2
$\begingroup$

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?

$\endgroup$

1 Answer 1

2
$\begingroup$

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 $1 < p < \infty$.

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – Konstantin
    Commented Mar 1, 2017 at 19:23
  • $\begingroup$ $\int_{0}^{2\pi}P(r,\theta-\theta')d\theta'=1$ for all $0 \le r < 1$ and $0 \le \theta\le 2\pi$. $\endgroup$ Commented Mar 2, 2017 at 3:31
  • 1
    $\begingroup$ @Konstantin : I added a bit more at the end, plus the comment above. $\endgroup$ Commented Mar 2, 2017 at 5:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .