space of analytic functions

For what follows suppose that $G\subset ℂ$ is a region. We wish to take the set of all holomorphic functions on $G$, denoted by $𝒪(G)$, and make it into a metric space. We will define a metric such that convergence in this metric is the same as uniform convergence on compact subsets of $G$. We will call this the space of analytic functions on $G$.

It is known that thereexists a sequence of compact subsets $K_n\subset G$ such that $K_n\subset K_{n+1}^{\circ }$ (interior of $K_{n+1}$), such that $\bigcup K_n^{\circ }=G$ andsuch that if $K$ is any compact subset of $G$, then $K\subset K_n$ for some $n$.Now define the quantity ${\rho }_n(f,g)$ for $f,g\in 𝒪(G)$ as

$${\rho }_n(f,g):=\underset{z\in K_n}{sup}\{|f(z)-g(z)|\}.$$

We define the metric on $𝒪(G)$ as

$$d(f,g):=\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{2}\right)}^n\frac{{\rho }_n(f,g)}{1+{\rho }_n(f,g)}.$$

This can be shown to be a metric. Furthermore, it can be shown that the topology generated by this metric is independent of the choice of $K_n$, even thoughthe actual values of the metric do depend on the particular $K_n$ we have chosen.Finally, it can be shown that convergence in $d$ is the same as uniform convergence on compact subsets. It is known that if you have a sequence ofanalytic functions on $G$ that converge uniformly on compact subsets, then the limit is in fact analytic in $G$, and thus $𝒪(G)$ is a complete space.

Similarly, we can treat the functions that are meromorphic on $G$, and define$M(G)$ to be the space of meromorphic functions on $G$. We assume that thefunctions take the value $\mathrm{\infty }$ at their poles, so that they are defined atevery point of $G$. That is, they take their values in the Riemann sphere, or the extended complex plane. We just need to replacethe definition of ${\rho }_n(f,g)$ with

$${\rho }_n(f,g):=\underset{z\in K_n}{sup}\{\sigma (f(z),g(z))\},$$

where $\sigma$ is either the spherical metric on the Riemann sphere, or alternatively the metric induced by embedding the Riemann sphere in ${ℝ}^3$. Both of those metrics produce the same topology, and that is all that we care about. The rest of the definition is the same as that of $𝒪(G)$. There is, however, one small glitch here. $M(G)$ is not a complete metric space. It is possible that functions in $M(G)$ go off to infinity pointwise, but this is the worst that can happen. For example, the sequence $f_n(z)=n$ is a sequence of meromorphic functions on $G$, andthis sequence is Cauchy in $M(G)$, but the limit would be $f(z)=\mathrm{\infty }$ andthat is not a function in $M(G)$.

Remark.

Note that $𝒪(G)$ is sometimes denoted by $H(G)$ in literature. Also note that $A(G)$ is usually reserved for functions which areanalytic on $G$ and continuous on $\overline{G}$ (closure of $G$).

Remark.

We can similarly define the space of continuous functions, and treat $𝒪(G)$ and $M(G)$ as subspaces of that. That is, $𝒪(G)$would be a subspace of $C(G,ℂ)$ and $M(G)$ would be a subspaceof $C(G,\widehat{ℂ})$.