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So far in complex analysis books I have studied about Uniqueness theorem: If $f$ is analytic in a domain $D$ and if its set of zeroes has a limit point in $D$ then $f\equiv 0$ on $D$, I want to know is this result holds for harmonic functions?

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It's not true, if you consider real valued functions, see my reply. – user20266 Jun 10 '12 at 15:38
up vote 5 down vote accepted

In general harmonic functions are functions $u:\mathbb{R}^n\rightarrow \mathbb{R}$. If $0$ is a regular value of such a function (which is true for almost every real number) then $u^{-1}(0)$ is locally an $(n-1)$ dimensional submanifold. Actually this is true for every $C^1$-function, regardless whether it is harmonic or not. So the question makes only sense for $u: \mathbb{R}\rightarrow \mathbb{R}$. In this case the equation is $u^{''}=0$ which implies $u$ is linear.

So the answer is yes, in one dimension, cause there the solution is explicitly known, no otherwise, since the zero set is usually locally a submanifold of dimension $\ge1$.

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Yes, since it is a direct consequence of the maximum principle, which holds also for harmonic functions.

edit: I seem to have misread the question. This uniqueness theorem doesn't seem to follow. (There is uniqueness, of course, when boundary conditions are specified).

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The maximum principle for holomorphic functions and harmonic functions are two different pairs of shoes. Holomorphic functions are complex valued, harmonic functions (usually) real valued. – user20266 Jun 10 '12 at 15:36
The idea in both cases is the same - it follows because the value inside the curve is the average of the value on it. – yaakov Jun 10 '12 at 15:45
This works only if the source is one dimensional. – user20266 Jun 10 '12 at 15:50

Assume $u$ is harmonic on $\Omega \subset \mathbb{R}^n$, ($n > 1$) and let $Z = \{ x : u(x) = 0 \}$. (I will leave the one-dimensional case as an exercise; note that for $n=1$ harmonic is the same as affine.)

If $Z$ contains an open set, then $Z = \Omega$, so in this sense the uniqueness theorem holds. However, $Z$ may very well have limit points (contrary to the case of holomorphic functions) without $Z = \Omega$. A very simple example would be $u(x, y) = x$. (With an obvious generalization to higher dimension.)

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awsomesalaaaaaaaaaaa – Un Chien Andalou Jun 10 '12 at 18:23

I don't think this problem is true at all.

The real or imaginary part of an analytic function is always an harmonic function.

$f:\mathbb{C}\rightarrow\mathbb{C}:z\mapsto z$ is an analytic function thus $g:\mathbb{R^2}\rightarrow\mathbb{R}:(x,y) \mapsto \operatorname{Im}(g(x+iy))=y$ is a harmonic function. $g(x,y)=0$ for every point on the real line (points with $y=0$). But $g$ is not the zero function on any open subset of the complex plane (including the unit disc).

Gotta hate harmonic functions.

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Why the down votes? Did I make a mistake in my reasoning? – Bobby Ocean Jun 8 '13 at 21:44
I think not each harmonic function is a real or imaginary part of an analytic function, this is just locally true. I edit your answer to make this clear (It took me a while to understand your counterexample but I also did not find a mistake). I hope, this is okay for you. – tampis Jan 19 '14 at 21:24
I think you got the down votes for the sentence „Gotta hate harmonic functions“. I think there are a lot of people on who love harmonic functions... ;-) – tampis Jan 19 '14 at 21:38

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