(more info)

Math for the people, by the people.

Main Menu

sections Encyclopædia
Papers
Books
Expositions

meta Requests (200)
Orphanage
Unclass'd
Unproven (490)
Corrections (6)
Classification

talkback Polls
Forums
Feedback
Bug Reports

downloads Snapshots
PM Book

information News
Docs
Wiki
ChangeLog
TODO List
Legalese
About

continuous functional calculus

(Feature)

Let be the algebra of bounded operators over a complex Hilbert space . Let $N \in B(H)$ be a normal operator.

The continuous functional calculus is a functional calculus which enables the expression

$\displaystyle f(T)$

to make sense as a bounded operator in

, for continuous functions

More generally, when $\mathcal{A}$ is a -algebra with identity element , and is a normal element of $\mathcal{A}$ , the continuous functional calculus allows one to define $f(x) \in \mathcal{A}$ when is a continuous function.

More precisely, if $\sigma(x)$ denotes the spectrum of and $C(\sigma(x))$ denotes the -algebra of complex valued continuous functions on $\sigma(x)$ , we will define a continuous homomorphism

$C(\sigma(x)) \longrightarrow \mathcal{A}$

$f \mapsto f(x)$

that satisfies the functional calculus properties.

There are several reasons to require the continuity of on the spectrum $\sigma(x)$ .

For example, suppose $\lambda_0 \in \sigma(x)$ . The function $f(\lambda) = \frac{1}{\lambda - \lambda_0}$ is clearly not continuous in $\lambda_0$ . By the functional calculus properties we would obtain

$\displaystyle f(x)= \frac{1}{x-\lambda_0 e} = (x - \lambda_0 e)^{-1}$

but $x - \lambda_0 e$ is not invertible since $\lambda_0 \in \sigma(x)$ .

The abstraction towards -algebras is almost necessary. Indeed, -algebras are the appropriate object where to state and prove the continuous functional calculus. The conclusions towards then follow as a particular case.

Preliminary construction

Let $\mathcal{A}$ be a unital -algebra and a normal element in $\mathcal{A}$ . Let $\mathcal{A}[x] \subseteq \mathcal{A}$ be the -subalgebra generated by and the identity of $\mathcal{A}$ .

Thus, $\mathcal{A}[x]$ is the norm closure of the algebra generated by , and .

Moreover, since is normal, , it follows that $\mathcal{A}[x]$ is commutative. Thus, $\mathcal{A}[x]$ consists of elements $y \in \mathcal{A}$ that can be approximated by polynomials in and .

Recall the following facts:

Since $\mathcal{A}[x]$ is a commutative unital -algebra, the set $\bigtriangleup$ of multiplicative linear functionals on $\mathcal{A}[x]$ is a compact Hausdorff space.
Let $C(\bigtriangleup)$ denote the -algebra of complex valued continuous functions on $\bigtriangleup$ . The Gelfand transform $\;\widehat{.} \;\; : \mathcal{A}[x] \longrightarrow C(\bigtriangleup)$ is a -isomorphism.

The following result is perhaps the key for the definition of the continuous functional calculus.

Theorem - $\bigtriangleup$ and $\sigma(x)$ are homeomorphic topological spaces.

Proof : Define $F: \bigtriangleup \longrightarrow \sigma(x)$ by

$\displaystyle F(\phi) := \widehat{x}(\phi)= \phi(x), \;\;\;\;\;\; \phi \in \bigtriangleup$

We need to check that is well defined, i.e. $\phi(x) \in \sigma(x)$ for all $\phi \in \bigtriangleup$ .

From the identity

$\displaystyle \phi(x-\phi(x)e) = \phi(x)-\phi(x)\phi(e) = \phi(x)-\phi(x)=0$

follows that $x-\phi(x)e$ cannot be invertible in $\mathcal{A}[x]$ (recall that $\phi$ is a multiplicative linear functional on $\mathcal{A}[x]$ ).

Thus, $\phi(x) \in \sigma_{\mathcal{A}[x]}(x)$ . By the spectral invariance theorem, we see that $\phi(x) \in \sigma(x) = \sigma_{\mathcal{A}[x]}(x)$ , and so is well defined.

is continuous - Suppose $\phi_{\alpha}$ is a net in $\bigtriangleup$ such that $\phi_{\alpha} \longrightarrow \phi$ . Recall that the topology in $\bigtriangleup$ is the weak-* topology, so $\phi_{\alpha}(x) \longrightarrow \phi(x)$ .
Thus, $F(\phi_{\alpha}) \longrightarrow F(\phi)$ and so is continuous.
is injective - Suppose $F(\phi_1) = F(\phi_2)$ . Then, $\phi_1(x) = \phi_2(x)$ . Since
$\displaystyle \phi_i(x^*)=\widehat{x^*}(\phi_i) = \overline{\widehat{x}(\phi_i)}=\overline{\phi_i(x)}$
we must also have $\phi_1(x^*)=\phi_2(x^*)$ .
This clearly implies that

$\displaystyle \phi_1(p(x,x^*))= \phi_2(p(x,x^*))$
for every polynomial in two variables .
Recall that the "polynomials" are dense in $\mathcal{A}[x]$ . So we must have $\phi_1(y)=\phi_2(y)$ for every $y \in \mathcal{A}[x]$ , i.e. $\phi_1 = \phi_2$ .
is surjective - Let $\lambda \in \sigma(x) = \sigma_{\mathcal{A}[x]}(x)$ . Then $x - \lambda e$ is not invertible.
Since $\mathcal{A}[x]$ is commutative, $x - \lambda e$ is contained in a maximal ideal $\mathcal{M}$ .

As $\mathcal{M}$ is maximal ideal, the quotient $\mathcal{A}[x]/\mathcal{M}$ is a division algebra, and so by the Gelfand-Mazur theorem, $\mathcal{A}[x]/\mathcal{M}$ must ne isomorphic to $\mathbb{C}$ .

Therefore the quotient homomorphism

$\displaystyle \phi : \mathcal{A}[x] \longrightarrow \mathcal{A}[x]/\mathcal{M}=\mathbb{C}$
is a multiplicative linear functional such that $\phi(x-\lambda e) = 0$ , i.e. $\phi(x) = \lambda$ , i.e. $F(\phi) = \lambda$ .
Therefore, is surjective.

Since is a continuous bijective function from the compact Hausdorff space $\bigtriangleup$ to $\sigma(x)$ , it follows that it must be a homeomorphism. $\square$

Definition of the continuous functional calculus

Since the Gelfand transform $\;\widehat{.} \;\; : \mathcal{A}[x] \longrightarrow C(\bigtriangleup)$ is a -isomorphism, and $\bigtriangleup$ and $\sigma(x)$ are homeomorphic topological spaces, we obtain a -isomorphism

$\displaystyle \Gamma : \mathcal{A}[x] \longrightarrow C(\sigma(x))$

by setting $\Gamma (y)= \widehat{(y)} \circ F^{-1}$ , where

is the homeomorphism between $\bigtriangleup$ and $\sigma(x)$ .

Definition - Suppose is a normal element in a unital -algebra $\mathcal{A}$ . For every $f \in C(\sigma(x))$ we define

$\displaystyle f(x):=\Gamma^{-1}(f)\;\; \in \mathcal{A}[x] \subseteq \mathcal{A}$

The mapping $\Gamma^{-1}$ (such that $f \mapsto f(x)$ ) is called the continuous functional calculus for .

We now prove the functional calculus properties for the continuous functional calculus:

We have already seen that $\Gamma$ is a continuous -isomorphism. Thus, so is $\Gamma^{-1}$ .
If is a polynomial in two variables then, from the definitions of $\Gamma$ and , we obtain

$\displaystyle \Gamma(p(x,x^*))(\lambda)$ $\displaystyle =$ $\displaystyle \widehat{p(x,x^*)} \circ F^{-1}(\lambda)$

$\displaystyle =$ $\displaystyle \widehat{p(x,x^*)} (\phi_{\lambda})$

$\displaystyle =$ $\displaystyle p(\widehat{x},\widehat{x^*})(\phi_{\lambda})$

$\displaystyle =$ $\displaystyle p(\phi_{\lambda}(x),\phi_{\lambda}(x^*))$

$\displaystyle =$ $\displaystyle p(\lambda , \overline{\lambda})$

and therefore if $f(\lambda) = p(\lambda, \overline{\lambda})$ we have

$\displaystyle \Gamma^{-1} (f) = p(x,x^*)$

Notice that this includes the particular case when is a polynomial in one variable

$\displaystyle \Gamma^{-1} (p) = p(x)$

as desired.

Properties

The spectral mapping theorem assures that for $f \in C(\sigma(x))$
$\displaystyle \sigma(f(x)) = f(\sigma(x))$
When $\sigma(x) \subset \mathbb{R}^+$ the continuous functional calculus assures the existence of a square root $\sqrt{x}$ of , since $\sqrt{\lambda}$ is defined and continuous on $\lambda \in \sigma(x)$ .

Anyone with an account can edit this entry. Please help improve it!

"continuous functional calculus" is owned by asteroid.

(view preamble)