I know the question below is a known result but, I would need some help to prove it!
Well, I know that in the Poisson integral induces an isometric isomorphism between $L^p(\mathbb{T})$ and the Hardy space $h^p(\mathbb{D})$ for $p>1$. I'm reading Function Spaces and Partial Differential Equations: Classical analysis and this question is the remark 5.24 but it's not proved.
Now how can I do to prove this?
Thanks.
EDIT: The definition of $h^p$ that I have is this: $h^p(\mathbb{D})=\{u\in Har{\mathbb{D}: ||u||_{h^p}=sup_{0<r<1}M_{p}(u,r)< \infty}$}, where the $Har\{\mathbb{D}\}$ are the group of harmonic functions while $M_{p}(u,r)=(\int_{-\pi}^{\pi}|u(re^{it})|^p\frac{dt}{2 \pi})^\frac{1}{p}$.
It's a good thing I asked for the definition, because it turns out that what you mean by $h^p$ is not what I thought you meant! FYI the reason I was thrown off is that $h^p$ is not what's usually called a Hardy space; I could tolerate calling it a "harmonic Hardy space".
Say $u=P[f]$ and define $$u_r(t)=u(re^{it}).$$Then $u_r=f*P_r$ (where the $*$ denotes convolution), so if $f\in L^p(\Bbb T)$ then $$||u_r||_{L^p(\Bbb T)}\le||P_r||_1||f||_p=||f_p||.$$ So $||u||_{h^p}=\sup_r||u_r||_p\le ||f||_p$.
On the other hand, we know that $u_r\to f$ almost everywhere, so Fatou's Lemma shows that $$||f||_p\le\liminf_r||u_r||_p\le\sup_r||u_r||_p.$$