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Let $ f(z) $ be a holomorphic function over the unit disk $ \Delta:=\{ z\in \mathbb C: |z|<1 \} $. Suppose that $ |f(z)|\le 1 $ for $ z\in\Delta $. Show that for any positive integer $ n $, there is a polynomial $ p_n(z) $ of degree at most $ n $ such that $$ |f(z)-p_n(z)|\le (n+2)|z|^{n+1} $$ for any $ z\in \Delta $.


My attempt:

I notice that if we take $ p_n(z)=\sum_{k=0}^n\frac{f^{(k)}(0)}{k!}z^k $, then \begin{align} \left|f(z)-\sum_{k=0}^n\frac{f^{(k)}(0)}{k!}z^k\right|&=\left| \sum_{k=n+1}^\infty\frac{f^{(k)}(0)}{k!}z^k \right|\\ &\le\sum_{k=n+1}^\infty |z|^k\\ &\le\frac{|z|^{n+1}}{1-|z|} \end{align} If $ |z|\le\frac{n+1}{n+2} $, we have proved $ |f(z)-p_n(z)|\le (n+2)|z|^{n+1} $, but what about $ \frac{n+1}{n+2}<|z|<1 $ ?

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  • $\begingroup$ How did you get rid of $f^{k}(0)/k!$ in the first inequality (from first to second displayed line)? $\endgroup$ Commented Apr 1, 2019 at 14:30
  • $\begingroup$ @uniquesolution By Cauchy inequalities. $\endgroup$
    – Bach
    Commented Apr 1, 2019 at 14:52

1 Answer 1

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The way to do it is like this: let $f(z)=\sum{a_kz^k}, p_n(z)=\sum_{0\le k \le n}a_kz^k$; note that $$\frac{1}{2\pi}\int_0^{2\pi}{|f(re^{i\theta})|^2}d{\theta}=\sum{|a_k|^2r^{2k}},\ 0<r<1, $$ so $|f(z)| \le 1$ implies trivially $ |a_k|^2r^{2k}\le 1, 0<r<1 $, so taking $r \to 1$ we get $|a_k| \le 1$ for all $ k $.

Let $$g(z)=\frac{f(z)-p_n(z)}{z^{n+1}},$$ $g$ analytic in the unit disc; fix $0<r<1$, hence by maximum modulus for $|z|\le r,$ \begin{align} |g(z)| &\le \sup_{|w|=r}{\frac{|f(w)-p_n(w)|}{r^{n+1}}} \\ &\le \sup_{|w|=r}{\frac{|f(w)|+|p_n(w)|}{r^{n+1}}}\\ & \le \frac{1+\sum_{0\le k \le n}|a_k|}{r^{n+1}} \\ &\le \frac{n+2}{r^{n+1}}.\end{align} Letting $r \to 1$ we get $|g(z)| \le n+2$ in the whole (open) unit disc and we are done!

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  • $\begingroup$ Nice! Thank you! $\endgroup$
    – Bach
    Commented Apr 9, 2019 at 1:03
  • $\begingroup$ You are welcome $\endgroup$
    – Conrad
    Commented Apr 9, 2019 at 1:50

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