Is the inverse of a nonzero intertwining operator in Schur's lemma bounded? Definition. Let $G$ be any topological group. A representation of $G$ on a nonzero complex Hilbert space $V$ is a group homomorphism $\pi$ of $G$ into the group $GL(V)$ of bounded linear operators on $V$ with bounded inverses, such that the resulting map $G\times V\to V$ given by $(g,v)\mapsto \pi(g)v$ is continuous. 
Schur's lemma. Let $G$ be a topological group. Let $V_1$ and $V_2$ be nonzero complex Hilbert spaces. Suppose $(\pi_1,V_1)$ and $(\pi_2,V_2)$ are any two irreducible unitary representations of $G$. Let $T: V_1\to V_2$ be a bounded linear operator such that $T\pi_1(g)=\pi_2(g)T$ for all $g\in G$. Then if $T\neq 0$ then $T$ is invertible.
Well-known proof is as follows:
$Proof.$  If $x\in \operatorname{Ker}T$ then we have $T\pi_1(g)x=\pi_2(g)Tx=0$ for all $g\in G$. So the closed subspace $\operatorname{Ker}T$ of $V_1$ is invariant under $\pi_1$. By the irreducibility of $\pi_1$, we must have $\operatorname{Ker}T=0$ since $T\neq 0$. On the other hand, if $y\in \operatorname{Im}T$ then there is some $x\in T$ such that $Tx=y$. For $g\in G$, we get $\pi_2(g)y=\pi_2(g)Tx=T\pi_1(g)x$, which implies that $\operatorname{Im}T$ is invariant under $\pi_2$. By the irreducibility of $\pi_2$, we must have $\operatorname{Im}T=V_2$ since $T\neq 0$. So $T$ is invertible.
My question: Can we also conclude that $T^{-1}$ is a bounded linear operator? Thanks!
 A: The open mapping theorem implies that any bijective bounded linear operator between Banach spaces has a bounded inverse. This will answer your question.
But the proof you wrote is in fact wrong. You may only conclude that the image is dense, since the closure of the image is a closed invariant subspace. So what you need to show is that the image of $T$ is already closed. To do this, consider the bounded linear intertwiner $S:=T^{\ast} \circ T: V_1 \rightarrow V_1$, where $T^{\ast}: V_2 \rightarrow V_1$ is the adjoint operator. By what is also known as "Schur's Lemma", one gets $S = \lambda \text{id}_{V_1}$ for some $\lambda \in \mathbf{C}$. By the assumption that $T \neq 0$, there is a vector $v_1 \in V_1$ with $Tv_1 \neq 0$, which satisfies
$$
\lambda \langle v_1, v_1 \rangle_{V_1} = \langle Sv_1 , v_1 \rangle_{V_1} = \langle Tv_1, Tv_1 \rangle_{V_2} > 0\,.
$$
Hence $\lambda > 0$ and it follows that $\frac{1}{\sqrt{\lambda}}T: V_1 \rightarrow V_2$ is an isometric $G$-morphism. Therefore, $\frac{1}{\sqrt{\lambda}}T$ and $T$ have closed images.
