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


2 Answers 2


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 $|1/R|$.

  • $\begingroup$ 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 $\endgroup$
    – Jytug
    Commented May 12, 2015 at 13:10

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

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


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