you leave out a hypothesis? As stated the answer is obviously no; the function could do more or less anything in the set where it's negative. For that matter, any strictly negative function satisfies your hypothesis... $\endgroup$ – David C. Ullrich Aug 18 '15 at 17:44
  • $\begingroup$ @DavidC.Ullrich Let's take a continuous and nonnegative function. $\endgroup$ – user103411 Aug 18 '15 at 17:45

No. Let $U=\Bbb R^2$, $f(x,y)=|x|$.

  • $\begingroup$ okay. Does the property hold for subharmonic functions? $\endgroup$ – user103411 Aug 18 '15 at 18:58
  • $\begingroup$ It's trivial for non-negative subharmonic functions, just from the definition. $\endgroup$ – David C. Ullrich Aug 18 '15 at 19:00
  • $\begingroup$ Instead of considering that, consider the definition of subharmonic: For every p there exists $\rho>0$ such that for every $r\in(0,\rho)$, $u(p)$ is no larger than the average of $u$ over a circle with center $p$ and radius $r$. That's clear for $p$ in the set $f>0$ because $f$ is harmonic there. And it's clear if $f(p)=0$ as well... $\endgroup$ – David C. Ullrich Aug 18 '15 at 19:16
  • $\begingroup$ okay, got it. But what I've meant above is the situation given $\Delta f\geq-1$ in $\{f>0\}$, it follows $\Delta f\geq -1$ in U, so even more general? $\endgroup$ – user103411 Aug 18 '15 at 19:23
  • 2
    $\begingroup$ Could be, dunno. $\endgroup$ – David C. Ullrich Aug 18 '15 at 21:16

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