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Let $G$ be a connected, non-compact Lie group and let $\rho: G \to GL_n(\mathbb C)$ be a complex representation of $G$.

Definition: $\rho$ is quasi-unitary if there exists some positive hermitian matrix $A \in GL_n(\mathbb C)$ such that \begin{equation}\sup_{g \in G}||\rho(g) \cdot A \cdot \rho(g)^*|| < C \end{equation} for some $C > 0$, where $\rho(g)^*$ denotes the conjugate-transpose of $\rho(g)$, and $|| \cdot ||$ is some matrix-norm.

If $\rho$ is unitary, one can chose a positive $A \in GL_n(\mathbb C)$ such that $\rho(g) \cdot A \cdot \rho(g)^* = A$ for all $g \in G$, hence $\rho$ is quasi-unitary.

Question: Is every quasi-unitary representation unitary ?

I believe that this is the case, but I can't quite prove it. My idea is to consider the conjugate-action of $G$ on $GL_n(\mathbb C)$ induced by a quasi-unitary representation $\rho$, given by $g.B := \rho(g) \cdot B \cdot \rho(g)^*$. Let $A \in GL_n(\mathbb C)$ be such that $\sup_{g \in G} ||g.A|| < C$. Then $G.A$ (the $G$-orbit of $A$) is bounded (hopefully with compact closure). Now it is somehow intuitive to assume that a fixed point $C \in \overline{G.A}$ of the $G$-action must exist. Unfortunaltey, all fixed point Theorems that I know do not quite apply to this situation.

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