# Equivalence of the Hardy spaces on the unit disk and the unit circle

The Hardy space $H^p$ is the vector space of holomorphic functions $f$ on the open unit disk that satisfy: $$\sup_{0<r<1}\left(\frac{1}{2\pi} \int_0^{2\pi}\left|f \left (re^{i\theta}\right )\right|^p \; \mathrm{d}\theta\right)^\frac{1}{p}<\infty.$$

The Poisson kernel for the unit disc is: $$P_{r}(\theta )=\sum _{n=-\infty }^{\infty }r^{|n|}e^{in\theta }={\frac {1-r^{2}}{1-2r\cos \theta +r^{2}}}=\operatorname {Re} \left({\frac {1+re^{i\theta }}{1-re^{i\theta }}}\right),\ \ \ 0\leq r<1.$$ Question: How is it possible to demonstrate the equivalence of the Hardy spaces on the unit disk and the unit circle using the Poisson kernel?

You cannot demonstrate such an equivalence for $p=1$ or for $p=\infty$ as an $L^p$ radial limit. If $f$ is a function that is Harmonic on the unit disk with $L^1$ norms uniformly bounded on concentric cirlces, then $f$ corresponds with a finite Borel measure on the unit circle. For $p=\infty$, you can allow $L^{\infty}$ boundary functions, but you don't get $L^{\infty}$ convergence of the functions on concentric circles unless the boundary function is continuous. But you do get the isometric correspondence between the $L^{\infty}$ boundary function and the $H^{\infty}$ Hardy class function.
For $1 < p < \infty$, a function $f$ that is harmonic on the open unit disk and satisfies $\sup_{0 < r < 1} \int_{0}^{2\pi}|f(re^{i\theta})|^pd\theta < \infty$ is the Poisson integral of a unique $L^p$ function. As you might expect, this requires a weak argument, which is why the endpoint cases $p=1,p=\infty$ don't work out the same way. For $1 < p < \infty$, there is a weakly convergent subsequence $g_n(\theta) = f(r_n e^{i\theta})$ that converges to some $g\in L^p[0,\pi]$. Because the Poisson kernel is in $L^p$ for $1 \le p \le \infty$ on any interior circle, you obtain $$f(rse^{i\theta}) = \frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-r^2}{1-2r\cos(\theta-\theta')+r^2}f(se^{i\theta'})d\theta', \\ 0 < r,s < 1,\;\;\; 0 \le \theta \le 2\pi.$$ Using the weak limit gives $$f(re^{i\theta})=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-r^2}{1-2r\cos(\theta-\theta')+r^2}g(\theta')d\theta'.$$ By writing the kernel as $$P(r,\theta) = \frac{1}{2\pi}\frac{1-r^2}{1-2r\cos(\theta)+r^2}$$ and applying Holder's inequality for $1 < p < \infty$ with $P(r,\theta)=P(r,\theta)^{\frac{1}{p}}P(r,\theta)^{\frac{1}{q}}$, you obtain $$|f(re^{i\theta})|^p \le \left(\int_{0}^{2\pi}P(r,\theta-\theta')d\theta'\right)^{\frac{p}{q}} \int_{0}^{2\pi}P(r,\theta-\theta')|g(e^{i\theta'})|^pd\theta' \\ |f(re^{i\theta})|^p \le \int_{0}^{2\pi}P(r,\theta-\theta')|g(e^{i\theta'})|^pd\theta'$$ Integrating in $\theta$ gives $$\int_{0}^{2\pi}|f(re^{i\theta'})|^pd\theta' \le \int_{0}^{2\pi}|g(e^{i\theta'})|^p d\theta'$$ Hence, $\|f\|_{H^p(\mathbb{D})} \le \|g\|_{L^p(\mathbb{T})}$. Because $g$ is a weak limit of radial slices of $f$, the opposite inequality can be shown, and that's how you end up with an isometric correspondence between the radial boundary function and the hardy space function. If $f_r(e^{i\theta})=f(re^{i\theta})$, then $\|f_r\|_{L^p} \le \|g\|_{L^p}$ and $g$ is the weak $L^p$ limit of $f_{r_n}$ for some sequence $\{ r_n \}$ tending up to $1$. Consequently, \begin{align} \|g\|_{L^p} & = \sup_{\|h\|_{L^{q}}=1}|\langle g,h\rangle| \\ & =\sup_{\|h\|_{L^{q}}=1}\lim_n |\langle f_{r_n},h\rangle| \\ & \le \sup_{\|h\|_{L^{q}}=1}\lim_n \|f_{r_{n}}\|_{L^p}\|h\|_{L^q} \\ & \le \sup_{\|h\|_{L^{q}}=1}\|f\|_{H^p}\|h\|_{L^q} \\ & = \|f\|_{H^p}\sup_{\|h\|_{L^{q}}=1}\|h\|_{L^q} = \|f\|_{H^p}. \end{align} Therefore $\|f\|_{H^{p}} = \|g\|_{L^p}$ for $1 < p < \infty$.
• Thank you for the great reply! Could you please explain: - why we care only about weak convergence for $1<p<\infty$?, - how do we know that we can cut $(\int_{0}^{2\pi}P(r,\theta-\theta ')d\theta ')^\frac{p}{q}$, is it less than 1? - could you elaborate a bit on "Because $g$ is a weak limit of radial slices of $f$, the opposite inequality can be shown"? Sorry for so many additional questions. Commented Mar 1, 2017 at 19:23
• $\int_{0}^{2\pi}P(r,\theta-\theta')d\theta'=1$ for all $0 \le r < 1$ and $0 \le \theta\le 2\pi$. Commented Mar 2, 2017 at 3:31