# Converse to Runge's approximation theorem

I'm trying to prove the following fact

If $K$ is a compact set whose complement is not connected, then there exists a function $f$ holomorphic in a neighborhood of $K$ which cannot be approximated uniformly by polynomials in $K$.

This is an exercise from Stein & Shakarchi's complex analysis, and the book gives the following hint

Pick a point $z_0$ in a bounded component of $K^{c}$, and let $f(z)=\displaystyle\frac{1}{(z-z_0)}$. If $f$ can be approximated uniformly by polynomials on $K$, show that there exists a polinomial $p$ such that $\left |{(z-z_0)p(z)-1}\right |<1$. Use the maximum modulus principle to show that this inequality continues to hold for all $z$ of $K^{c}$ that contains $z_0$.

I don't know how to begin the problem, any hint will be appreciated, please don't give answers.

Thanks

You can begin by considering the case where $K=\{z\in \Bbb C:|z|=1\}$ and $z_0=0$, which is simpler but uses the same ideas. These ideas are described here.

• ok, i hadn't seen it right away because i hadn't seen that inmediate consequence to the maximum modulus principle, you need only to check that if $\Omega$ is the bounded component containing $z_0$, then $\bar{\Omega}-\Omega\subseteq{K}$. Thanks!!! Oct 14, 2012 at 16:26

Partition $$K^c$$ into disjoint path connected components and pick a component $$S$$ which is bounded (this exists since $$K$$ is bounded) and open since $$K$$ is closed. Observe that the boundary of $$S$$, $$\partial S \in K$$.

Now pick any $$z_0 \in S$$ and define $$f(z):= \frac{1}{z-z_0}$$. I claim this function can't be uniformly approximable by polynomials in $$K$$. Assume the contray, and pick any $$\epsilon < \frac{.005}{|\partial S|}$$ and a polynomial $$p(z)$$ such that $$|p(z)-f(z)| < \epsilon$$ for all $$z \in K$$.

Then $$|2 \pi| = \left| \int_{\partial S} f(z) dz \right| = \left| \int_{\partial S} p(z) dz + \int_{\partial S} f(z)-p(z) dz \right| = \left| \int_{\partial S} f(z)-p(z) dz \right| < |\partial S| \epsilon < .005$$, a contradiction.

Directly mimicking the proof for the case $$K$$ is "nice" one ends up in the above proof easily, but after searching on MO I find this is wrong since the boundary of $$S$$ can behave very badly.

Can this idea be adapted to give a proof of this, or the only solution proceeds with Maximum Modulus principle ?