Bounded holomorphic function over unit disk uniquely determines a bounded function on the boundary by taking the limit

Question

Now given $f \in H^\infty(D)$, I want to show by functional analysis that there exists $f^\star \in L^\infty(\partial D)$ such that $f$ is determined by $f^\star$ via cauchy integral over the boundary of the disk and that $\lim_{r \to 1^{-}}f(re^{i\theta}) =f^\star(e^{i\theta})$ for a.e. $\theta \in (-\pi, \pi]$.

My progress so far:

Given $|z| < r < 1$, we have by cauchy integral formula,

$$f(z) = \frac{1}{2\pi i}\int_{\partial D(0,r)} \frac{f(w)}{w-z}dw = \frac{1}{2\pi }\int_{-\pi}^\pi f(re^{i \theta})\frac{re^{i \theta}}{re^{i \theta}-z}dz$$

Consider $f_r \in L^\infty(\partial D)$ such that $f_r(\theta) = f(re^{i \theta})$ and let $f^\star \in L^\infty(\partial D)$ denote the weak-$\star$ limit of $f_r$ as $r$ goes $1$ with respect to functions in $L^1(\partial D)$. Therefore, $$ f(z) = \frac{1}{2\pi i}\int_{\partial D} \frac{f^\star(w)}{w-z}dw$$

Thus it remains to show that the limit is $f^\star$ and then I'm got stuck. My professor hints that I can use Lebesgue differentiation theorem to derive that

$$f(re^{i\theta}) = \int_{-\pi}^\pi f^\star(e^{i\theta}) \frac{e^{i \tau}}{e^{i \tau}-re^{i\theta}} d\tau$$

which thus converges to $f^\star(e^{i\theta})$ as $r \to 1$, but I'm not really sure how to get this step. Any suggestion?

Answer 1

This is known as Fatou’s theorem. In fact, for complex borel measure $\mu$ which support on $\mathbb{T}$. The Poisson integral of $\mu$, defined by $P[d\mu](z)=\int_{\mathbb{T}}\frac{1-|z|^2}{|1-\overline{z}\zeta|}dm(\zeta)$, has radical limit almost every where(respect to Lebesgue measure $dm=\frac{d\theta}{2\pi}$), which is $\frac{d\mu}{dm}$. Then for $L^{p}(\mathbb{T})$ case, $1\leq p\leq +\infty$, $f^*dm$ defined a complex measure on $\mathbb{T}$. Because $f$ is holomorphic, the Fourier coefficients of $f_r$ supports on $\mathbb{Z_{\geq 0}}$. Also note that “Take Fourier coefficients” is w*-continuous, this shows the Fourier coefficients of $f^*$ also supports on $\mathbb{Z}_{\geq 0}$. This deduce the Poisson integral and Cauchy integral of $f^*$ are coincide.

For the proof of Fatou’s theorem, see Hoffman’s Banach Spaces of Analytic Functions page 34, which is use the method of “good kernel”. For another proof using Littilewood maximal function, see Rudin’s Function Theory in Polydiscs theorem 2.3.1 page 24.