# Topology on the set of analytic functions

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.

• Completeness (or quasi/local-completeness, more generally). (And "of course", local convexity.) – paul garrett Aug 26 '13 at 13:07
• The topology you describe is just the subspace topology, for the subspace of analytic functions inside the space of all continuous mappings. For all mappings this is called "compact-open topology", and it is indeed natural in the sense of category theory: it is the unique inner hom functor for the category of compactly-generated hausdorff spaces. – Anton Fetisov Aug 26 '13 at 16:38
• I don't know if there is any reasonable definition of hom functor for analytic category. The problem is to define complex structure on an infinite-dimensional space. – Anton Fetisov Aug 26 '13 at 16:41
• "a limit in our topology of analytic functions" Erm... What kind of animal is this? The topology is not even defined outside $H(D)$... – fedja Aug 26 '13 at 21:51
• @fedja: you are right. I was thinking of some kind of completeness property, but this assumes that there is a metric, which I would not like to assume a priori. – Alexandre Eremenko Aug 27 '13 at 3:45

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.
• Does the same argument also applies to the spaces of continuous functions in a region (and characterizes the topology of uniform convergence on compact subsets) ? – Alexandre Eremenko Aug 28 '13 at 18:38
• Hmm, I think yes. The closed (poly)disks will have to be replaced by arbitrary compact neighborhoods, though. This observation is probably related to the earlier comment of Anton Fetisov. – Igor Khavkine Aug 28 '13 at 19:51
• This argument applies basically to any function space you can think of. The reason is that lifting universally topology from stalks to sections is a general sheaf-theoretic construction. It says nothing about defining topology on stalks in the first place. In fact, for that reason people usually move the other way, from topology on sections to topology on stalks. – Anton Fetisov Aug 28 '13 at 23:16
• I think what is special about holomorphic functions is that the stalks and germs have a particularly simple structure. – Igor Khavkine Aug 29 '13 at 6:44

Here is a method of recovering the topology of $H(U)$ from general considerations. The idea is that the dual of $E$ of $H(U)$ has the following universal property: $E$ is a complete locally convex space (even a so-called nuclear Silva space, i.e., an inductive limit of a sequence of Banach spaces with nuclear intertwining mappings) and $U$ embeds in $E$ in such a manner that every holomorphic mapping from $U$ into a Banach space lifts in a unique manner to a continuous linear mapping on $E$. We now forget the topology on $H(U)$ and note that the existence of such a universal space can be proved without recourse to this duality (this is a standard construction as a closed subspace of a suitable large product of Banach spaces---analogous to the construction of the free locally convex space over a completely regular space or a uniform space---see, e.g. Raikov, Katetov, etc.) Such a free object is always unique in a suitable sense. Now it follows from the universal property (applied to scalar-valued functions) that $H(U)$ is, as a vector space, naturally identifiable with the dual of the universal space. It can then be provided with the corresponding strong topology which is thus intrinsic. But this is precisely the standard Fréchet space topology (the fact that we are dealing with a symmetric duality between a nuclear Fréchet space, resp. Silva space is relevant here).

Added as an edit after Alexandre's comment since I am not entitled to comment.

One way to construct the universal space is to take the free vector space over $U$ and provide it with the finest locally convex topology such that the embedding of $U$ is holomorphic, then take the completion.

I doubt that you will find the fact that the dual of $H(U)$ has the universal property in the literature (such considerations were never fashionable---too much category theory for the analysts, too much hard analysis for the category theorists perhaps). It follows very easily from the theory of duality for $H(U)$ (Köthe, Crelle (191)). An accessible version in english is in the book "Complex Analysis: a functional analysis approach" by Ruecking and Rubel. The vector-valued case is in the seminal follow-up papers to Köthe's by Grothendieck in Crelle, 192.

I should note that the duality mentioned above was originally developed by the portuguese mathematician J. Sebastião e Silva in a sadly forgotten article in Port. Math. 9 (1950) 1-130 and this again has its source in work by Cacciopoli and Fantappié. The universal property mentioned above has many analogues---e.g., the distributions on the unit interval, unversal for smooth mappings into Banach spaces (with obvious generalisations), Radon measure on the unit interval or a compact space, universal for continuous mappings, bounded Radon measures on a completely regular space (bounded, continuous mappings), uniform measures on a uniform space (bounded, uniformly continuous mappings). See, for example Raikov, Math. Sb. 63 (1964) 582-590, Tomašek, Czech. Math. J. 20 (1970) 1-18, 19-33.

• berlin: could you give a more precise reference, for non-specialists on topological vector spaces, on the a) existence proof of such $E$ and b) that the dual of $H(U)$ has this property ? – Alexandre Eremenko Mar 24 '14 at 22:45
• The universal property of $E$ follows from Schwartz' $\varepsilon$-product (which in many cases coincides with the completed injective tensor product): $H(U,X) = H(U) \varepsilon X = L(H(U)'_{co}, X)$ where the last equality is the definition and $F'_{co}$ is the dual of the locally convex space $F$ endowed with uniform convergence on all absolutely convex compact sets. Since $H(U)$ is a Montel space, in our case this is the same as the strong dual of $H(U)$. – Jochen Wengenroth Mar 25 '14 at 8:42
• @Alexandre. More on the universal property of $H(U)'$. This was known to Fantappiè in 1943 (not in this language, of course)---see the concept of Fantappiè indicatrix of an operator--- and so before the Schwartzian theory of distributions and long before Grothendieck and Schwartz developed the connection between tensor products of lcs's and operators, in particular for nuclear spaces. A succinct discussion of this and the path from the original work of Fantappiè to its incorporation into modern function analysis by the actors mentioned above is in a review by Horvath, BAMS 25 (1991) p. 162 – berlin Mar 27 '14 at 16:20
• @berlin: Thanks. Fortunately the book is available in our library:-) – Alexandre Eremenko Mar 28 '14 at 0:09

If $\mathcal S$ is a locally convex topology on $H(D)$ such that all evaluations are continuous then the identity $(H(D)$,compact open) $\to (H(D),\mathcal S)$ has closed graph. Therefore, whenever $\mathcal S$ is good for the closed graph theorem, the compact open topology will be finer. The most general class of locally convex spaces which are good for the closed graph theorem as range spaces is that of webbed spaces introduced by de Wilde (see e.g. the functional analysis book of Meise and Vogt). This class contains all Banach spaces and is stable with respect to countable inductive (=direct) and projective (=reverse) limits. In particular, it contains all Frechet spaces, LF-spaces, projective limits of LF-spaces,...

Moreover, since there is no strictly coarser barrelled locally convex topology on a Frechet space, we get a possible answer to your question: If $\mathcal S$ is a locally convex topology on $H(D)$ so that $H(D)$ is webbed and barrelled and all evaluations are continuous then is coincides with the compact open topology.

In the article http://www.zentralblatt-math.org/zbmath/search/?q=an%3A1199.32010 the topology of the exponential convergence on compacts was introduced. Let $\Omega$ be a domain in $\mathbb{C}^n$ and $PSH(\Omega)$ be the set of functions plurisubharmonic on $\Omega$. A sequence of $u_n \in PSH(\Omega)$ exponentially and uniformly convergerges on compacts to the function $u$ if $\exp u_n$ converges to $\exp u$ uniformly on compacts.The exponential uniform convergence on compacts is a generalization of the uniform convergence on compacts. It should be noted that $u \in PSH(\Omega)$ as well as $\exp u \in PSH(\Omega)$. The topology of the exponential uniform convergence on compacts is metrizable as follows. Let $C_n$ be a seqquence of compacts exhausting $\Omega$. We put $d_n(u,v):=\sup\{|\exp u(z)-\exp v(z)|: \,z \in C_n\}$ and $$d(u,v):=\sum\limits_{n=1}^\infty \frac{2^{-n}d_n(u,v)}{1+d_n(u,v)}.$$ Then $PSH(\Omega)$ is a complete metric space. M. Girnyk proved that the set $\log|A|(\Omega)$ of the logarithms of the moduli of functions holomorphic on $\Omega$ is nowhere dense in $PSH(\Omega)$ with that metrics. PS. The author proved the last statement in the cases $\Omega=\mathbb{C}^n$ and $\Omega=\mathbb{D}^n$.

• Thanks. But I asked about topologies on the space on analytic (not PSH) functions. – Alexandre Eremenko Aug 28 '13 at 18:35

An obvious obstruction to the proposed topological characterization is the following: Endow $H(D)$ with the discrete topology, call it $\tau$. Then the space $\langle H(D) , \tau\rangle$ has the following properties: $f \mapsto f(z)$ is continuous for every $z$, $\langle H(D), \tau\rangle$ is a topological ring (as in, point-wise addition and point-wise multiplication of functions are continuous), and $\langle H(D), \tau\rangle$ is metrizable. The obvious down-side to this topology is that it is not separable and not natural in the same way but it does offer a counter-example to the statement

If $\tau$ is a (Hausdorff) topology on $H(D)$ for which point-evaluation is continuous, the group and ring operations are continuous, and $H(D)$ is metrizable, then $\tau$ must be the compact-open topology.

Given any topological space $$(T,\tau)$$ and a mapping $$\varphi :T\to T$$, it is natural to ask

What is the coarsest topology, finer than $$\tau$$, such that $$\varphi$$ is continuous.

This topology is the supremum of all $$\tau_n$$, which is constructed by adding the inverse images, through $$\varphi$$, of all open subsets of $$\tau_{n-1}$$. Note it $$\hat{\tau}$$.

If one takes the space $$T_0$$ of infinitely differentiable functions on $$\mathbb{R}$$, $$\tau_0$$, the topology of local uniform convergence (which is reasonable to preserve continuity) and $$\varphi=\frac{d}{dx}$$, then $$\hat{\tau_0}$$ is the topology of compact uniform convergence of functions and all their derivatives (one can check that the sequence $$\tau_{n}$$ is strictly increasing). The process is stationary iff $$\varphi$$ is already continuous (obvious). What is remarkable is that, due to Cauchy, $$\varphi$$ is continuous for $$\tau_0$$. The process can be applied to a set $$(\varphi_i)_{i\in I}$$ of functions, $$\tau_n$$ being constructed by adding the inverse images, through $$\varphi_{i}$$ of all open subsets of $$\tau_{n-1}$$ and taking their unions and finite intersections. If, on $$C^{\infty}(D;\mathbb{C})$$, one takes $$\varphi_1=\frac{\partial}{\partial x}$$ and $$\varphi_2=\frac{\partial}{\partial y}$$, one get the topology of compact uniform convergence of functions and all their (partial) derivatives. The process described above is, again, strictly increasing, but the restriction of it to $$H(D)$$ is stationary which is IMHO remarkable.

Let $\mathcal T$ be the topology of uniform convergence on compact subsets.

Can we do these? ...

(a) $H(D)$, with addition, is a Polish group in $\mathcal T$... separable, completely metraizable.

(b) If $A, B$ are Polish groups, and $\phi : A \to B$ is a Borel measurable homomorphism, then $\phi$ is continuous.

(c) The sigma-algebra on $H(D)$ generated by the point-evaluations is the same as the Borel sigma-algebra for $\mathcal T$.

Would these be enough to prove...

(z) If $\mathcal S$ is a topology on $H(D)$ making it a Polish group such that the point-evaluations are continuous, then $\mathcal S = \mathcal T$.