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I'm going crazy with this: $\mathbb{H}$ is the upper half plane. $f\in L^p(\mathbb{R})$, $u(z)$ is the poisson integral of $f$. $u(z)=(P_y*f)(x)$ with $z=(x,y)\in\mathbb{H}$ and $P_y(x)=\frac{y}{\pi(x^2+y^2)}$ the poisson kernel. $\mu$ a positive measure on $\mathbb{H}$ I want to prove that those are equivalent:

$(i)$ For $1<p<+\infty$ and $\forall f\in L^p(\mathbb{R})$, $u(z)\in L^p(\mathbb{H};\mu)$

$(ii)$ For $1<p<+\infty$ $$\int_{\mathbb{H}}|u(z)|^p\, d\mu\le C_p\int_{\mathbb{R}}|f|^p\, dt$$ with $f\in L^p(\mathbb{R})$ and $C_p$ costant.

It's clear that $(ii)\Rightarrow(i)$, but i can't see why $(i)\Rightarrow (ii)$

The book "Bounded analytic function" by John Garnett reccomend to use the closed graph theorem, but i can't see how.

Davide Vecchi
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When you say you don't see how to use CGT here it's not clear whether you don't see what needs to be done to apply CGT to the problem or you understand what needs to be done but don't see how to do it. Regarding what needs to be done:

Assume (i). We can then define a linear map $T:L^p(\Bbb R)\to L^p(\mu)$ by $$Tf(x+iy)=f*P_y(x).$$According to CGT, to show $T$ is bounded you can assume that $f_n\to f$ in $L^p$ and that $Tf_n\to u$ in $L^p(\mu)$, and you then need to show that $Tf=u$.

Hint regarding how to show $Tf=u$: If $1/p+1/p'=1$ then $P_y\in L^{p'}(\Bbb R)$. So $f_n\to f$ in $L^p$ implies...

Edit: Having revealed all in a comment I may as well give the argument here: We have $$|Tf_n(x,y)-Tf(x,y)|\le||f_n-f||_p||P_y||_{p'};$$hence $Tf_n\to Tf$ pointwise. Since $Tf_n\to u$ in $L^p(\mu)$ some subsequence tends to $u$ a.e.$[\mu]$; hence $Tf=u$ a.e.$[\mu]$.

David C. Ullrich
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