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Maybe this is rather obvious, but I'm stuck. Let's consider the Laplace equation in the upper half plane with boundary condition $g$, $i.e.$ $$ \Delta u(x,y)=0, u(x,0)=g(x). $$ Then the solution is given by the Poisson integral, $$ u(x,y)=P_y*g. $$ Then I know some pointwise bounds if $g$ is good enough. The question is:

If $g\in L^p(\mathbb{R}),$ $1<p<\infty$, can I say $u\in L^p(\mathbb{R}\times\mathbb{R}^+)$?

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Equation $\Delta u=0$ is called the Laplace equation, btw.

Edit. The answer to your question is no. Consider $f(z)=1/(z+i)$. On the real line it belongs to $L^p$ with any $p>1$. In the upper half-plane it does not belong to any $L^p$ if $p\leq 2$. If you want a real function, take a real or imaginary part of this.

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  • $\begingroup$ Is your $f=P_y*g$ for some $g$? Maybe the result is not true for every $p$. Is the result true for $p=2$, i.e., given $g\in H^1(\mathbb{R})$, is $P_y*g\in L^2(\mathbb{R}\times\mathbb{R}^+)$? $\endgroup$
    – guacho
    Commented Mar 6, 2014 at 16:26

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