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Is it possible to construct a $\bf{harmonic}$ function $u:\mathbb{R}^N\to \mathbb{R}$ satisfying:

I - There exist a sequence $x_n\in \mathbb{R}^N$ such that $|x_n|\to\infty$ and $u(x_n)\to\infty$,

II - There exist a sequence $y_n\in\mathbb{R}^N$ such that $|x_n-y_n|\leq\delta_n$, where $\delta_n\to 0$ and $u(y_n)\to 0$?

I think that such function does not exist, but I was unable to prove it.

Thank you.

Tomás
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  • Is this not true whenever $u$ is continuous? – Jack M Jun 18 '13 at 21:56
  • I dont think so @JackM. At least untuitively looks like that a example can be constructed. – Tomás Jun 18 '13 at 22:09
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    @JackM Definitely not. Here's an example on $\mathbb R^2$. Let the function be zero on the line $y=0$ and go to infinity on the line $y=1/x$ as $x$ grows large. Linearly interpolate to get a continuous function. – Potato Jun 18 '13 at 22:33
  • By using @Potato idea, instead of interpolating linearly , we can use a partition of unity to constructo a $C^\infty$ function. – Tomás Jun 18 '13 at 22:37
  • @Tomás Are the harmonic functions dense in the $C^\infty$ functions, or something like that? If so, you could get a counterexample that way, by using the $C^\infty$ example you found. – Potato Jun 18 '13 at 22:40
  • @Potato I dont think this is true. At least I never heard about it. – Tomás Jun 18 '13 at 22:46
  • @Tomás Since you have a $C^\infty$ counterexample, you might try to see if such a thing is true (or something like it) and use an approximation argument. Just a thought. – Potato Jun 18 '13 at 22:47

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The function $z\mapsto \sin(z^2)$ is an entire function on $\mathbb C$, so its real part is a harmonic function $\mathbb C\to\mathbb R$. Along the real axis it oscillates between $-1$ and $1$, more and more rapidly as you move away from $0$. So taking $x_n=\sqrt{(2n+\frac12)\pi}$ and $y_n=\sqrt{(2n-\frac12)\pi}$ seems to give what you were looking for.

Andreas Blass
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