Let $A(D)$ be the space of holomorphic functions on the open unit disk $D$ and continuous on the closed disk $\bar{D}$. Then $A(D)$ is a Banach space if we set $\|f\|=\sup\{|f(z)|:z\in\bar{D}\}$. For $f\in A(D)$ write $$ f(z)=\sum_{n=0}^{\infty}a_nz^n. $$ a) Let $\{c_n\}_{n=0}^{\infty}$ be a sequence of complex numbers with the property that if $f\in A(D)$ represented as above, and $$ f^*(z)=\sum_{n=0}^{\infty}c_na_nz^n, $$ then $f^*\in A(D)$. Prove that there exists $C>0$ such that $\|f^*\|\leq C\|f\|$ for all $f\in A(D)$.
I am thinking that I need to prove $T:A(D)\to A(D)$ with $T(f)=f^*$ is bounded (or continuous). Do you have any other ideas?
b) Let $\omega\in D$. Prove that there exists a Borel measure $\mu_{\omega}$ on $\partial D=\{z:|z|=1\}$ such that $$ f(\omega)=\int_{\partial D}f(t)d\mu_{\omega}(t),\;f\in A. $$
I am thinking that we can use the Maximum Modulus principle, but is the measure $\mu_{\omega}$ unique? Could you please give me some ideas?
Thank you so much.
For a), use closed graph theorem to show $T$ is bounded. It suffices to show if $f_k \to f$ and $Tf_k = f_k^* \to g$ in $A(D)$, then $g= Tf = f^*$. It follows from Cauchy's integral formula $$ a_n = \frac{1}{2\pi i}\int_{|z|=r} \frac{f(z)}{z^{n+1}}dz $$ and uniform convergence of $(f_k)$ and $(f^*_k)$. (Hint: compare coefficients of power series expansions of $f^*$ and $g$.)
For b), let $B(\mathbb{T})$ be a family of $g\in C(\mathbb{T})$ such that $g = f|_\mathbb{T}$ for some $f\in A(D)$. (Here, $\mathbb{T}=\partial D$.) Then, $B(\mathbb{T})$ is a subspace of $ C(\mathbb{T})$. Let $$ \phi_\omega: g\in B(\mathbb{T}) \mapsto f(\omega)\in \mathbb{C}. $$ Then, $\phi_\omega$ is a well-defined linear functional (since there is unique $f\in A(D)$ such that $g = f|_\mathbb{T}$) with $\|\phi_\omega\|_{B(\mathbb{T})\to \mathbb{C}}\leq 1$ by maximum modulus theorem. By Hahn-Banach theorem, there exists an extension $\Phi_\omega : C(\mathbb{T}) \to \mathbb{C}$ of $\phi_\omega$ with $\|\Phi_\omega\|_{C(\mathbb{T})\to \mathbb{C}}=\|\phi_\omega\|_{B(\mathbb{T})\to \mathbb{C}}$. Now, by Riesz representation theorem, we know there exists a finite Borel measure $\mu_\omega$ on $\mathbb{T}$ such that $$ \Phi_\omega(f|_\mathbb{T})=f(\omega) = \int_\mathbb{T} f|_\mathbb{T}(t)d\mu_\omega(t)=\int_\mathbb{T} f(t)d\mu_\omega(t) $$ for all $f \in A(D)$.