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.