Bourbaki

Proposition 15

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

Théories spectrales, chapter I, § 6. Algèbres stellaires

Let A be a star-algebra. Let xx be a normal element of A.

a) Suppose that A is unital and let ff be a continuous function from SpA(x)_A(x) into C\mathbf{C}. In order that f(x)f(x) be positive, it is necessary and sufficient that the image of ff be contained in R+\mathbf{R}_+;

b) Let ff be a continuous function from SpA′(x)'_A(x) into C\mathbf{C} such that f(0)=f(0) = 0. In order that f(x)f(x) be a positive element of A, it is necessary and sufficient that the image of ff be contained in R+\mathbf{R}_+.

Assertion a) follows from assertion a) of cor. 2 of I, p. 111, and assertion b) follows from prop. 13 of I, p. 114.

Let xx be a hermitian element of the star-algebra A. Its spectrum is contained in R\mathbf{R} (prop. 4 of I, p. 106). Consider the continuous functions from SpA′(x)'_A(x) into R\mathbf{R} defined by

f1:t↦f_1:t\mapsto sup(t,0t,0),f2:t↦,f_2:t\mapsto sup(−t,0-t,0),f3:t↦∣t∣,f_3:t\mapsto |t|.

We put

x+=f1(x),x−=f2(x),∣x∣=f3(x)(3) x^+=f_1(x),x^-=f_2(x),|x|=f_3(x) \tag{3}

Since the functions f1,f2,f3f_1,f_2,f_3 take positive real values and vanish at 0, the elements x+,x−x^+,x^- and ∣x∣|x| are positive elements of A (Prop. 15, a)) which belong to the stellar subalgebra of A generated by xx.

One has f1(t)−f2(t)=tf_1(t)-f_2(t) =t for every t∈Rt\in \mathbf{R}, as well as the relations f1+f2=f3f_1+f_2=f_3 and f1f2=0f_1f_2= 0. Hence the relations:

x=x+−x−,∣x∣=x++x−,x+x−=x−x+=0(4) x=x^+-x^-,|x|=x^++x^-,x^+x^-=x^-x^+= 0 \tag{4}

Since the functional calculus mapping is isometric, one has

∥∣x∣∥=∥x∥,∥x+∥⩽∥x∥,∥x−∥⩽∥x∥ \| |x| \|=\|x\|,\|x^+\|\leqslant \|x\|,\|x^-\|\leqslant \|x\|

Let xx be a positive element of A. It is hermitian, hence normal. Let α∈R+∗\alpha \in \mathbf{R}^*_+, and let gg be the restriction to SpA′(x)'_A(x) of the function t↦tαt\mapsto t^{\alpha}; one writes xα=g(x)x^{\alpha}=g(x). It is a positive element of the stellar subalgebra of A generated by xx. Let α\alpha and β\beta be in R+∗\mathbf{R}^*_+. Since the functional calculus mapping is an algebra morphism, and by Prop. 13 of I, p. 114, one has

xαxβ=xα+β(xα)β=xαβ(5) x^{\alpha}x^{\beta}=x^{\alpha+\beta}(x^{\alpha})^{\beta}=x^{\alpha \beta} \tag{5}
√/ \surd_/

One shall also write x‾=x12\overline{x}=x^{12}.

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-5-6-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.