Alternative definition of Hardy spaces

Question

Classically, Hardy spaces $H^p$on the disk are introduced as set of functions analytic on $\mathbb{D} = \{z \in \mathbb{C}: |z|<1\}$, which has bounded $H^p$ norm: $$ \|f\|_{H^p} = \sup_{0\leq r < 1} \left( \frac{1}{2\pi} \int_{0}^{2\pi} |f(r \, e^{\mathrm{i} \theta})|^p \, d \theta \right)^{1/p}, \text { for } 1 < p \leq \infty, \quad \|f\|_{H^\infty} = \sup_{z \in \mathbb{D}} |f(z)| .$$

Now for the case $p=\infty$ the $\|\cdot\|_{H^\infty}$ norm coincides with $\sup_{|z|=1} |f(z)|$ for all analytic functions on $\mathbb{D}$ (due to the maximum modulus principle).

Also, for $p<\infty$ and it can be shown for $f \in H^p$ (see e.g. the book by Duren) $$ \|f\|_{H^p} = \sup_{0\leq r < 1} \left( \frac{1}{2\pi} \int_{0}^{2\pi} |f(r \, e^{\mathrm{i} \theta})|^p \, d \theta \right)^{1/p} = \lim_{ r \rightarrow 1} \left( \frac{1}{2\pi} \int_{0}^{2\pi} |f(r \, e^{\mathrm{i} \theta})|^p \, d \theta \right)^{1/p} = \left( \frac{1}{2\pi} \int_{0}^{2\pi} |f( \, e^{\mathrm{i} \theta})|^p \, d \theta \right)^{1/p}. \quad (*)$$

Shouldn't it be possible to define the Hardy spaces rightaway as those analytic functions in $\mathbb{D}$ that have bounded $L_p$ norm on $\{|z|=1\}$, i.e. $(*) \leq \infty$.?

If yes, can you give me a reference where this definition is used?

Answer 1

Consider the function $f(z) = \exp{(1+z)/(1-z)}.$ This function has modulus $1$ on the boundary except at $1.$ So therefore $f\in H^\infty?$ Hardly. Look at $f(r)$ as $r\to 1^-.$

Answer 2

Any counterexample for $p=\infty$ is automatically a counterexample for all $p>0$. You have $f$ which is not $H^\infty$, but which has boundary values in $L^\infty$. Hence the boundary values are in $L^p$. And hence $f$ is not in $H^p$, because of the result that says if $f$ is in $H^p$ and the boundary values are in $L^\infty$ then $f$ is in $H^\infty$.

Answer 3

An alternative definition of $H^{p}(D)$ for $1 \le p \le\infty$ is as the set of $f \in L^{p}(\mathbb{T})$ for which $$ \int_{0}^{2\pi}f(e^{it})e^{int}dt=0,\;\;\; n = 1,2,3,\cdots. $$ The Poisson integral of $f$ then gives a function $F$ that (a) is holomorphic in $D$ because of the absence of the above terms; and (b) has the required boundary continuity that comes from the Poisson integral. That's very close to what you want.