If $\int_V f \;dV = 0$ when can we say that $f=0$ everywhere

If $\int\limits_V f \; \mathrm dV = 0$ can we say that $f=0$ everywhere? Or what conditions are there on concluding this.

In particular I want to solve the PDE $\nabla^2 f=f^3$ on the region $$D=\{(x,y)\in\mathbb{R}|x^2+y^2<1\}$$ with $f=0$ on the boundary.

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Seb note that $f$ could change sign ! But if in addition you have that $f$ is positive then you right – Leandro Oct 31 '11 at 2:56
If this holds for only one given domain $V$, no chance unless you add the condition that $f\geqslant0$ everywhere. If this holds for every subdomain $V$, you win. In both cases the conclusion is that $f=0$ except on a set of measure zero--unless $f$ has some regularity such as being continuous, and then this set of measure zero is in fact empty. – Did Oct 31 '11 at 2:58
Multiply both sides by $f$ and integrate by parts to get $\int_V \|\nabla f\|^2 dx = -\int_V f^4 dx$. Since $\|\nabla f\|$ and $f^4$ are both non-negative, this implies that $f=0$ is the only solution. – Jeff Nov 4 '11 at 1:06

Multiply both sides by $f$ and integrate by parts to get

$\int_V \|\nabla f\|^2 dx = -\int_V f^4 dx$.

Since $\|\nabla f\|$ and $f^4$ are both non-negative, this implies that $f=0$ is the only solution.

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Or, both of those could be negative. – Dhaivat Pandya Dec 4 '11 at 2:47
$\lVert \nabla f\rVert$ tends to be non-negative rather than negative, Dhaivat. – Mariano Suárez-Alvarez Dec 4 '11 at 6:59

Let me extend the boundary beyond its original limit, filling up the newly added space by f=0.

$\int\int\int \nabla^2 f dV=\int\int\int \nabla.(\nabla f) dV$

$=\int\int (\nabla f).n dS=0$

Therefore, $\int\int\int \nabla^2 f dV=0$

But $\int\int\int \nabla^2 f dV$ for the additional space is zero. Therefore the integral is zero for the original space also.

We may now consider a sub-volume[V'] inside the original 3D region. Let the surface/boundary of this sub-volume be S'. We may try to perform the same trick as before but the difficulty is that f in general is not zero on the boundary.A continuous extension with f=0 will not be possible. This time we may carry out an infinitesimally small extension so the gradf becomes zero on the extended boundary. Again the volume integral is zero.

The integral seems to vanish for an arbitrary volume

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