# $|f(z)-p_n(z)|\le (n+2)|z|^{n+1}$ with $\deg(p_n(z))\le n$ over the unit disk

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 $$\frac{n+1}{n+2}<|z|<1$$ ?

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

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 $$\frac{1}{2\pi}\int_0^{2\pi}{|f(re^{i\theta})|^2}d{\theta}=\sum{|a_k|^2r^{2k}},\ 0 so $$|f(z)| \le 1$$ implies trivially $$|a_k|^2r^{2k}\le 1, 0, 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 $$0, hence by maximum modulus for $$|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!