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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 $A$.

2 Answers 2

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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 $f(z) =z^{-2}$?2012-12-20
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    @Fabian: sure. Replace $f_n, f$ above with $zf_n, zf$.2012-12-20
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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 $$|1/z - P(z)| Thus, $$|1-zP(z)|

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 $z=0$?2014-05-20
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    @WilsonofGordon Yes, why not? The obtained inequality $|1-zP(z)|<1$ holds for all $z$ in a disc containing zero.2014-05-20
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    But the set $A$ on which our discussion is based is an annulus!2014-06-05
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    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 $A$.2014-06-05