Let $h^{p}(D)=\{u ;u \ $harmonic in D and bounded means of order p$ \}$
Its well known that the Hardy space "sits nice" in $L^{2}$ by considering Fourier series. Do we have analagous result for $h^{p}$ with $p>1$?
I.e can we go back and forth between $h^{p}(D)$ and ALL of $L^{p}(T)$? (Not just a subset as in the Hardy space case.) I understand so far that all of $h^p$ are possion integrals of its non tangential limit functions, which given $u \in h^p$ exists unique in $L^p$. On the other hand any possion integral of a $L^p$ function is harmonic in the disc and has bounded means of order $p$ i.e an $h^p$ functions.
Also the $L^{p}$ norm of the boundardy function and the hardy norm of the function coincide which make me think we have some kind of isometry.
This has a rather easy answer I think, after one is beginning to grasp this theory.
Let $Pu$ be the possion integral of $u\in L^{p}$. Then $P$ is surjective isometry, i,e a bijection from $L^p$ to $h^p$, this is Theorem 11.30 in Rudin's Real and complex analysis.
Furthermore we have Fatou's theorem providing us with the fact that its inverse is given by the non-tangential limits. Fatous theorem is 11.24 in the same book.
Since this is my own answer please downvote and comment its not correct.