About irreducible subgroups of $GL(V)$, $V$ a vector space of infinite dimension Let $F$ be a field, let $V$ be a $F$-vector space, let us say that a subgroup $G$ of $GL(V)$ is irreducible if $V > 0$ and there is no $G$-invariant subspace $W$ of $V$ such that $0 < W < V$.
If I'm not wrong, D.J.S. Robinson, A Course in the Theory of Groups, 8.1.5, p. 219, implies what follows : if $F$ is an algebraically closed field, if $V$ is a $F$-vector space with finite dimension, if $G$ is an irreducible subgroup of $GL(V)$, if $h$ is a $F$-endomorphism of $V$ commuting with every automorphism in $G$, then $h$ is scalar, i.e. there exists an element $f$ of $F$ such that $h$ is the multiplication by $f$.
Could anybody give a counterexample in the case where the dimension of $V$ is infinite (if such a counterexample exists) ? In other words, could anybody give an example of the following situation : $F$ is an algebraically closed field, $V$ is a $F$-vector space with infinite dimension, $G$ is an irreducible subgroup of $GL(V)$, $h$ is a $F$-endomorphism of $V$ commuting with every automorphism in $G$ and $h$ is not scalar ?
If I'm not wrong, $G$ must be infinite and cannot be cyclic. Now, life is short, so I prefer not to spend time on an already solved question. Thanks in advance for the answers. 
 A: In more standard language you are asking whether Schur's lemma holds for infinite-dimensional irreducible representations. It continues to hold in the following form:

If $G$ is a group and $V$ is an irreducible representation of $G$ over a field $F$, then $\text{End}_G(V)$ is a division algebra over $F$.

If $V$ is finite-dimensional and $F$ is algebraically closed, the only such division algebra is $F$ itself, so we get the more typical statement of Schur's lemma. But this fails once either of these conditions is relaxed.
Explicitly, even if $F$ is algebraically closed, you can take $V = F(t)$ and $G = F(t)^{\times}$ acting on $V$ by left multiplication. Then $V$ is irreducible (and infinite-dimensional, despite the fact that $G$ is abelian!), and $\text{End}_G(V) \cong F(t)$. 
A: At least in the context of Banach spaces, continuous group actions and closd invariant subspaces, the answer is negative. Namely, there are invertible bounded operators $T: B\to B$ of Banach spaces without proper closed invariant subspaces (Enflo, Read, et al), see this wikipedia article for the references. Now, take the group $G$ generated by $T$. Such group commutes with $T$ itself which is not a scalar operator. I am not sure if you ask a purely linear algebra question without continuity assumptions and closedness requirement for a subspace. 
