A question related to the approximation of an analytic function

Question

Suppose $\Omega:=\mathbb{C}\setminus \{0\}.$ Given a function $f$ that is analytic on $\Omega,$ does there always exist a sequence of polynomials or rational functions in $C(\Omega)$ (space of all continuous functions on $\Omega$) whose limit in $C(\Omega)$ is $f?$

Thanks in advance.

Edit: (from comments below)

• The notion of convergence we're interested in is uniform convergence on compact subsets.
• This problem comes from an old qualifying exam, which the OP is practising.
• The OP has not come across Runge's theorem, and would like to know if this can be solved without Runge's theorem.

Answer 1

First, I'll remark that the statement we're trying to prove is precisely the statement of Runge's theorem as applied to $\Omega := \mathbb C \setminus \{ 0 \}$. That makes me suspect that you're meant to know Runge's theorem for this qualifying exam that you're preparing for.

Runge's theorem can be stated in more than one way, but the version that's most easily applicable here is:

Let $\Omega$ be an open subset of $\mathbb C$, and let $A \subset \mathbb C$ be a set that contains at least one point in each connected component of $S^2 \setminus \Omega$. (Here, $S^2 := \mathbb C \cup \{ \infty \}$ is the Riemann sphere.)

Then for every holomorphic function $f : \Omega \to \mathbb C$, there exists a sequence of rational functions $r_n: S^2 \to S^2$, whose poles lie only in $A$, such that $r_n$ converges to $f$ uniformly on compact subsets of $\Omega$.

In your example, $S^2 \setminus \Omega = \{ 0, \infty \}$, so the set $A := \{ 0, \infty \}$ contains one point in each connected component of $S^2 \setminus \Omega$. A "rational function" $S^2 \to S^2$ with poles lying only in $\{ 0, \infty \}$ is the same thing as a rational function on $\mathbb C$ with possibly a pole at $0$; this rational function would also have a pole at $\infty$ if the degree of the polynomial in its numerator is greater than the degree of the polynomial in its denominator. So your statement follows immediately from Runge's theorem.

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However, as @Jose27 points out, we can also prove your statement by more elementary means.

Let $f$ be holomorphic on $\Omega := \mathbb C \setminus \{ 0 \}$. Consider the Laurent expansion $f(z) = \sum_{n = -\infty}^{\infty} a_n z^n$. The reasoning is:

• The series of positive-power terms $\sum_{n = 0}^{\infty} a_n z^n$ has an infinite radius of convergence, so it converges uniformly on $\{ z \in \mathbb C : |z| \leq R\}$, for any $R \in (0, \infty)$.
• Similarly, the series of negative-power terms $\sum_{n = -\infty}^{0} a_n z^n$ converges uniformly on $\{ z \in \mathbb C : |z| \geq r\}$, for any $r \in (0, \infty)$. (Perform the substitution $w := 1 / z$, then apply the same logic as above.)
• Therefore, the sequence of rational functions $r_k(z) := \sum_{n = -k}^{k} a_n z^n$ converges uniformly on $\{ z \in \mathbb C : r \leq |z| \leq R \}$ as $k \to \infty$, for any $0 < r < R < \infty$.
• Finally, any compact subset $K \subset \Omega$ lies inside an annulus of the form $\{ z \in \mathbb C : r \leq |z| \leq R \}$, for some suitable choices of $r$ and $R$.