Corollary 5

Let $\varrho$ be a unitary representation of a topological group G in a Hilbert space E. Let $\pi$ be an irreducible unitary representation of G in a Hilbert space F.

a) For every nonzero G-morphism $u$ of $\pi$ into $\varrho$, there exists $\lambda \in \mathbf{R}^*_+$ such that $\lambda u$ is isometric;

b) Every G-morphism $u$ of $\pi$ into $\varrho$ has closed image;

c) Every nonzero G-morphism $v$ of $\varrho$ into $\pi$ is surjective.

Let us prove assertion a), which implies assertion b) (Lemma 8 of I, p. 107). Since the morphism $u$ is nonzero, it is injective (Lemma 2 of V, p. 378). The formula $q(x, y) =\langle u(x)|u(y)\rangle$ for $x$ and $y$ in F then defines a continuous sesquilinear form on F. Since $u$ is a G-morphism, one has $q(\pi (g)x, \pi (g)y) =q(x, y)$ for all $g$ in G and $(x, y)\in F\times F$, hence $q$ is G-invariant. By Cor. 3, there exists $\alpha \in \mathbf{R}^*_+$ such that

for every $(x, y)\in F\times F$, and therefore $\alpha^{-1/2}u$ is isometric.

Let us finally prove assertion c). Since $\pi$ is irreducible, the image of $v$ is dense in F (Lemma 2 of V, p. 378), hence the adjoint $v^*$ is injective (TVS, V, p. 41, Prop. 4). By b), the image H of $v^*$ is a closed subspace of E; it is therefore a subrepresentation of $\varrho$, and $v^*$ induces by passing to the subspaces a G-isomorphism of F onto H. The restriction $w$ of $v$ to H defines a G-morphism of H into F, which is injective since its kernel is the intersection of H and Ker($v$) $= H^{\circ}$ (loc. cit.). The mapping $w$ is therefore an isomorphism (Corollary 2); in particular, $v$ is surjective.