[Math] Topology on the set of analytic functions

cv.complex-variablesgn.general-topology

Let $H(D)$ be the set of all analytic functions in a region $D$ in $C$ or in $C^n$.
Everyone who worked with this set knows that there is only one reasonable topology
on it: the uniform convergence on compact subsets of $D$.

Is this statement correct?

If yes, can one state and prove a theorem of this sort:

If a topology on $H(D)$ has such
and such natural properties, then it must be the topology of uniform convergence on
compact subsets.

One natural property which immediately comes in mind is that the point
evaluatons must be continuous. What else?

EDIT: 1. On my first question, I want to add that other topologies were also studied.
For example, pointwise convergence (Montel, Keldysh and others). Still we probably all feel
that the standard topology is the most natural one.

  1. If the topology is assumed to come from some metric, a natural assumption would be
    that the space is complete. But I would not like to assume that this is a metric space
    a priori.

  2. As other desirable properties, the candidates are continuity of addition and multiplication. However it is better not to take these as axioms, because the topology
    of uniform convergence on compacts is also the most natural one for the space of
    meromorphic functions (of one variable), I mean uniform with respect to the spherical metric.

Best Answer

I think the following characterization is true:

The standard topology on $H(D)$ is the initial topology with respect to the projections $f \mapsto [f]_z$ for each $z\in D$, where $[f]_z$ is the germ of $f$ at $z$.

For this statement to make sense, we need to endow the space $\mathcal{O}_z$ of germs of holomorphic functions at $z$ with a topology. I think it is reasonable to give it the topology of the inductive limit $\mathcal{O}_z = \bigcup_{r>0} P_{z,r}$, where $P_{z,r}$ is the space of power series absolutely convergent on a (poly)disk of radius $r$ centered at $z$. The topology on each $P_{z,r}$ coincides with the subspace topology of the $\sup$-norm topology on bounded continuous functions on the closed (poly)disk. Depending on one's preference, there might be different, equivalent ways to define the same topology.

  1. The standard topology contains the initial one. That's because the projections $f \mapsto [f]_z \mapsto f(z)$ are continuous and the evaluations maps $f \mapsto f(z)$ are continuous in the standard topology.
  2. The initial topology contains the standard one. Consider a compact set $K\subset D$, an $\epsilon > 0$ and a standard neighborhood $V(f,K,\epsilon)$ of $f$ in $H(D)$. Then $V(f,K,\epsilon)$ contains an intersection of finitely many initial topology neighborhoods of $f$. These neighborhoods could, for instance, be generated by a finite subcover of a cover of $K$ by open (poly)disks such that $f$ is continuous on the closure of each.
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