Bourbaki

Corollary 1

Tag 0396, permanent. Cite it as [Bourbaki, Tag 0396].

Théories spectrales, chapter V, § 1. Représentations unitaires

Let π\pi be an irreducible unitary representation of a topological group G in a Hilbert space E. Let uu be a closed partial operator on E. Suppose that uu has dense domain, that dom(uu) is stable under π\pi and that u∘π(g)=π(g)∘uu\circ \pi (g) =\pi (g)\circ u for all g∈Gg\in G. Then dom(uu) =E= E and uu is a homothety.

The partial operator u∗∘uu^*\circ u is a positive self-adjoint partial operator on E (Prop. 12 of IV, p. 241); the same is true of v=1E+u∗∘uv= 1_E+u^*\circ u, and the latter is injective since −1∉-1\notin Sp(u∗∘uu^*\circ u) (Prop. 17 of IV, p. 248). We have u∗∘π(g)=π(g)∘u∗u^*\circ \pi (g) =\pi (g)\circ u^* for all g∈Gg\in G (Lemma 6 of V, p. 381), whence v∘π(g)=π(g)∘vv\circ \pi (g) =\pi (g)\circ v for all g∈Gg\in G. Since vv is injective, it follows that v−1∘π(g)=π(g)∘v−1v^{-1}\circ \pi (g) =\pi (g)\circ v^{-1} for all g∈Gg\in G. But v−1v^{-1} belongs to L\mathscr{L} (E), hence by Prop. 6 there exists λ∈C\lambda \in \mathbf{C} such that v−1=λ1Ev^{-1}=\lambda 1_E. Necessarily λ≠0\lambda \not = 0, which implies that E = Im(v−1v^{-1}) == dom(vv)⊂\subset dom(uu), whence dom(uu) =E= E. Since uu is closed, we have u∈L(E)u\in \mathscr{L}(E) (TVS, I, p. 19, Cor. 5), hence u∈u\in HomG(π,π)_G(\pi , \pi ) and uu is a homothety by Prop. 6.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Machine translation of the English, by gpt-5.4, gpt-5-6-mini, against glossary version 34. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.