If$\int_0^{2\pi} |f^*(e^{i\theta})|^{1/3}d\theta <\infty$ then $f\in H^{1/3}(D) $

Question

Question If$\int_0^{2\pi} |f^*(e^{i\theta})|^{1/3}d\theta <\infty$ then $f\in H^{1/3}(D) $ where $f$ is analytic and $f^*$ is non tangential limit of $f$ $H^{1/3}(D)$ is Hardy Space.

Answer 1

This is wrong at least twice. First, $H^{1/3}$ is a space of holomorphic functions, and the functions in $L^{1/3}(D)$ are not holomorphic. So $L^{1/3}\subset H^{1/3}$ is simply not true, hence any "proof" must be wrong. (A more sensible version of the problem would be to show that $L^{1/3}\cap H(D)\subset H^{1/3}$; that's also false, not quite so obviously.)

You ask whether the second-last inequality is right. In fact the very first inequality is wrong or worse. Starting with $f\in L^{1/3}(D)$ you say $\int_0^{2\pi}|f(e^{it})|^{1/3}\,dt<\infty$. That's not so; in fact that doesn't make any sense, because there's no such thing as $f(e^{it})$ in the first place.

(If the "second-last" inequality is $\lim_{r\to 1^-} \int_0^{2\pi} |f(re^{i\theta})|^{1/3}d\theta \leq \int_0^{2\pi} |f(e^{i\theta})|^{1/3}d\theta$$ $, that's also wrong - no reason it should hold, unless for example $f$ is holomorphic...)

Edit: Turns out that what the OP really wanted to know was why a proof in a certain post works.

Note first that in that other post the author uses the same letter, $f$, for the holomorphic function $f$ and also for its non-tangential maximal function. Very bad notation, leading to confusion over whether the author is or not assuming that $f\in L^p$.

So, if $f$ is a function defined in $D$ we let $f^*(e^{it})$ denote the non-tangential maximal function of $f$.

Trivial Lemma. Suppose $f\in H(D)$ and $p>0$. Then $f\in H^p(D)$ if and only if $f^*\in L^p(\Bbb T)$.

Note that's $L^p(\Bbb T)$, not $L^p(D)$.

Proof. Well, one direction is not actually trivial: If $f\in H^p$ then it's a standard result, towards the start of any treatment of Hardy spaces, that $f^*\in L^p(\Bbb T)$.

Suppose now that $f^*\in L^p(\Bbb T)$. If $0<r<1$ the definition of $f^*$ shows that $$|f(re^{it})|\le f^*(e^{it}).$$Hence $$\int_0^{2\pi}|f(re^{it})|^p\,dt\le \int_0^{2\pi} f^*(e^{it})^p\,dt;$$so $\sup_r\int_0^{2\pi}|f(re^{it})|^p\,dt<\infty$.