continuous functional calculus
to make sense as a bounded operator in , for continuous functions .
More generally, when is a -algebra (http://planetmath.org/CAlgebra) with identity element , and is a normal element of , the continuous functional calculus allows one to define when is a continuous function.
More precisely, if denotes the spectrum of and denotes the -algebra of complex valued continuous functions on , we will define a continuous homomorphism
that the functional calculus (http://planetmath.org/FunctionalCalculus) .
There are several reasons to require the continuity of on the spectrum .
For example, suppose . The function is clearly not continuous in . By the functional calculus we would obtain
but is not invertible since .
The abstraction towards -algebras is almost . Indeed, -algebras are the appropriate where to and prove the continuous functional calculus. The conclusions towards then follow as a particular case.
1 Preliminary construction
Thus, is the norm closure of the algebra generated by , and .
Recall the following facts:
The following result is perhaps the for the definition of the continuous functional calculus.
is such an homeomorphism.
: We need to check that is well defined, i.e. for all .
From the identity
follows that cannot be invertible in (recall that is a multiplicative linear functional on ).
Thus, . By the spectral invariance theorem, we see that , and so is well defined.
is continuous - Suppose is a net in such that . Recall that the topology in is the weak-* topology, so .
Thus, and so is continuous.
is - Suppose . Then, . Since
we must also have .
This clearly implies that
for every polynomial in two variables .
Recall that the ”polynomials” are dense in . So we must have for every , i.e. .
is - Let . Then is not invertible.
Therefore the quotient homomorphism
is a multiplicative linear functional such that , i.e. , i.e. .
Therefore, is surjective.
Since is a continuous bijective function from the compact Hausdorff space to , it follows that it must be a homeomorphism.
2 Definition of the continuous functional calculus
If is the homeomorphism between and defined as above, then the mapping is a *-isomorphism between the algebras and . Since the Gelfand transform is a also a *-isomorphism, we obtain a *-isomorphism
by setting .
Definition - Suppose is a normal element in a unital -algebra . For every we define
The mapping , such that , is called the continuous functional calculus for .
We now prove the functional calculus (http://planetmath.org/FunctionalCalculus) for the continuous functional calculus and show its uniqueness:
Theorem 2 - Let be a unital -algebra, a normal element and the identity function in . The continuous functional calculus for is the unique unital *-homomorphism between and which sends to . In particular, for every polynomial in of the form , we have .
: We have seen that the continuous functional calculus for is a *-homomorphism between and . Recall that was defined by . It is clear by the definition that is unital. Also, for every . Taking the identity function we obtain that for every
Since the Gelfand transform is a *-isomorphism, we must have .
Now, let be a polynomial of the form . Notice that . If is any unital *-homomorphism such that , then one must have . Thus all such unital *-homomorphisms coincide on the subspace of polynomials of the above form. By the Stone-Weierstrass theorem (http://planetmath.org/StoneWeierstrassTheoremComplexVersion), this subspace is dense in . Thus, all such unital *-homomorphisms coincide in , and uniqueness is proven.
The spectral mapping theorem assures that for
When the continuous functional calculus assures the existence of a square root of , since is defined and continuous on .
|Title||continuous functional calculus|
|Date of creation||2013-03-22 17:30:02|
|Last modified on||2013-03-22 17:30:02|
|Last modified by||asteroid (17536)|
|Defines||continuous functions of normal operators|