I have the following problem:
We have $K\subset \Bbb{C}$ a compact subset and $f:K\rightarrow \Bbb{C}$ a continuous function. We take $a\in \Bbb{C}\setminus K$. We assume that there exists a sequence of rational functions $R_n$ with poles in $a$ such that $$\max_{z\in K} |f(z)-R_n(z)|\rightarrow 0$$ Now i fix $k\in \Bbb{N}$ I need to show that there is a sequence of rational functions $Q_n$ with poles in $a$ s.t. $\max_{z\in K} |f(z)^k-Q_n(z)|\rightarrow 0$$
I somehow don't see where to start. I first thought about Runges theorem but there I need a connected set doesn't I?
I also thought about proving it by induction. I‘m really sorry that I can‘t give you more work, but I have some trubles in working with this approximation stuff. It always seems extreamly frightening to me…
I had the following version of runges theorem:
Let $f:\Omega\rightarrow \Bbb{C}$ be analytic. Suppose $K\subset \Omega$ is compact. Assume $\Bbb{C}\setminus K=U_0 \cup \left(\bigcup_{i\geq 1}U_i\right)$ where the $U_i$'s are open and connected, $U_0\supset\{|z|>R\}$ and $U_i$ are bounded for all $i\geq 1$. Then pick $a_i$ in each connected component $U_i$ for $i\geq 1$ then there exists a rational function $R_n$ with poles in $(a_i)$ s.t.$$\max_{z\in K} |f(z)-R_n(z)|\rightarrow 0$$
Thanks for your help
Fix $M>\max_K |f|$. Then for some $n_0=n_0(M)$, we have $\max_K |R_n|<M$ for all $n>n_0$. Now $Q_n=R_n^k$ satisfies, for all $n>n_0$ and $z \in K$,
\begin{eqnarray} |Q_n(z) -f(z)^k| &=& \Bigl|\sum_{j=0}^{k-1} R_n(z)^j \cdot f(z)^{k-1-j}\Bigr| \cdot |R_n(z)-f(z)| \\ &\le& kM^k \cdot|R_n(z)-f(z)| \,. \end{eqnarray} Therefore, $$\max_{z \in K} |Q_n(z) -f(z)^k| \le kM^k \cdot \max_{z \in K}|R_n(z)-f(z)| \to 0 $$ as $n \to \infty.$