By Kalnin, Robert Avgustovich

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Proof. 5): rλ (a) ≤ a = rσ (a). 5). Ì ÓÖ Ñº A closed ∗-subalgebra of a Hermitian Banach ∗-algebra is Hermitian as well. Proof. One uses the fact that a Banach ∗-algebra C is Hermitian if and only if for every a ∈ C one has rλ (a) ≤ rσ (a). 6). Ä ÑÑ º Let B be a Hermitian closed ∗-subalgebra of a Banach ∗-algebra A. If an element of B is invertible in A, then it is invertible in B. (Here we use the canonical imbedding of B in A, cf. ) Proof. Let b ∈ B be invertible in A. The element b∗ b is invertible ∗ in A (with inverse b−1 (b−1 ) ).

We have to show that b∗ = b. 7). So b = b∗ by the uniqueness statement of the preceding theorem. 5). Ì ÓÖ Ñº Let A be a Banach algebra, and let x be an element of A with sp(x) ⊂ ]0, ∞[. ) Then x has a unique square root y in A with sp(y) ⊂ ]0, ∞[. Moreover x has no other square root with spectrum in [0, ∞[. The element y belongs to the closed subalgebra of A generated by x. If A is a Banach ∗-algebra, and if x is Hermitian, so is y. Proof. We may assume that rλ (x) ≤ 1. The element a := x − e then has sp(a) ⊂ ] − 1, 0], whence rλ (a) < 1.

It is used that the continuation is isometric, which implies that its image is complete, hence closed in A. 2). ÓÖÓÐÐ ÖÝº Let a = a∗ be a Hermitian element of a C*-algebra A. If f ∈ C sp(a) o satisfies f ≥ 0 on sp(a) then f (a) is Hermitian and sp f (a) ⊂ [0, ∞[. § 12. AN OPERATIONAL CALCULUS 39 Proof. Let g ∈ C sp(a) o with f = g 2 and g ≥ 0 on sp(a). Then g is real-valued, hence Hermitian in C sp(a) o . This implies that also g(a) is Hermitian, which thus has real spectrum. It follows that f (a) = 2 g(a) 2 is Hermitian and that sp f (a) = sp g(a) ⊂ [0, ∞[ by the Rational Spectral Mapping Theorem.

### Algebra y funciones elementales by Kalnin, Robert Avgustovich

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