Let $0<r<R$ and let $A=\{z\in\mathbb{C}:r\leq|z|\leq R\}$.
Show that there is a positive number $\epsilon>0$ so that for any polynomial $p$,
$sup\{|p(z)-\frac{1}{z}|:z\in A\}\geq\epsilon$
I don't really know how to approach this problem.
My first approach was to work by contradiction:
BWOC,suppose that $\forall\epsilon>0$, $sup\{|p(z)-\frac{1}{z}|:z\in A\}<\epsilon$
Then, I tried getting a circular contour in the region between the two circles:
$\gamma(t)=r'e^{it}$ for $[0,2\pi]$ where $r<r'<R$
and then examining the 2 contour integrals:
$\int_{\gamma}p(z)dz$ and $\int_{\gamma}\frac{1}{z}dz$
But I don't know how to proceed to reach a contradiction.
I then tried applying the Maximum Modulus principle since $p(z)-\frac{1}{z}$ is holomorphic.
I could then conclude that $|p(z)-\frac{1}{z}|$ reaches its maximum on the boundaries of A. But then I also get stuck...
Any suggestions?
Let $\rho \in [r,R]$ and let $\gamma(t) = \rho e^{it}$, a circle around the origin of radius $\rho$.
Then $\int_\gamma p(z) dz = 0$, $\int_\gamma { 1 \over z} dz = 2 \pi i$ and so $|\int_\gamma (p(z)-{1 \over z}) dz | = 2 \pi$.
Since \begin{eqnarray} 2 \pi &=& |\int_\gamma (p(z)-{1 \over z}) dz | \\ &=& |\int_0^{2 \pi} (p(\rho e^{it})-{1 \over \rho e^{it}}) i \rho e^{it} dt| \\ &\le& \rho \int_0^{2 \pi} |p(\rho e^{it})-{1 \over \rho e^{it}}|dt \\ &\le& 2 \pi \rho \sup_{|z|=\rho} |p(z)-{1 \over z}| \end{eqnarray} we see that $\sup_{|z|=\rho} |p(z)-{1 \over z}| \ge {1 \over \rho}$.