Abelian Banach Algebras
[18/0%]Let be a Banach algebra with identity and let be the smallest closed two-sided ideal of containing ; is called the commutator ideal of .
Show that is an abelian Banach algebra.
If is a closed ideal of such that is abelian, then .
If is a homomorphism, then and induces a homomorphism such that , where is the natural map. Hence .
Let be the set of homomorphisms of and let be the set of homomorphisms of . Show that the map defined in (c) is a homeomorphism of onto .
Using the terminology of Exercises 2.6 and 2.7, let be an abelian Banach algebra without identity and show that if is a maximal modular ideal, then there is a homomorphism such that . Conversely, if is a nonzero homomorphism, then is a maximal modular ideal. Moreover, the correspondence is a bijection between homomorphisms and maximal modular ideals.
If is an abelian Banach algebra and is a homomorphism, then is continuous and . If has an approximate identity such that for all , then (see Exercise 2.8).
Let be an abelian Banach algebra and let be the set of nonzero homomorphisms of . Show that is locally compact if it has the relative weak* topology from (Exercise 3).
With the notation of Exercise 4, assume that has no identity and let be the algebra obtained by adjoining an identity. For in , let be the spectrum of as an element of and show that . Also, show that the maximal ideal space of , , is the one-point compactification of .
With the notation of Exercise 4, for each in define by . Show that and the map of into is a contractive homomorphism with kernel .
If is locally compact, show that is a homeomorphism of onto the maximal ideal space of .
Let be locally compact and for each open subset of let . Show that is a closed ideal of and that every closed ideal of has this form. Moreover, the map is a lattice isomorphism from the lattice of open subsets of onto the lattice of ideals of .
With the notation of the preceding exercise, show that is a modular ideal if and only if is compact.
If is an abelian Banach algebra and , say that is a rational generator of if is dense in . Show that if is a rational generator of , then is homeomorphic to .
Verify the statements made in the following example. If is the Volterra operator and is the closure in of , then is an abelian Banach algebra and . In other words, has a unique maximal ideal, .
Say that are generators of if is the smallest Banach algebra with identity that contains . Show that are generators of if and only if , and if is the maximal ideal space, then there is a homeomorphism of onto a compact subset of such that if is a polynomial in variables, then .
Verify the statements made in the following example. Let be the closure in of the polynomials in . If is the maximal ideal space of , then is homeomorphic to (here is the function whose value at in is ). Now . If , then the Maximum Modulus Theorem shows that has a continuous extension to that is analytic in ; also denote this extension by . The continuous homomorphisms on are of the form for some in .
(Zelazko) Let be an algebra and suppose is a linear functional such that for all in . Show that is a homomorphism.
Let be an abelian Banach algebra with identity that is semisimple [that is, ]. If is the norm on and is another norm on that also makes into a Banach algebra, then these two norms are equivalent.
Let be the closure in of the polynomials in and let . Show that .
Show that is a generator of but is not.
Show that does not have a single generator though it does have a single rational generator (that is, an element such that is dense).