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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?

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Hint: You can use closed graph theorem. That is, show that if $f_n\to f$ and $C_\varphi f_n\to g$, then $g=C_\varphi f$.

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  • $\begingroup$ 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) $\endgroup$ – JustDroppedIn Feb 23 at 23:27
  • $\begingroup$ @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 $$f(0)=\frac1{\pi r^2}\int_{x^2+y^2\le r^2}f(x+iy)\mathrm dx\mathrm dy.$$ $\endgroup$ – Song Feb 23 at 23:33

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