, so the spectrum is closed. Since the spectrum is also bounded, it is compact by Heine-Borel.
Proposition: Let be a Banach algebra and a (left, right, two-sided) ideal of . Then:
- The closure of is also a (left, right, two-sided) ideal.
- If is proper, then so is .
- If is two-sided, closed, and proper, the quotient is also a Banach algebra and the map is a morphism of Banach algebras.
- If is maximal, it is already closed.
Proof. : is the closure of a subspace, hence also a subspace, hence closed under addition and scalar multiplication. If is a Cauchy sequence converging to , then , hence is a Cauchy sequence converging to , so is closed under left multiplication by elements of if is and similarly for right multiplication.
: An ideal is proper if and only if it contains no invertible elements if and only if it does not contain the identity (and this is true for left, right, and two-sided ideals). Since the open ball of radius about the identity consists of invertible elements, it must be disjoint from any proper ideal , so cannot contain the identity and hence must also be proper.
: If is two-sided, closed, and proper, the quotient is a Banach space and an algebra, and the map is a weak contraction and an algebra homomorphism. It remains to show that the quotient norm on satisfies , but this follows from the fact that the cosets contain the possible products of elements in the cosets , hence the infimum of the norms in the former is at most the infimum of the norms in the latter.
This already shows that the quotient norm on satisfies , since implies and the quotient map is a weak contraction. A direct argument is to observe that the open ball of radius about consists only of invertible elements, hence is disjoint from .
: Follows directly from .
Corollary: Any algebra homomorphism is a morphism of Banach algebras.
This is one of several results in Banach algebra theory to the effect that an algebra homomorphism between certain Banach algebras is automatically continuous; see, for example, Section I.5 of Dales et al.
Basic properties assuming the spectrum is non-empty
The following is fundamental.
Black Box #1: The spectrum is non-empty.
The idea of the proof is that if were empty, then the resolvent would be a non-constant “bounded holomorphic function,” which contradicts Liouville’s theorem. This proof actually works with no modifications for complex matrices and shows that they have eigenvalues without going through either characteristic polynomials or the fundamental theorem of algebra, and in fact this is one way to prove the fundamental theorem of algebra. In general we need to develop some complex analysis for Banach space-valued holomorphic functions, which we defer to a later post.
For now, it it is more important to see what can be done with this result.
Corollary (Gelfand-Mazur): Any Banach algebra which is a division algebra is isometrically isomorphic to .
Proof. Let be such a Banach algebra and let be nonzero. is non-empty, so let be such that is not invertible. Since is nonzero, it must be invertible, so , hence is not invertible. But then it must be equal to , so .
Corollary: Let be a commutative Banach algebra. Then if and only if there exists a morphism such that .
Proof. : if there exists such an , then lies in a maximal ideal, so cannot be invertible. (This argument works in any algebra over .)
: if , then is not invertible, hence is contained in a maximal ideal , which is already closed. Since by Gelfand-Mazur, the quotient map is the desired .
(Note that we make essential use of the axiom of choice here.)
A digression into quantum mechanics
The non-emptiness of the spectrum, besides leading to the above results, also has the following corollary, which is apparently due to Wielandt.
Proposition: The Weyl algebra does not embed into a Banach algebra.
Proof. Suppose to the contrary that are elements of a Banach algebra satisfying . We make the elementary observation that , from which it follows that
But we know that , so must be invariant under addition by . Since is non-empty, it must contain arbit