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Suppose $S= \left\{x \in \mathbb{R}^3 : a <x_1< b \right\} $ is an infinite strip the three dimensional Euclidean Space. Is it true that the only $L^2$ harmonic function in this strip is the zero function?

(I ask this because as you know the only harmonic functions in the entire space (in the distributional sense) are the harmonic polynomials and therefore the answer would be true if S was replaced by the entire space)

thanks, ali

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    $\begingroup$ One notable difference to the case of the whole space is that there are bounded harmonic functions, like $\sinh(x_1)\sin(x_2)$. $\endgroup$ Jul 7, 2015 at 20:40
  • $\begingroup$ It's easy to build counterexamples that are actually harmonic in a half-space, by convolving the Poisson kernel with a compactly supported function. $\endgroup$ Jul 7, 2015 at 21:32
  • $\begingroup$ Would you clarify why you think that would give an L^2 function? Poisson kernel is not square integrable as it is like 1/r $\endgroup$
    – Ali
    Jul 7, 2015 at 21:43
  • $\begingroup$ I suspect that it might not be difficult to construct specific counter examples but I want to understand a bit better what's going wrong ... Specifically is this because of the euclidean metric and if so can one characterize different metrics for which L^2 harmonic functions in (S,g) would be the trivial function $\endgroup$
    – Ali
    Jul 7, 2015 at 21:46
  • $\begingroup$ @Ali: I'm not sure I understand your objections; maybe my comment was slightly misleading. I'm convolving with the Poisson kernel for upper half space. This gives me $\sim r^{-3}$ decay in the unrestricted variables; see here: en.wikipedia.org/wiki/Poisson_kernel#On_the_upper_half-space $\endgroup$ Jul 7, 2015 at 22:16

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For a concrete example, the dipole potential $$ \dfrac{x_1}{(x_1^2 + x_2^2 + x_3^2)^{3/2}}$$ is harmonic and $L^2$ on $a < x_1 < b$ if $0 < a < b$. Higher multipole potentials such as $$ \dfrac{2 x_1^2 - x_2^2 - x_3^2}{(x_1^2 + x_2^2 + x_3^2)^{5/2}}$$ are harmonic and $L^2$ on $a < x_1 < \infty$.

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  • $\begingroup$ very nice examples basically you just take the green function and differentiate it further and further to make it in L^2 or any L^p space. $\endgroup$
    – Ali
    Jul 8, 2015 at 17:13

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