Bourbaki

Proposition 6

Tag 02Z3, permanent. Cite it as [Bourbaki, Tag 02Z3].

Théories spectrales, chapter IV, § 2. Endomorphismes normaux

The mapping m:g↦m~gm:g\mapsto \widetilde{m}_g is an isometric unital morphism of the involutive algebra L∞(X,μ)L^{\infty}(X, \mu) into the involutive algebra L(L2(X,μ))\mathscr{L}(L^2(X, \mu)).

In particular, for every g∈L∞(X,μ)g\in L^{\infty}(X, \mu), the multiplication endomorphism m~g\widetilde{m}_g is normal; it is hermitian if and only if gg is locally μ\mu-almost everywhere real-valued.

By Prop. 5, the mapping m~\widetilde{m} is an injective isometric morphism of unital Banach algebras of L∞(X,μ)L^{\infty}(X, \mu) into L(L2(X,μ))\mathscr{L}(L^2(X, \mu)). Let g∈L∞(X,μ)g\in \mathscr{L}^{\infty}(X, \mu). For f1f_1 and f2∈L2(X,μ)f_2\in \mathscr{L}^2(X, \mu), one has

⟨f1∣m~g(f2)⟩=∫Xf1(x)‾g(x)f2(x)dμ(x)=⟨m~g‾(f1)∣f2⟩ \langle f_1|\widetilde{m}_g(f_2)\rangle =\int_X\overline{f_1(x)}g(x)f_2(x)d\mu(x) =\langle \widetilde{m}_{\overline{g}}(f_1)|f_2\rangle

whence it follows that m~g∗=m~g‾\widetilde{m}^*_g=\widetilde{m}_{\overline{g}}, which proves that mm is an involutive morphism. The final assertions follow from this (cf. I, p. 106, Prop. 5).

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.

Cited by

Nothing.

Cites

Nothing.

Machine translation of the English, by gpt-5-6-mini, gpt-5-mini, gpt-5.4, 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.