Is there a name for this extension of Runge's theorem?
Theorem: Let $K\subset\mathbb{C}$ be compact, and let $A\subset K^c$ be a set which intersects each component of $K^c$. Let $f$ be meromorphic on an open set $U$ containing $K$. Let $A_f$ denote the set of poles of $f$ in $K$. Then there is a sequence of rational functions $\{r_n\}$ such that $r_n\to f$ uniformly on $K$, and for each $n$, the poles of $r_n$ are contained in the set $A\cup A_f$.
I have posted a proof of this result below. It is such a natural generalization of Runge's theorem that I am sure it is known, but I would like a reference so that I can use it in a paper without proving it.
Thanks!
Best Answer
Proof: Since $K$ is compact, we may impose the following assumptions on $K$:
Let $z_1,z_2,\ldots,z_k$ be the poles of $f$ in $K$ with multiplicities $m_1,m_2,\ldots,m_k$. Define $q(z)=(z-z_1)^{m_1}\cdots(z-z_k)^{m_k}$, and define $g(z)=f(z)/q(z)$, which is analytic on some neighborhood of $K$. As in user zhw.'s solution here, we can find a sequence of rational functions $\{t_n\}$ such that $t_n\to g$ uniformly on $K$, and each $t_n$ has all of its poles in $A$, and for each $z_i$, each $t_n$ matches the value of $f$ and its first $m_k$ derivatives at $z_i$. That is,
$$\begin{array}{rcl} t_n(z_i)&=&g(z_i)\\ {t_n}'(z_i)&=&g'(z_i)\\ &\vdots\\ t_n^{(m_k)}(z_i)&=&g^{(m_k)}(z_i). \end{array}$$
Define $r_n=t_n/q$. Then each $r_n$ has all of its poles in $A\cup A_f$.
CLAIM: $r_n\to f$ uniformly on $K$.
Fix $\epsilon>0$, and define $M=m_1+m_2+\cdots+m_k$. Choose a $\delta>0$ small enough so that at each pole $z_i$ of $f$ in $K$, the disk $B_{z_i}(3\delta)$ is contained in $K^\circ$, and contains no other poles of $f$ other than $z_i$. Pick some $N>0$ such that for all $n>N$ and $z\in K$, $|t_n(z)-g(z)|<\epsilon\delta^M$.
Therefore if $z\in K$, with $|z-z_i|>\delta$ for all $i$, we have
$$|r_n(z)-f(z)|=\dfrac{|t_n(z)-g(z)|}{|q(z)|}<\dfrac{\epsilon\delta^M}{\delta^M}=\epsilon.$$
Now, for each $i\in\{1,\ldots,k\}$, the assumption on the function $t_n-g$ has a zero at $z_i$ of multiplicity $m_k$. Therefore $r_n-f$ is analytic on is analytic at $z_i$, and therefore on all of $K$. Now, for each $i\in\{1,\ldots,k\}$, $r_n-f$ is analytic on $B_{3\delta}(z_i)$, and has magnitude less than $\epsilon$on $\partial B_{\delta}(z_i)$, so by the maximum modulus theorem, $|r_n-f|<\epsilon$ on $B_{\delta}(z_i)$.
We have $|r_n-f|<\epsilon$ on $K$, so we conclude that $r_n\to f$ uniformly on $K$.