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This is a problem from my past Qual.

"Let $D$ denote the unit disk and $f:D\to D$ be analytic. Show that there exists a sequence $n_i$ s.t. $f^{n_i}(z)$ converges pointwise for all $z\in D$. Here $f^n=f\circ f\circ\ldots\circ f$ ($n$ times)."

I have no idea how to start. I have an analytic function, so I have its Taylor series in a small neighborhood, I know the Cauchy-Riemann equations. That's it. I mean usually when I deal with $f^n$, I study $f$. In this case it seems I don't have a lot of information to study $f^n$. So I'm stuck here.

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  • $\begingroup$ Did you mean $f^n(z)-(f(z))^n$? or $f^n(z)=(f^{n-1}\circ f)(z)$, $f^1=f$, In the former, isn't it trivial? $|f^n(z)|=|f(z)|^n\xrightarrow{n\rightarrow\infty}0$ since $|f(z)|<1$. $\endgroup$ – Oliver Diaz May 28 at 3:29
  • $\begingroup$ oh no this is composition. $\endgroup$ – T C May 28 at 3:30
  • $\begingroup$ That's what I thought. I edited your question to add composition symbols to avoid confusion. $\endgroup$ – Oliver Diaz May 28 at 3:39
  • $\begingroup$ This seems to be related to what is known as normal families. See pages 281-282 of Rudin's real and complex analysis. In particular Theorem14.6 $\endgroup$ – Oliver Diaz May 28 at 3:51
  • $\begingroup$ Do you mean that $f$ is a holomorphic function on the unit disc in $\mathbb{C}$? $\endgroup$ – WimC Jun 2 at 11:00
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Note that all coefficients of all functions $f^n$ have modulus at most $1$, by their Cauchy integral formula. Since the closed unit disc is compact, it follows that there is a subsequence $f^{n_k}$ such that their coefficients (viewed as functions $\mathbb{N}\to \overline{\mathbb{D}}$) converge pointwise to a sequence $a_0, a_1, \ldots$ in the closed unit disc. Set $g(z)=\sum_ka_kz^k$. This is a holomorphic function on the unit disc. Then show that the sequence of functions $f^{n_k}$ converges pointwise to $g$ on the unit disc. (For example, consider only the first $m$ terms in their series and derive a pointwise bound depending om $m$, then take $m \to \infty$.)

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  • $\begingroup$ Which coefficients are you referring to? Do you mean the Taylor coefficients? $\endgroup$ – Michael Burr Jun 2 at 10:13
  • $\begingroup$ @MichaelBurr Yes, the Taylor coefficients at $z=0$. $\endgroup$ – WimC Jun 2 at 10:56
  • $\begingroup$ The family described in the problem is a normal family since $f^{n}=f\circ\ldots\circ f$ ($n$ times) is in fact uniformly bound. One can either used a theorem that states that families of holomorphic functions in a region that are uniformly bounded in compact subsets are normal (meaning every subsequence has a uniform convergent subsequence etc) or use Cauchy estimates to prove Lipschitz condition and some versions of the Ascolli-Arzela theorem. $\endgroup$ – Oliver Diaz Jun 2 at 14:46

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