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I'm interested in sufficient conditions for a notion of sequential convergence to be induced by a topology. More precisely: Let $V$ be a vector space over $\mathbb{C}$ endowed with a notion $\tau$ of sequential convergence. When is there a topology $\mathcal{O}$ on $V$ that makes $V$ a topological vector space such that "sequences suffice" in $(V,\mathcal{O})$, e.g. $(V,\mathcal{O})$ is first countable, and convergence in $(V,\mathcal{O})$ coincides with $\tau$-convergence? Is the topology $\mathcal{O}$ uniquely determined?

By a notion $\tau$ of sequential convergence on a vector space $V$ I mean a "rule" $\tau$ which assigns to certain sequences $(v_n)_{n\in\mathbb{N}}\subset V$ (which one would call convergent sequences) an element $v\in V$ (a limit of $(v_n)_n$). One could write $v_n\stackrel{\tau}{\to}v$ in this case. This process of "assigning a limit" should satisfy at least that any constant sequence $(v,v,v,...)$ is convergent and is assigned the limit $v$. Also, given a convergent sequence $(v_n)_n$ with limit $v$ any subsequence $(v_{n_k})_k$ should have $v$ as a limit.
I would also like this concept of assigning a limit to be compatible with addition in $V$ and multiplication by a scalar.
Maybe one should include further restrictions. In fact I would like to know which further assumptions on this "limiting process" one has to assume in order to ensure that this limiting procedure corresponds to an actual topology on $V$ which makes $V$ a topological vector space in which a sequence converges if and only if it $\tau$-converges.

Let me give two examples. If we take for instance a topological vector space $(V,\mathcal{O})$ then we have a notion of convergence in $V$ based upon the set $\mathcal{O}$ of open sets of $V$. This notion of convergence clearly satisfies the above assumptions on $\tau$.
If on the other hand we consider $L^\infty([0,1])$ equipped with the notion of pointwise convergence almost everywhere, then there is no topology on $L^\infty([0,1])$ which makes $L^\infty([0,1])$ a TVS in which a sequence converges if and only if it converges pointwise almost everywhere. Still, convergence almost everywhere also satisfies the above assumptions on $\tau$.
So the above assumptions on this concept of convergence are necessary but not sufficient conditions for what I mean by a notion of convergence to correspond to an actual topology. The question is: Which further assumptions do I have to make?

On a less general level I'm particularly interested in the following case: Let $G\subset\mathbb{C}^d$ be a domain, $X$ a (complex) Banach space and let $H^\infty(G;X)$ denote the space of bounded holomorphic functions $f\colon G\to X$. Now consider the following notion $\tau$ of sequential convergence on $H^\infty(G;X)$: We say that a sequence $(f_n)_{n\in\mathbb{N}}\subset H^\infty(G;X)$ $\tau$-converges to $f\in H^\infty(G;X)$ if $\sup_{n\in\mathbb{N}}\sup_{z\in G} \|f_n(z)\|_X$ is finite and $f_n(z)$ converges in $X$ to $f(z)$ for every $z\in G$. Is there a topology $\mathcal{O}$ on $H^\infty(G;X)$ such that "sequences suffice" in $(H^\infty(G;X),\mathcal{O})$ and a sequence $(f_n)_{n\in\mathbb{N}}\subset H^\infty(G;X)$ converges w.r.t. the topology $\mathcal{O}$ if and only if it $\tau$-converges? Is this topology $\mathcal{O}$ unique if existent? What if we drop the "sequences suffice"-restriction? Is $(H^\infty,\mathcal{O})$ locally convex? Metrizable? What if we replace $X$ by a more general space like a LCTVS or a Frechet space?

Thank you in advance for any suggestions, ideas or references.

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$H^{\infty}(G;X)$ is not an algebra unless $X$ itself is. –  Plop Jan 31 '11 at 11:32
    
Thank you, I changed my post accordingly. I also forgot to mention, that by $H^\infty$ I denote, of course, the space of bounded holomorphic functions. –  lvb Jan 31 '11 at 11:55
    
May I ask precisely what a "notion $\tau$ of sequential convergence" is? Is this a "rule" (really, a set) saying which sequences converge and to what points they converge? –  Pete L. Clark Jan 31 '11 at 15:07
    
Yes, that's exactly what I mean. I included some changes in my post in order to be somewhat more precise on that point. –  lvb Jan 31 '11 at 16:24

2 Answers 2

up vote 4 down vote accepted

I am addressing only the first part of your question (i.e., nothing with the structure of vector space; only topology and limits of sequences). I will quote here part of Problems 1.7.18-1.7.20 from Engelking's General Topology. (It would be better if you could get the book. I believe it used to be here, but the links don't work now. Perhaps you'll find it in the Internet.)

L*-space is a pair $(X, \lambda)$, where X is a set and $\lambda$ a function (called the limit operator) assigning to some sequences of points of X an element of X (called the limit of the sequence) in such a way that the following conditions are satisfied:

(L1) If $x_i=x$ for $i = 1,2,\dots$, then $\lambda x_i = x$.

(L2) If $\lambda x_i = x$, then $\lambda x' = x$ for every subsequence $x'$ of $x$.

(L3) If a sequence $\{x_i\}$ does not converge to $x$, then it contains a subsequence $\{x_{k_i}\}$ such that no subsequence of $\{x_{k_i}\}$ converges to $x$.

These properties are sufficient to define a closure operator on $X$ (not necessary idempotent).

If $(X,\lambda)$ fulfills and additional condition

(L4) If $\lambda x_i = x$ and $\lambda x^i_j = x_i$ for $i = 1,2,\dots$, then there exist sequences of positive integers $i_1, i_2,\dots$ and $j_1, j_2, \dots$ such that $\lambda x_{j_k}^{i_k} = x$.

L*-space $X$ satisfying (L4) is called an S*-space. The closure operator given by S*-space is idempotent.

Using this closure operator we get a topology, such that the convergence of the sequences is given by $\lambda$. A topology can be obtained from a L*-space (S*-space) if and only if the original space is sequential (Frechet-Urysohn).

References given in Engelking's book are Frechet [1906] and [1918], Urysohn [1926a], Kisynski [i960].

Frechet [1906] Sur quelques points du calcul fonctionnel, Rend, del Circ. Mat. di Palermo 22 (1906), 1-74.

Frechet [1918] Sur la notion de voisinage dans les ensembles abstraits, Bull. Sci. Math. 42 (1918), 138-156.

Kisynski [1960] Convergence du type L, Coll. Math. 7 (1960), 205-211.

Urysohn [1926a] Sur les classes (L) de M. Frechet, Enseign. Math. 25 (1926), 77-83.

NOTE: Some axioms for convergence of sequences are studied in the paper: Mikusinski, P., Axiomatic theory of convergence (Polish), Uniw. Slaski w Katowicach Prace Nauk.-Prace Mat. No. 12 (1982), 13-21. I do not have the original paper, only a paper which cites this one; it seems that the axioms are equivalent to (L1)-(L3) and the uniqueness of limit. (But I do not know, whether some further axioms are studied in this paper.)

EDIT: In Engelking's book (and frequently in general topology) the term Frechet space is used in this sense, not this one. I've edited Frechet to Frechet-Urysohn above, to avoid the confusion.

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That sounds pretty much like what I was looking for. I'll go get myself a copy of Engelking's book. Thank you very much! –  lvb May 7 '11 at 17:32

First note that your notion of $\tau$-convergence imposes to consider only bounded holomorphic functions.

If $X$ is finite-dimensional, Montel's theorem tells you that $f_n$ $\tau$ converges to $f$ iff $f_n$ converges uniformly to $f$ on any compact subset, and this "notion" is metrizable.

If $X'$ is separable, the "notion of convergence" is still metrizable (compose $f$ and the $f_n$ with $\Lambda$ for all $\Lambda$ in an enumerable and dense family in $X'$).

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Thank you. In the finite dimensional case Montel obviously implies that $\tau$-convergence implies compact convergence, but I'm having difficulties convincing myself that the converse also holds, i.e. that compactly convergent sequences of bounded holomorphic functions are in fact uniformly bounded. Could you please elaborate on that point? –  lvb Jan 31 '11 at 14:29
    
You are absolutely right, it is quite easy to build an example of an unbounded sequence however convergeing uniformly to 0 on any compact. But I think it is too early to surrender. Assume the $f_n$ are bounded but not uniformly bounded, and converge to 0 uniformly on any compact of G. Then for every $n$, $|f_{\phi(n)}(z_n)| \geq 3^n$ for some $z_n \in G$. Then $\cdot \mapsto \sum_k 2^{-k}|\cdot (z_k)|$ is a seminorm on $H^{\infty}(G;X)$ for which $(f_n)_n$ is unbounded. So the "notion of convergence" is still given by a family of seminorms, and thus a topology. –  Plop Jan 31 '11 at 15:38

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