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$M$ 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 $M$?

Summer
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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 $M$.

Bombyx mori
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  • $M$ might not be compact. Maximum principle could not be applied. –  Nov 01 '13 at 04:37
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    Use maximum principle on all compact subsets. – Neal Nov 01 '13 at 04:50
  • I cannot see where to use maximal principle. This is essentially a local problem so locally we need to prove that on $\mathbb R^n$, the solution $f$ of $Lf=0$ where $L$ is a second-order elliptic operator has the unique continuation property. But I do not know if this is true for general $L$. – Summer Nov 02 '13 at 00:48
  • @Hezudao: I suggest you review your own question instead. – Bombyx mori Nov 02 '13 at 02:52
  • Can you write the details of the proof of the statement above? I'm referring to a proof that uses trivially only the maximum principle. Thank you –  Nov 02 '13 at 10:26
  • @user55449: The proof is trivial. Use maximal principle on closure of open neighborhoods and use path connectedness. – Bombyx mori Nov 02 '13 at 16:48
  • @user32240: Say we are considering the problem on the unit ball $B$ and $f$ vanishes on some open set $U$ of B. Where do you use the maximal principle since the function $f$ may take its maximum and minimum on $\partial B$? – Summer Nov 02 '13 at 17:36
  • @Hezudao: the function $f$ must vanish on the whole ball. Assume it does not, try to prove by contradiction. – Bombyx mori Nov 02 '13 at 17:44
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    thanks for your edit. Can you explain better the part of the argument where you take the boundary of $ U \cap V $. Clearly in this set the function is null up to th boundary. How do you arrive to conclude that $ f $ is null on whole $ V$ ? Thanks –  Nov 02 '13 at 20:01
  • @user55449: The function is $0$ up to the boundary. But it cannot be nonzero on the boundary. At the boundary since we know it is $0$, choosing a small enough neighborhood on a fixed point $p'$ on the boundary. Then by mean value principle the function's mean value on the $V$ half of the ball must be 0 as well. By continuity there must be some point valued $0$ as well. Connect $p$ and $p'$, by maximal principle on the ball, $f$ must vanish on the path identically. But this violates the assumption $f$ can be nonzero on the neighborhood if we choose the path carefully. – Bombyx mori Nov 02 '13 at 20:34