I'm trying to prove the following:
Suppose that $\varphi:\mathbb{D}\rightarrow\mathbb{D}$ is an analytic function (where $\mathbb{D}$ is the open unit disk) with the property that $f\in L_a^{2}(\mathbb{D})$ implies $f\circ\varphi\in L_a^{2}(\mathbb{D})$. Define $C_{\varphi}:L_a^{2}(\mathbb{D})\rightarrow L_a^{2}(\mathbb{D})$ by $C_{\varphi}(f)=f\circ\varphi$. Show that the composition operator $C_{\varphi}$ is a bounded linear operator on $L_a^{2}(\mathbb{D})$.
Theorem 2.1.1 of this paper shows a similar proof of general measure space, and also this MSE post is similar but in $[0,1]$ instead of $\mathbb{D}$. However, I was not able to fully figure out how to find the proof of the above statement. How does one prove this?
Best Answer
Assuming that $L^2_a(\mathbb{D})$ is the Bergman space, i.e. $$\{f:\mathbb{D}\to\mathbb{C}\vert f\text{ is holomorphic and }\int_\mathbb{D}|f(z)|^2dA<\infty\} $$ then this is a direct consequence of the closed graph theorem. We have that $L^2_a(\mathbb{D})$ is complete, so in order to show that the well-defined linear operator $C_\varphi$ is bounded, we can apply the closed graph theorem.
Here is the standard application: Let $\{f_n\}\subset L^2_a(\mathbb{D})$ be a sequence with $f_n\to0$ and assume that $C_\varphi(f_n)\to g\in L^2_a(\mathbb{D})$. If we can show that $g=0$, then we will have that $C_\varphi$ is continuous.
Note that since $C_\varphi(f_n)\to g$ in $L^2$-norm, there exists a subsequence $\{C_\varphi(f_{n_k})\}$ that converges pointiwse (almost everywhere, but these are all continuous, hence everywhere) to $g$, so $$g(z)=\lim_{k\to\infty}f_{n_k}\circ\varphi(z)$$ for all $z\in\mathbb{D}$. But since $f_n\to0$ in $L^2$ norm we also have that $f_{n_k}\to0$ in $L^2$ norm, so there exists a further subsequence $f_{n_{k_j}}$ that converges almost everywhere to $0$ (which again, since everything here is continuous is the same as "everywhere"), so $$\lim_{j\to\infty}f_{n_{k_j}}(w)=0$$ for all $w\in\mathbb{D}$. In particular, for $w=\varphi(z)$ we have that $$\lim_{j\to\infty}f_{n_{k_j}}(\varphi(z))=0$$ but $\{f_{n_{k_j}}(\varphi(z))\}$ is a subsequence of $\{f_{n_k}(\varphi(z))\}$ which converges to $g(z)$. Therefore $\{f_{n_{k_j}}(\varphi(z))\}$ also converges to $g(z)$, and this shows that $g(z)=0$. As $z$ was arbitrary, this shows that $g=0$.
By the closed graph theorem, we can conclude that $C_\varphi$ is a bounded operator.