Spectrum property

Question

Can someone please provide good demonstration for this theorem:

Theorem 1.9. Let $A$ be a Banach algebra. If the elements $a, b \in A$ satisfy $ab = ba$, then

$\sigma (a + b) ⊆ \sigma (a) + \sigma (b)$, and

$\sigma(ab) ⊆ \sigma(a) \cdot \sigma(b)$

Answer 1

Sketch: 1) We do it in the commutative case 2) We restrict to the commutative case by considering the bicommutant of C[x,y].

Assume that A is commutative first. Then by Gelfand, for every x∈A, we have σ(x)={ϕ(x);ϕ∈A^} where A^ denotes the set of characters (nonzero algebra homomorphisms from A to C).

It follows readily that for all x,y∈A: σ(x+y)⊆σ(x)+σ(y)andσ(xy)⊆σ(x)σ(y). Now back to the general case where A is not assumed to be commutative, take x,y∈A such that xy=yx.

Let B=C[x,y] be the unital algebra generated by x,y. And denote B′ its commutant, that is the set of all z in A which commute with every element of B (equivalently, with x and y). Note that for any sets S⊆T, we have T′⊆S′. Also S is commutative if and only if S⊆S′.

Since B is commutative, we have: B⊆B′⇒B′′⊆B′⇒B′′⊆(B′′)′. So B′′ is commutative. Moreover, one checks easily that it is closed (hence a unital Banach algebra) and contains B.

We need to compare the spectra relative to B′′ and to A for elements of B′′. So take z in B′′. Of course, if z is invertible in B′′, it is invertible in A. Now conversely, assume that z is invertible in A. Now for all u in B′, we have zu=uz so uz−1=z−1u. Hence z−1 belongs to B′′. Therefore zinvertible in A⇔zinvertible inB′′ for all z∈B′′. Applying this to z=b−λ1, we see that σA(b)=σB′′(b)∀b∈B′′. This shows that we can restrict to the commutative unital Banach algebra B′′ and apply the above commutative case to x,y,x+y,xy which all belong to B, hence to B′′.

Note: to see why this result is natural, and why the inclusions are not equalities in general, consider the case of A=Mn(C). If x and y commute, they can be triangularized simulatneously. Then the result is clear. You see also why you don't necessarily have equalities.

Finally, note that it would not work to restrict to B, since we may not have the spectral permanence we want. We really need to move up to the bicommutant B′′.