Is there an "easy" proof of the existence of radial limits of functions in the hardy spaces $H^p$?

Question

Let $HOL(\mathbb{D})$ be the analytic functions defined on the unit disk and for $1\leq p \leq \infty$, $$H^p = \{f\in HOL(\mathbb{D}) : \lim_{r\nearrow 1} \int_{rC}|f(z)|^p \frac{dz}{2\pi i} < \infty \} $$

We know that for $f\in H^p$ the radial limits, $\lim_{r\nearrow 1} f(re^{i\theta})$ exists almost everywhere. Is there an "easy" proof for this?

I think the steps are

1. $f\in H^p \implies f\in H^1$ and $f$ is harmonic (ok, no problem here).

2. There is a unique measure $\mu$ on the circle $C$ such that $f(re^{i\theta}) = \int_\theta P(r,\theta) d\mu$ where $P$ is the Poisson Kernel. (can someone help me with this?)

3. Write $\mu(\theta) = g(\theta) d\theta + d\mu_\perp $ and use this work to show that as $r\nearrow 1$, $f(re^{i\theta})$ converges to $g(\theta)$ on supp$(\mu_\perp)^c$.

Is this correct and is there an easier proof?

Answer 1

As far as I know - from reading several texts on Hardy spaces - all proofs of this fact utilize Fatou’s theorem (or a variation of).

Using weak/weak-* compactness of the unit ball in $L^p(C)$, one can show directly that “the boundary function” $f^*$ exists, i.e.

1. $f^*$ is in $L^p(C)$,

2. $f^*$ is a limit (in the norm for $1 \leq p < \infty$, and weak-* for $p = \infty$) of the functions $f_r(w) := f(rw); \; w \in C, \; 0 < r < 1$,

3. $\widehat{f^*}(n) = 0$, for all $n \in \mathbb{Z}_{< 0}$,

4. $\|f^*\|_{L^p(C)} = \|f\|_{H^p(\mathbb{D})}$, and

5. $f_r = f^* * P(r, \cdot)$.

See Theorem 2.2.2 from Hardy Spaces, by N. Nikolski for a proof of this fact. As Nikolski then mentions in the beginning of Section 2.5, this is not sufficient to obtain radial limits a.e. on $C$.

See also this math stackexchange discussion, and note that for $1 \leq p < \infty$ we get using $L^p(C)$ convergence that some subsequence $f_{r_k}$ converges to $f^*$ a.e. On $C$.

Of course, this is not to say that an “easier” proof cannot/does not exist.

Answer 2

Based on the previous answer's sketch, I tried to fill in the details. I think this works. If anyone sees an error, please correct it.

Recall $H_p(D)\subseteq H_1(D),\ \forall \ 1\leq p\leq \infty.$ Fix $f \in H_{1}(D)$ and assume $f_{r} \in L_{1}$. Now set $$ \lambda_{r}(g) := \int f(re^{i\theta})g(e^{i\theta})\frac{d\theta}{2\pi} $$ a linear functional on $L_1([0, 2\pi])$. Next, $|\lambda_{r} g| \leq \|g\|_{\infty} \|f_{r}\|_{1},$ and so $\|\lambda\|\leq \|f\|_1$. This implies that $\lambda_{r} \in \{\lambda \in L_{\infty}^*(C): \|\lambda\| \leq \|f\|_1 \} \subseteq L_{\infty}^{\ast}(C)$, a closed neighborhood of the dual space of $L_{\infty}$. The Banach-Alaoglu theorem implies that $\{ \|\lambda\| \leq \|f\|_1 \}$ is weak$^*$ compact, which furnishes a subsequence $r_n\nearrow 1$ and $\lambda^* \in \{ \|\lambda\| \leq \|f\|_1 \}$ such that $\lambda_{r_n}(g) \to \lambda^*(g) \ \forall \ g\in L_\infty$.
Now from the Riesz representation theorem, $L_{\infty}^{\ast}$ is the set of bounded, finitely additive Borel measures on $L_{\infty}$ citation Dunford-Schwartz, Linear Operators I, Theorem IV.8.16, page 296. So we have a bounded, finitely additive measure, $\mu$ such that $$ \int f_{r_{k}}(e^{i\theta}) g(e^{i\theta})\tfrac{d\theta}{2\pi} = \lambda_{r_{k}}(g)\to \lambda(g) = \int g(e^{i\theta}) d\mu(\theta). $$ Now from the Radon-Nikodym theorem, $d\mu(\theta) = f^{\ast}(e^{i\theta})\frac{d\theta}{2\pi} + d\mu^{\perp}$ where $d\mu^{\perp} \perp d\theta$ ($d\mu^\perp$'s support is Lebesgue measure 0 and conversely).
But then it must be the case that $d\mu^{\perp}$ is the zero measure. If this were not the case, then $$ g = \begin{cases} |d\mu^{\perp}|(\theta) & \theta \in supp(d\mu^{\perp}), \\ 0 & \text{elsewhere} \end{cases} \quad \in L_{\infty}, $$ as $g = 0$ a.e. $d\theta$. Consequently, $ 0 = \lambda(g) = 0 + |\mu^{\perp}([0, 2\pi])|^{2}$ and $$ \lambda(g) = \int g d\mu = \int gf^{\ast} \tfrac{d\theta}{2\pi} = \lim_{r_{n} \rightarrow 1} \int gf_{r_{n}}\tfrac{d\theta}{2\pi}. $$ This implies that $$ \lim_{r_{n} \rightarrow 1} \left| \int g(f_{r_{n}} - f^{\ast}) d\mu \right| = 0. $$

Fix $r_{0}e^{i\theta_{0}} \in D$ and we have $$ f_{r_{0}}(e^{i\theta_{0}}) = \int f_{r}(e^{i\varphi}) P_{r_{0}}(\theta_{0} - \varphi) \frac{d\varphi}{2\pi} $$ for any $r$ with $r_{0} < r < 1$. In particular, for any $r_k$ such that $\lim_{k \rightarrow \infty} r_k = 1$ \begin{equation} \label{eqn:f*-poisson} f_{r_{0}}(e^{i\theta_{0}}) = \lim_{k \rightarrow \infty} f_{r_{0}}(e^{i\theta_{0}}) = \lim_{k \rightarrow \infty} \int f_{r_{k}}(e^{i\varphi}) P_{r_{0}}(\theta_{0} - \varphi) \frac{d\varphi}{2\pi} = \int f^{\ast}(e^{i\varphi})P_{r_{0}}(\theta_{0} - \varphi)\frac{d\varphi}{d\pi}. \end{equation} Fix $\epsilon > 0$. Then Luzin's theorem implies that there exists a compact subset $E_{\epsilon} \subseteq C$ such that $|E_{\epsilon}| > 2\pi - \epsilon$ and $f^{\ast}|_{E_{\epsilon}}$ is continuous. Seeking the radial limit we let $r_0\to 1$ and observe $$ \lim_{r_0\to 1} f_{r_{0}}(e^{i\theta_{0}}) = f^{\ast}(e^{i\theta_{0}}) \qquad \text{ for all } \theta \in E_{\epsilon}, $$ since $P_{r_{0}}$ is an approximation to the identity. Since $\epsilon$ was determined arbitrarily, $|\{\theta \in [0,2\pi] : \lim f_{r}(e^{i\theta}) \neq f(e^{i\theta}) \}| = 0.$ This shows the radial limits exist a.e., property (c) of the theorem follows from Eq. \ref{eqn:f*-poisson}.

To prove (a), $$ \hat{f}^{\ast}(k) = \int f^{\ast}(e^{i\theta})e^{-ik\theta} d\mu = \lim_{n \rightarrow \infty} \int f_{r_n}(e^{i\theta})e^{ik\theta} = \hat{f}(k)\ . $$

To verify (b), extend $f^*$ analytically to $D$ via $$ f^{\ast}(z) = \int_{\omega \in C_{1}(0)} \frac{f^{\ast}(\omega)}{\omega - z}\frac{d\omega}{2\pi i} $$ and note that $$ f^{\ast}(z) = \lim_{n \rightarrow \infty} \int_{\omega \in C_{1}(0)} \frac{f_{r_{n}}(\omega)}{\omega - z}\frac{d\omega}{2\pi i} = \lim_{n \rightarrow \infty}{f_{r_n}(z)}. $$