The Fell-Doran problem is a problem in functional analysis. It goes as follows: Let $A$ be a complex unital algebra, $X$ a locally convex space, and $L(X)$ the algebra of all continuous endomorphisms of $X$. Suppose that we have a representation of $A$ on $X$, by which we simply mean an algebra homomorphism
$$
 T : A \rightarrow L(X)
$$
which is irreducible (no proper closed invariant subspace) and has trivial commutant (any bounded operator commuting with all the $T\_a$ must be a multiple of the identity). The Fell-Doran problem is: Is $T(A)$ dense in $L(X)$ in the strong operator topology? 


My question is: Is this a problem having to do with the fact that we didn't require a topology on our algebra? In other words, what can be said about the case when $A$ is actually a 'topological algebra' and the map $T$ is required to be continuous in some sense? Does that make the problem trivial, i.e. is the answer then automatically yes?


By the way, I have heard that so far there is almost no progress on the Fell-Doran problem in general; not even for Hilbert spaces! The only thing that is known is that there exists a certain concrete space where the answer is affirmative.



Is the Fell-Doran problem trivial in a topological setting?
I don't agree with Andrew: more specifically, if $A$ is the algebra of compact operators on a Hilbert space $H$, then let $T\in L(H)$. If, say, $H$ is separable, then let $(e\_n)$ be an orthonormal sequence, and let $P\_n$ be the orthogonal projection on the span of $e\_1,\cdots,e\_n$. Then $P\_n$ is compact, and $P\_n(x)\rightarrow x$ in norm for each $x\in H$. In particular, $P\_nT\in A$ for each $A$ and $P\_nT\rightarrow T$ is the strong operator topology.


More generally, the Kaplansky Density Theorem (see your favourite book on Operator Algebras, or Wikipedia) shows that if $A\subseteq L(H)$ is a $\*$-closed algebra, then the unit ball of $A$ is strong operator dense in the unit ball of the von Neumann algebra which $A$ generates. So if $A$ has trivial commutant, then the von Neumann bicommutant theorem shows that $A$ generates all of $L(H)$. In particular, $A$ is strong operator dense in $L(H)$.


Of course, the question is: what happens if $A$ isn't $\*$-closed...


