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Let $u$ and $v$ be real-valued harmonic functions on $U=\{z:|z|<1\}$. Let $A=\{z\in U:u(z)=v(z)\}$. Suppose $A$ contains a nonempty open set. Prove $A=U$.

Here is what I have so far: Let $h=u-v$. Then $h$ is harmonic. Let $X$ be the set of all $z$ such that $h(z)=0$ in some open neighborhood of $z$. By our assumptions on $A$, $X$ is not empty. Let $z\in X$. Then $h(z)=0$ on some open set $V$ containing $z$. If $x\in V$, then $h(w)=0$ in some open set containing $x$, namely $V$. So $X$ is open.

I want to show $X$ is also closed but I am having trouble doing so. Any suggestions:

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You could instead show that $h$ is identically 0. –  rotskoff Jun 2 '12 at 22:05
    
A little more effort shows that it is sufficient that $A$ is uncountable. –  AD. Jun 6 '12 at 19:49

2 Answers 2

up vote 4 down vote accepted

Each real harmonic function $h$ on a simply connected domain defines unique up to the constant holomorphic function $f\in\mathcal{O}(U)$ such that $$ \mathrm{Im}(f)=h $$ $$ \mathrm{Re}(f)= \int\limits_{(x_0,y_0)}^{(x,y)}\left(\frac{\partial h}{\partial y}dx-\frac{\partial h}{\partial x}dy\right)+C $$

If $h=0$ on some ball $B\subset A$, then respective function $f=C$ on $B$. Since $A$ is open, by uniqueness principle $f=C$ on $U$. Hence $h=\mathrm{Im}(f)=0$ on $U$.

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Harmonic functions are continuous, and closed under addition and scalar multiplication. Therefore $u-v$ is harmonic (and continuous), so $\{z\in\mathbb{C}:u(z)=v(z)\}$ is closed.

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Why $X=[u=v]$ ? –  no identity Jun 2 '12 at 22:28
    
You showed it was open and closed. Since the domain is connected, $X = [u=v]$. –  ncmathsadist Jun 2 '12 at 22:38
    
We are talking about your proof, how could you refer to my proof in your solution? Could you explain me in detail wny $X$ is connected and why it is closed? –  no identity Jun 2 '12 at 22:41
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This "proof" does not use the fact that $u-v$ is harmonic, just that it is continuous. But the result is not true for arbitrary continuous functions. –  Robert Israel Jun 6 '12 at 19:19

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