Composition operator is bounded on Bergman space for $p=2$

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?

Hint: You can use closed graph theorem. That is, show that if $$f_n\to f$$ and $$C_\varphi f_n\to g$$, then $$g=C_\varphi f$$.
• I also tried this but can't see how! I even assumed that $f_n\to0$ and that $C_\varphi(f_n)\to g$ and tried to show that $g=0$, which seems even easier (and by linearity is of course allowed) – JustDroppedIn Feb 23 at 23:27
• @JustDroppedIn Ah, I'm not sure where you were stuck... but I guess a good start is to show that $f_n\to f$ in $L^2_a$ implies that $f_n\to f$ locally uniformly. Additional hint is to apply Gauss mean value theorem to get a pointwise bound in terms of its $L^2$-norm. (cf: Gauass mean value theorem says, for example $$f(0)=\frac1{\pi r^2}\int_{x^2+y^2\le r^2}f(x+iy)\mathrm dx\mathrm dy.$$ – Song Feb 23 at 23:33