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Let $u$ be a non-constant harmonic function on $\mathbb{R}$. Show that $u^{-1}(c)$ is unbounded.

I am not getting what theorem or result to apply. Could anyone help me?

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    Are you sure there is not something missing here? The harmonic functions on $\mathbb{R}$ are just the linear functions, and then the result is definitely false.2012-06-24

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Let $u(x)=x$, for all $x \in \mathbb{R}$. Then $u''(x)=0$ for all $x$. But $u^{-1}(\{c\}) = \{c\}$. I think somebody is cheating you :-)

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    yes I was thinking so, one girl gave me the problem :P2012-06-24
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    You might like to consider whether this was the intended question: If $u:\mathbb{R}^n\to\mathbb{R}$ is a non-constant harmonic function, then $u^{-1}(c)$ is unbounded in $\mathbb{R}^n$ for any $c\in\mathbb{R}$ - and then either prove it or find a counter-example.2012-06-24
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    well, what will happen then? @ Old John, dear sir?2012-06-24
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    @TaxiDriver If you still want to know the answer for $n>1$ (which is quite a different story from $n=1$), you should post another question with a link to this one.2013-06-15