# Lower boundary for $|f(z) - 1/z|$, where $f(z)$ is holomorphic

I've been trying to prove the following statement:

Let $f:U \rightarrow \mathbb{C}$ be holomorphic with $\overline{B(0, R)} \subset U$. Suppose $r < R$. Prove that $$\sup\limits_{r \leq |z| \leq R}\left(\left|f(z) - \frac{1}{z}\right| \right) \geq \frac{1}{R}$$

It's a way of proving that $\frac{1}{z}$ is not a uniform limit of a sequence of polynomials on a ,,ring'' $r \leq |z| \leq R$.

I tried to use Cauchy's integral theorem to evaluate $f(z)$, however it didn't lead me anywhere close to the solution.

I would appreciate some help with this exercise

Integrate $$\oint_C \left|f(z) - \frac{1}{z}\right|$$ where $C$ is the path $$|z|=R$$ by two techniques:

• (1) Contour integration. Since $f(z)$ is holomorphic, the only contribution is from the pole in $\frac{1}{z}$

• (2) ordinary integration, using the fact that $|f(z)|$ is always less than or equal to its sup value.

You then get an inequality with the $2\pi R$ times the sup value on one side, and $2\pi$ on the other, which leads to the sup value being no greater than $|1/R|$.

• I don't get the first part - the function $|f(z) - 1/z|$ might not be holomorphic away from that pole also, since $|z|$ is not holomorphic May 12, 2015 at 13:10

Hint: Write $f(z) = \sum_{n=0}^{\infty}a_nz^n.$ For $r\le s \le R,$ consider

$$\frac{1}{2\pi}\int_0^{2\pi} |f(se^{it})-\frac{1}{se^{it}}|^2\,dt.$$