Bourbaki

Lemma 8

Tag 036G, permanent. Cite it as [Bourbaki, Tag 036G].

Théories spectrales, chapter IV, § 5. Opérateurs partiels normaux et théorème spectral

Let E be a separated topological vector space. Let (E1,∥⋅∥1)(E_1,\|\cdot \|_1) and (E2,∥⋅∥2)(E_2,\|\cdot \|_2) be dense subspaces of E endowed with Hilbert space structures such that the canonical injections of E1E_1 and E2E_2 into E are continuous. Let F be a subspace of E1∩E2E_1\cap E_2, dense in E1E_1 and E2E_2 for these Hilbert space structures. If ∥x∥1=∥x∥2\|x\|_1=\|x\|_2 for every x∈Fx\in F, then E1=E2E_1= E_2 and ∥⋅∥1=∥⋅∥2\|\cdot \|_1=\|\cdot \|_2.

Let x∈E1x\in E_1. There exists a sequence (xn)n∈N(x_n)_{n\in\mathbf{N}} in F such that xnx_n converges to xx in the Hilbert space E1E_1. Since ∥xn−xm∥2=∥xn−xm∥1\|x_n-x_m\|_2=\|x_n-x_m\|_1 for all integers nn and mm, the sequence (xn)(x_n) is a Cauchy sequence in E2E_2. Let yy be its limit. We have ∥y∥2=\|y\|_2= lim∥xn∥2=∥x∥1\|x_n\|_2=\|x\|_1. Since the canonical injections of E1E_1 and E2E_2 into E are continuous, we have xn→xx_n\rightarrow x in E and analogously xn→yx_n\rightarrow y in E. Thus it follows that x=yx=y and ∥x∥1=∥x∥2\|x\|_1=\|x\|_2; in particular, E1⊂E2E_1\subset E_2. We conclude by symmetry.

Let E be a complex Hilbert space. Let uu be a self-adjoint positive partial operator on E. For every α∈R+\alpha \in \mathbf{R}_+, we set uα=f(u)u^{\alpha}=f(u), where ff is the continuous mapping x↦xαx\mapsto x^{\alpha} of R+\mathbf{R}_+ into R\mathbf{R}. This is a self-adjoint positive partial operator (Cor. 1, a) of IV, p. 273). When u∈Lu\in \mathscr{L} (E), this notation is compatible with the notation relative to the involutive algebra L(E)(\mathscr{L}(E) (cf. Prop. 16 of I, p. 118). If α\alpha is a positive integer, the partial operator uαu^{\alpha} coincides with the partial operator defined by the composition u∘⋯∘uu\circ \cdots \circ u (Cor. 1, b) of IV, p. 273).

Let β∈R+\beta \in \mathbf{R}_+. We have uαβ=(uα)βu^{\alpha \beta}= (u^{\alpha})^{\beta} (Cor. 4 of IV, p. 274). In particular, if α>0\alpha >0, then the partial operator u1/αu^{1/\alpha} is the unique self-adjoint positive partial operator vv on E such that vα=uv^{\alpha}=u.

Suppose that 0⩽α⩽β0\leqslant \alpha \leqslant \beta. Then dom(uβu^{\beta})⊂\subset dom(uαu^{\alpha}) :: in fact, for every real number x⩾0x\geqslant 0, one has xα⩽1+xβx^{\alpha}\leqslant 1 +x^{\beta}, and the result then follows from the definition of the domain of uα(u^{\alpha}(cf. Prop. 3 of IV, p. 271).

Let uu be a closed operator with dense domain from E into a complex Hilbert space F. The partial operator u∗uu^*u is self-adjoint and positive (Prop. 12 of IV, p. 241). We write ∣u∣=(u∗u)1/2|u|= (u^*u)^{1/2}.

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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Cites

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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.