functional calculus

What is called *functional calculus* or *function calculus* are operations by which for $f$ a function (on the complex numbers , for instance) and $a$ a suitable linear operator (on a Hilbert space, for instance) one makes sense of the expression $f(a)$ as a new operator. Usually one requires that the assignment $f \mapsto f(a)$ is an algebra homomorphism, but not always, namely in some contexts as quantization the ordering effects may not respect the homomorphism property, see Weyl functional calculus.

Let $A$ be a C-star algebra (possibly non-commutative) and $a \in A$ a normal operator. With $sp(a)$ the operator spectrum of $a$ write $C(sp(a))$ for the commutative $C^\ast$-algebra of contrinuous complex-valued functions on $sp(a)$. Finally write $\iota : \in C(sp(A))$ for the function $\iota : x \mapsto x$.

There is a unique star-algebra homomorphism

$\phi_a : C(sp(a)) \to A$

such that $\phi(\iota) = a$.

For all $f \in C(sp(a))$ we have that $\phi(f) \in A$ is a normal operator.

This appears for instance as (KadisonRingrose, theorem 4.4.5).

Let $\langle a \rangle \subset A$ be the $C^\ast$-algebra generated by $a$, or in fact any commutative $C^\ast$-subalgebra of $A$ containing $a$.

Then by Gelfand duality there is a compact topological space $X$ and an isomorphism $\psi : \langle a \rangle \stackrel{\simeq}{\to} C(X)$.

Define a morphism

$(-) \circ \psi(a) : C(sp(a)) \to C(X)$

by $f \mapsto f \circ \psi(a)$. This is a continuous $\ast$-algebra homomorphism. Therefore so is the composite

$\phi
:
C(sp(a)) \stackrel{(-)\circ \psi(a)}{\to}
C(X)
\stackrel{\psi^{-1}}{\to}
\langle a\rangle
\hookrightarrow
A
\,.$

And this satisfies $\phi(\iota) = \psi^{-1}(\iota \circ \psi(a)) = \psi^{-1}\psi (a) = a$.

This establishes the existence of $\phi$. To see uniqueness, notice that any other morphism with these properties coincides with $\phi$ on all polynomials in $\iota$ and $\bar \iota$. By the Stone-Weierstrass theorem such polynomials form an everywhere-dense subset of $C(sp(a))$. Since moreover one can see that the two morphisms must be isometric (…) it follows that they in fact agree.

A standard textbook is for instance

- Richard Kadison, John Ringrose,
*Fundamentals of the theory of operator algebra*, Academic Press (1983)

See also

Revised on October 12, 2013 21:51:28
by Toby Bartels
(98.19.41.253)