Harmonic function on Riemannian manifold

Question

$M$ is a connected Riemannian manifold with $\Delta$ its Laplacian and $f$ is smooth function on $M$ such that $\Delta f=0$ and $f$ vanishes on some open set $U$ of $M$, then is $f$ identically $0$ on $M$?

Answer 1

Yes, I think you can prove this by maximal principle.

Edit:

Here is some detail as requested. We have to assume $M$ is connected. Now assuming for the moment that we have a point $x\in M$ on which $f$ is not zero. Then there exist a neighborhood around that point on which $f$ does not vanish identically. Then a string of such neighborhoods connects $x$ to the open set $U$ we had earlier.

In the most trivial case, we have two open sets $U,V$ in $M$ such that $f\equiv 0$ on $U$ but unknown on $V$. We consider the boundary of the intersection $U\cap V$, of which $f$ must obtain its maximum or minimum value. But we know in the inside $f$ is a constant $0$. So we can "push" $f$ off the boundary towards the interior of $V$. And this way we can prove $f$ must be vanishing inside of $V$.

The above argument proved $f$ must vanish inside of $M$. But on the boundary, the same argument as above still holds. So $f$ must vanish on all of $M$.