Application of Runge theorem

Question

Let $\mathbb D = \{z \in \mathbb C : \vert z \vert < 1\}$ and $\mathcal P$ the set of all polynomials with

$$ P(z) = 1 + \sum_{i=1}^n a_i z^i, \ a_n \neq 0,\ n \in \mathbb N.$$

Let $\emptyset \neq K \subset \partial \mathbb D$ compact set with $K \neq \partial \mathbb D$. Then for all $\epsilon > 0$ there exists a polynomial $P \in \mathcal P$ with $\max_{z \in K} |P(z)| < \epsilon$.

I want to use Runge's theorem to show that. But I can't figure out how to apply it. I know that $\mathbb C \setminus K$ is connected, so I can approximate every holomorphic function $f$ by polynomials, i.e. for every $\epsilon > 0$ there exists a polynomial $Q$ with $\max_{z \in K} |f(z) - Q(z)| < \epsilon$. But I don't get necessarily that $Q \in \mathcal P$. I tried to use the triangle equality but I couldn't come up with a solution. Maybe I just use the wrong Runge theorem and should use the one for rational functions. I would appreciate some hints. Thanks in advance!

Answer 1

Let $\epsilon > 0$. Since $K \neq \partial \mathbb D$, there exists an open set $\Omega \subset \mathbb C$ with $K \subset \Omega$ and $0 \not \in \Omega$. Thus $f: \Omega \to \mathbb C,\ z \mapsto \frac{1}{z}$ is holomorphic. Due to $K \neq \partial \mathbb D$ $\mathbb C \setminus K$ is connected. So due to the Runge theorem there exists a polynomial with \begin{align*} Q(z) = \sum_{i=0}^n a_i z^i \quad \text{with w.l.o.g.} \quad a_n \neq 0 \quad \text{and} \quad \Vert f-Q \Vert_{\infty, K} < \epsilon. \end{align*} Consider now the polynomial $P(z) = 1 - z Q(z)$. Then we have $P \in \mathcal{P}$ and we get that \begin{align*} \Vert P \Vert_{\infty, K} = \max_{z \in K} \vert P(z) \vert = \max_{z \in K} \vert 1 - z Q(z) \vert = \max_{z \in K} \left\vert \frac{1}{z}- Q(z) \right\vert = \Vert f-Q \Vert_{\infty, K} < \epsilon, \end{align*} since $K \subset \partial \mathbb{D}$ and thus $\vert z \vert = 1$ for $z \in K$.

I think that is a pretty clever way to prove it :)

Answer 2

Let $f(z) \equiv 1 $ for $z\in K .$ From Muntz-Sasz theorem there exist a polynomial $Q$ of form $Q(z)=\sum_{k =1}^n a_k x^k $ such that $\sup_{z\in K} |1- Q(z) |\leq \varepsilon .$ So you can take $P(z) =1-Q(z).$

Answer 3

Here's something easy: If $K \subset \{ e^{it}: -\pi/3 < t < \pi/3\},$ then $p_n(z) = (1-z)^n$ converges uniformly to $0$ on $K$ and $p_n(0)=1$ for all $n.$ Proof (sketch): Verify that for $t$ in this range, $|1-e^{it}| <1.$ It follows that $|1-e^{it}| < a$ for some $a\in (0,1)$ for every $e^{it}\in K.$ Since $a^n \to 0,$ we get the result.

Unfortunately I couldn't turn the above into a proof for the general $K.$ Runge solves it easily however: Choose disjoint open sets $U,V$ with $0\in U, K \subset V.$ The function $f=1$ on $U,$ $f=0$ on $V$ is holomorphic on $U\cup V.$ Since $\{0\} \cup K$ does not separate $\mathbb C,$ Runge gives us a sequence of polynomials that does the job.