# Prove that it's impossible to approximate $1/z$ with polynomials on an annulus

I heard a nice problem, presumably from an old qual, that I thought I'd share.

Problem: Let A be the annulus (in the complex plane) $A=\{z: r_1 \leq |z|\leq r_2\}.$ Prove that $f(z) = 1/z$ cannot be approximated uniformly by polynomials on $A$.

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If $f_n$ is a sequence of holomorphic functions converging uniformly to a holomorphic function $f$, then the contour integrals $\oint f_n$ of the $f_n$ around a circular contour converge to the contour integral $\oint f$. For $f = \frac{1}{z}$ the latter integral is nonzero, but it is zero for any sequence of polynomials $f_n$.

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+1: is it possible to extend your argument such that it works also for $f(z) =z^{-2}$? – Fabian Dec 20 '12 at 1:19
@Fabian: sure. Replace $f_n, f$ above with $zf_n, zf$. – Qiaochu Yuan Dec 20 '12 at 1:32

Here is another simple argument that only requires the maximum principle. Suppose that $1/z$ can be approximated uniformly on $A$ by polynomials. Let $m:=\operatorname{min} \{|1/z| : z \in A\}>0$. Then by assumption, there exists some polynomial $P$ such that $$|1/z - P(z)|<m \qquad (z \in A).$$ Thus, $$|1-zP(z)|<m|z|\leq1 \qquad (z \in A).$$

By the maximum principle, $$|1-zP(z)| < 1$$ for every $z$ in the disk whose boundary is the outer circle of $A$. But with $z=0$ we get a contradiction.

Essentially the same proof shows that for every compact set $K$ in the plane whose complement contains more than one component, there exists some function $f$ holomorphic on a neighborhood of $K$ that cannot be approximed uniformly on $K$ by polynomials.

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It is legit to set $z=0$? – Wilson of Gordon May 20 '14 at 13:55
@WilsonofGordon Yes, why not? The obtained inequality $|1-zP(z)|<1$ holds for all $z$ in a disc containing zero. – Kalim May 20 '14 at 14:16
But the set $A$ on which our discussion is based is an annulus! – Wilson of Gordon Jun 5 '14 at 0:42
the function $1-zP(z)$ is holomorphic everywhere in the plane. By the argument in the answer, it is less than $1$ in modulus on the exterior boundary of the annulus $A$. Hence, by the maximum modulus principle, it is less than one in the whole disk whose boundary is the outer circle of $A$. – Kalim Jun 5 '14 at 12:16