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Consider the function $f(z)=\frac{1}{z}$ on the annulus $A=\{z \in \mathbb{C} : \frac{1}{2} \lt{\vert{z}\vert} \lt2\}$.

Which of the following are true?

  1. There is a sequence $\{P_n(z)\}_n$ of polynomials that approximate $f(z)$ uniformly on compact subsets of $A$.

  2. There is a sequence $\{r_n(z)\}_n$ of rational functions, whose poles are contained in $\mathbb{C}\setminus A$ and which approximates $f(z)$ uniformly on compact subsets of $A$.

  3. No sequence $\{P_n(z)\}_n$ of polynomials that approximate $f(z)$ uniformly on compact subsets of $A$.

  4. No sequence $\{r_n(z)\}_n$ of rational functions, whose poles are contained in $\mathbb{C}\setminus A$ and which approximate $f(z)$ uniformly on compact subsets of $A$.

I know that every continuous function can be approximated by sequence of polynomials and given function is analytic in annulus A. But how to construct such sequence of functions?

Can we use identity theorem?

Robert Z
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dipali mali
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  • You are supposed to start from $\int_{|z|=1} \frac{1}{z}dz = 2i\pi$ and $\int_{|z|=1} P(z)dz = 0$. Also 2. and 4. are indeed trivial. – reuns Jun 10 '17 at 06:42
  • Question 5. same questions with $A = {z \in \mathbb{C} : \frac{1}{2} \lt{\vert{z}\vert} \lt2, \Re(z) > 0}$ and $A' = {z \in \mathbb{C} : \frac{1}{2} \lt{\vert{z}\vert} \lt2, z \not \in (\frac12,2)}$ (hint : simply connected) – reuns Jun 10 '17 at 06:45

1 Answers1

2

Hint. Consider the Runge's Theorem and its corollary about polynomials.

As regards 1) note that if $\{P_n(x)\}_n$ is a sequence of polynomials converging uniformly to $1/z$ on the compact set $K=\{|z|=r\}\subset A$, for $r\in (1/2,2)$, then we get the contradiction $$0=\int_{|z|=r}P_n(z)\, dz\to \int_{|z|=r}\frac{1}{z}\,dz=2\pi i\not=0.$$ On the other hand, by Runge's Theorem, the approximation by polynomials to $1/z$ can be done on the compact sets $K$ of $A$ such that $\mathbb{C}\setminus K$ is connected.

Robert Z
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