The exercise I have reads as follow:
- Let $f$ be an entire function. Compute \begin{equation} \frac1{2\pi i}\int_{|z|=1}\frac{1-zf(z)}{z}dz \end{equation}
- Prove that all $f$ entire functions verify that \begin{equation} \max_{|z|=1}|z^{-1}-f(z)|\ge1 \end{equation} Deduce that there doesn't exist any sequence of polynomials $P_n$ such that $P_n\to\frac1z$, uniformly in $\{|z|=1\}$.
I know that 1. can be done with $f(z) = \sum_{n=0}^\infty a_nz^n$: \begin{equation} \frac{1-zf(z)}{z} = \frac{1}z-f(z) \end{equation} So \begin{equation} \frac1{2\pi i}\int_{|z|=1}\frac{1-zf(z)}{z}dz = \frac1{2\pi i}\left(\int_{|z|=1}\frac{1}{z}dz - \int_{|z|=1}f(z)dz\right) = \frac1{2\pi i}(2\pi i-0) = 1 \end{equation}
The second exercise is also simple enough, since we got the following:
\begin{equation} 1 = \left| \frac1{2\pi i}\int_{|z|=1}\frac{1-zf(z)}{z}dz \right|\le \frac1{2\pi} \operatorname{len}(\{|z|=1\})\max_{|z|=1}|g(z)|\le \frac1{2\pi}2\pi\max_{|z|=1}\left|\frac{1-zf(z)}{z}\right| = \max_{|z|=1}\left|\frac{1}{z}-f(z)\right| \end{equation}
But how do I deduce from this that there doesn't exist a sequence of polynomials $P_n$ such that $P_n\to\frac1z$ uniformly in $\{|z|=1\}$?
Hint: Since every polynomial is an entire function you would have $$\lVert z^{-1}-P_n(z)\rVert_\infty \to 0$$ in $|z|=1$. Do you see the contradiction?
Full solution: Suppose such sequence of polynomials existed. Then you would have $$\lVert z^{-1}-P_n(z)\rVert_\infty \to 0$$ But from 2. for each $n\in \mathbb{N}$ and $z\in \mathbb{C}$ such that $|z|=1$ you would have $$\lVert z^{-1}-P_n(z)\rVert_\infty \geq 1$$ which is a contradiction.