TITLE: Is the Fell-Doran problem trivial in a topological setting?
QUESTION [5 upvotes]: 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.

REPLY [4 votes]: Actually, digging a bit deeper, I've found this paper by Zelazko: Colloquium Mathematicum which states the problem as: If $A\subseteq L(X)$ is topologically irreducible (that is, the orbit under $A$ of any non-zero vector is dense) and the commutant of $A$ is trivial, is $A$ totally irreducible.  This means that for each $n$, if $x=(x_1,\cdots,x_n)$ is a vector in $X^n$ with linearly independent co-ordinates, then $x$ should be cyclic for $A^n$ (where $A^n$ acts on $X^n$ is the obvious way).
Off hand, I don't see what, if anything, this has to do with $A$ being strong operator topology dense in $L(X)$.  Or is there more than one conjecture??