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I) (ii) (iii) (iv) Ø ÓÒº Let S, T be subsets of an algebra. We then have S ⊂ T ⇒ T ′ ⊂ S′, S ⊂ T ⇒ S ′′ ⊂ T ′′ , S ⊂ S ′′ , and S ′ = S ′′′ , the elements of S commute ⇔ S ⊂ S ′ . 5). Ò Ø ÓÒ (the second commutant)º Let S be a subset of an algebra A. The subset S ′′ is called the second commutant of S. It is a subalgebra of A containing S. It follows from the preceding proposition that S ′′ is commutative if the elements in S commute. The following result is useful sometimes. 6). ÈÖÓÔÓ× Ø ÓÒº Let S be a subset of an algebra A.

The following result is useful sometimes. 6). ÈÖÓÔÓ× Ø ÓÒº Let S be a subset of an algebra A. Let B denote the second commutant of S in A. For b ∈ B ∩ A we then have spB (b) = spA (b). § 13. ODDS AND ENDS: QUESTIONS OF IMBEDDING 41 Proof. Let b ∈ B ∩ A. We know that spA (b) = spAe(b), cf. 6). We also have spAe(b) ⊂ spB (b), cf. 10). So assume that λe−b is invertible in A with inverse d ∈ A. It is enough to prove that d ∈ B. So let c belong to the commutant of S in A. We have to show that d commutes with c.

There then exists τ ≥ 0 such that τ ≥ | an | ≥ rλ (an ) for all n. It follows that also τ ≥ | a | ≥ rλ (a). 9), we have rλ (τ e − a) = lim rλ (τ e − an ) ≤ τ. 11). Ì ÓÖ Ñº If A is a Hermitian Banach ∗-algebra then A+ is a closed convex cone in Asa . Proof. 10). 12). Ì ÓÖ Ñº Let A be a Banach ∗-algebra. Let α ≥ 2 and put β := α + 1. The following statements are equivalent. (i) A is Hermitian, (ii) for each a ∈ A one has sp(a∗ a) ⊂ [−λ, λ], where λ := max ( sp(a∗ a) ∩ R+ ) ∪ { 0 } , (iii) for all a ∈ A with rσ (a) < 1, one has rσ (c) ≤ 1, where c := α −1 (βa − aa∗ a), (iv) sp(a∗ a) does not contain any λ < 0 whenever a ∈ A.