Is the Fell-Doran problem trivial in a topological setting? 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.
 A: 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??
A: I don't know, but I would be very surprised if a topology made the problem any easier since I would imagine that you could always use the induced topology from $L(X)$ on $A$ so if you can answer it in the topological setting then you can answer it in the discrete setting.  More specifically, the question is about the image of $T$ but imposing a topology on $A$ only messes around with the domain so I'd be surprised if it had a significant effect.
However I suspect that I haven't understood the problem very well since it seems as though a simple example would be where $A$ was the compact operators together with the unit.  That seems to fit the conditions but it certainly isn't dense in the strong operator topology (though it is in the weak topology).
Edit: As I suspected, I hadn't.  As Matthew points out, the topology is not the strong operator topology but the weak operator topology with respect to the strong topology on $X$.  That is, $T_\gamma \to T$ if $T_\gamma x \to Tx$ for all $x$.
This, then, sounds a lot like the approximation question which is known to be false for arbitrary Banach spaces (paper of Eno, though I forget the exact reference) since in a space where the approximation property fails one could take the subalgebra of finite rank operators (plus the identity to make it unital).
However, my first point still seems valid.
A: 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...
