Is the subspace of open maps dense in the space of bounded operators between a Hilbert space and itself? I'm trying to prove Schur's lemma for unitary representations of a topological group. I've already read the standard proof, but I am trying to generalize the argument used for finite-dimensional representations (which is apparently not the standard approach).
I was able to prove that, if the set $\mathcal{O}(V)$ of open bounded linear operators $V \to V$ is dense in the Banach space $\mathcal{B}(V)$ of bounded operators $V \to V$ then Schur's lemma holds. However, I'm not convince $\mathcal{O}(V)$ is always dense in $\mathcal{B}(V)$. My question is: is this the case? In other words, is the space of open bounded operators always dense in the space of bounded operators between a Hilbert space and itself?
If this is case, is there a name for this result? If this is not the case generally, is there some condition one could impose on $V$ to get the desired result?
Thanks in advance ️
If you want more context: One can adapt the argument used for finite-dimensional representations to show every open representation homomorphism between an irreducible unitary representation $V$ and itself is a scalar multiple of the identity. The question then is: is every other representation homomorphism $V \to V$ a scalar multiple of the identity? If $\mathcal{O}(V)$ is dense in $\mathcal{B}(V)$ then the general result would follow almost immediately from the (not necessarilly true) fact that we can approximate every bounded operator with open bounded operators.
 A: Assuming you want to use the norm topology on $B(V)$ then this is not true when $V$ is infinite-dimensional.  Let me note first that by the open mapping theorem, a bounded operator $V\to V$ is an open map iff it is surjective.  Now suppose $T:V\to V$ is any non-surjective operator that has a left inverse $S$ (for instance, $T$ could be the right shift operator $\ell^2\to\ell^2$ which has the left shift $S$ as a left inverse).  Note that $S$ cannot be injective, since $ST=1$ means that $S$ must be surjective already on the image of $T$ but that image is not all of $V$.  Now note that if $E$ is any operator of sufficiently small norm, $S(T+E)=1+SE$ is invertible, since the set of invertible operators is open in the norm topology.  This implies that $T+E$ cannot be surjective, since if $T+E$ were surjective, then the non-injectivity of $S$ would imply that $S(T+E)$ is not injective.  Thus, every operator sufficiently close to $T$ (in the norm topology) fails to be surjective.
(If you want to use a weaker topology then surjective operators are probably trivially dense.  For instance, a basic open set in the strong operator topology on $B(V)$ only constrains your operator on a finite-dimensional subspace, so you can freely define it to be surjective on the orthogonal complement of that subspace.)
