Define the Bergman space as $L_a^2(\mathbb{D})=\{f\in L^2(\mathbb{D})| f\text{ is holomorphic in the disk}\}$, where the measure is the normalized Lebesgue measure $dA/\pi$. Equip this space with the classical $L^2$ norm; it is known that this is a Banach space. Let $\varphi:\mathbb{D}\to\mathbb{D}$ be a holomorphic function such that for all $f\in L_a^2(\mathbb{D})$ it is $f\circ\varphi\in L_a^2(\mathbb{D})$. Define the composition operator $C_\varphi:L^2_a(\mathbb{D})\to L_a^2(\mathbb{D})$ by $C_\varphi(f)=f\circ\varphi$ and prove that this linear operator is bounded.
I have no idea on how to proceed. I simply write down the norm of $C_\varphi(f)$ and have no idea how to estimate it. I believe that the key lies in the information "$f\in L_a^2(\mathbb{D})\implies f\circ\varphi\in L_a^2(\mathbb{D})$" and that for suitable choices of $f$ I can get some info on $\varphi$, but still I wouldn't know how to use it, as it appears only as an argument in $f$. Could someone give me a hint?
Best Answer
Hint: You can use closed graph theorem. That is, show that if $f_n\to f$ and $C_\varphi f_n\to g$, then $g=C_\varphi f$.