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Call an analytic function $f(z)=\sum_{n\geq 0}a_n z^n$ defined on unit circle $\mathbb{D}$ Bloch function if and only if $$||f||_B=\sup_{|z|<1}(1-|z|^2)|f'(z)|<\infty$$ Then the problem is to find a constant independent of $f(z)$ such that $$\sup_{n\geq 1}|a_n|\leq C ||f||_B$$

Intuitively I tried Cauchy's formula for $f'(z)=\sum_{n\geq 0} (n+1)a_{n+1}z^n$ which implies $$(n+1)a_{n+1}={1 \over {2\pi i}}\int_{|z|=\delta<1}\frac{f'(z)}{z^{n+1}}\ dz, \forall n=1,2...$$ then $$|a_{n+1}|\leq \frac{1}{(1-\delta^2)\delta^n (n+1)}||f||_B$$ which seems to be too sparse. But I do not how I can improve the estimate. Maybe there is some kind of inequality about $f'(z)$ and $(1-|z|^2)$ that can be derived by Schwarz's lemma but I still have no idea of where to go.

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Substituting $\delta:=\sqrt{1-\frac 1 {n+1}}$ in $$|a_{n+1}|\leq \frac{1}{(1-\delta^2)\delta^n (n+1)}||f||_B,$$ we come to the estimate $$ |a_{n+1}| \leq \frac 1 {\left(1- \frac 1 {n+1}\right)^{n/2}}||f||_B \le e^{\frac 1 2 }||f||_B . $$ Also see the paper.

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  • $\begingroup$ Thank you! I did not realize that I could take different $\delta$ with respect to $n$ which makes me doubt if this approach would work. Thank you ! $\endgroup$ – Roy Han Aug 9 '13 at 14:22

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