# continuous functional calculus

 $f(T)$

to make sense as a bounded operator in $H$, for continuous functions  $f$.

More generally, when $\mathcal{A}$ is a $C^{*}$-algebra (http://planetmath.org/CAlgebra) with identity element  $e$, and $x$ is a normal element  of $\mathcal{A}$, the continuous functional calculus allows one to define $f(x)\in\mathcal{A}$ when $f$ is a continuous function.

More precisely, if $\sigma(x)$ denotes the spectrum of $x$ and $C(\sigma(x))$ denotes the $C^{*}$-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 the functional calculus (http://planetmath.org/FunctionalCalculus) .

There are several reasons to require the continuity of $f$ 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 we would obtain

 $f(x)=\frac{1}{x-\lambda_{0}e}=(x-\lambda_{0}e)^{-1}$

but $x-\lambda_{0}e$ is not invertible  since $\lambda_{0}\in\sigma(x)$.

The abstraction towards $C^{*}$-algebras is almost . Indeed, $C^{*}$-algebras are the appropriate where to and prove the continuous functional calculus. The conclusions  towards $B(H)$ then follow as a particular case.

## 1 Preliminary construction

Let $\mathcal{A}$ be a unital $C^{*}$-algebra and $x$ a normal element in $\mathcal{A}$. Let $\mathcal{B}\subseteq\mathcal{A}$ be the $C^{*}$-subalgebra generated by $x$ and the identity      $e$ of $\mathcal{A}$.

Moreover, since $x$ is , $x^{*}x=xx^{*}$, it follows that $\mathcal{B}$ is commutative   and $\mathcal{B}$ consists of those elements $y\in\mathcal{A}$ that can be approximated by polynomials   $p(x,x^{*})$ in $x$ and $x^{*}$.

Recall the following facts:

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

$\,$

Theorem 1 - $\bigtriangleup$ and $\sigma(x)$ are homeomorphic  topological spaces  . Moreover, the mapping $S:\bigtriangleup\to\sigma(x)$ defined by

 $\displaystyle S(\phi):=\phi(x)$

is such an homeomorphism.

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

From the identity

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

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

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

• $S$ 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, $S(\phi_{\alpha})\longrightarrow S(\phi)$ and so $S$ is continuous.

• $S$ is - Suppose $S(\phi_{1})=S(\phi_{2})$. Then, $\phi_{1}(x)=\phi_{2}(x)$. Since

 $\phi_{i}(x^{*})=G(x^{*})(\phi_{i})=\overline{G(x)(\phi_{i})}=\overline{\phi_{i% }(x)}$

we must also have $\phi_{1}(x^{*})=\phi_{2}(x^{*})$.

This clearly implies that

 $\phi_{1}(p(x,x^{*}))=\phi_{2}(p(x,x^{*}))$

for every polynomial in two variables $p$.

Recall that the ”polynomials” $p(x,x^{*})$ are dense in $\mathcal{B}$. So we must have $\phi_{1}(y)=\phi_{2}(y)$ for every $y\in\mathcal{B}$, i.e. $\phi_{1}=\phi_{2}$.

• $S$ is - Let $\lambda\in\sigma(x)=\sigma_{\mathcal{B}}(x)$. Then $x-\lambda e$ is not invertible.

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

Therefore the quotient homomorphism

 $\phi:\mathcal{B}\longrightarrow\mathcal{B}/\mathcal{M}=\mathbb{C}$

is a multiplicative linear functional such that $\phi(x-\lambda e)=0$, i.e. $\phi(x)=\lambda$, i.e. $S(\phi)=\lambda$.

Therefore, $S$ is surjective.

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

## 2 Definition of the continuous functional calculus

If $S$ is the homeomorphism between $\bigtriangleup$ and $\sigma(x)$ defined as above, then the mapping $f\mapsto f\circ S$ is a *-isomorphism between the algebras $C(\sigma(x))$ and $C(\bigtriangleup)$. Since the Gelfand transform $G:\mathcal{B}\longrightarrow C(\bigtriangleup)$ is a also a *-isomorphism, we obtain a *-isomorphism

 $\Gamma:C(\sigma(x))\longrightarrow\mathcal{B}$

by setting $\Gamma(f):=G^{-1}(f\circ S)$.

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

 $f(x):=\Gamma(f)\;\;\in\mathcal{B}\subseteq\mathcal{A}$

The mapping $\Gamma$, such that $f\mapsto f(x)$, is called the continuous functional calculus for $x$.

We now prove the functional calculus (http://planetmath.org/FunctionalCalculus) for the continuous functional calculus and show its uniqueness:

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Theorem 2 - Let $\mathcal{A}$ be a unital $C^{*}$-algebra, $x\in\mathcal{A}$ a normal element and $\mathrm{id}$ the identity function in $\mathbb{C}$. The continuous functional calculus for $x$ is the unique unital *-homomorphism between $C(\sigma(x))$ and $\mathcal{A}$ which sends $\mathrm{id}$ to $x$. In particular, for every polynomial $p$ in $\mathbb{C}$ of the form $p(\lambda):=\sum c_{n,m}\,\lambda^{n}\overline{\lambda}^{m}$, we have $p(x)=\sum c_{n,m}\,x^{n}(x^{*})^{m}$.

: We have seen that the continuous functional calculus $\Gamma$ for $x$ is a *-homomorphism between $C(\sigma(x))$ and $\mathcal{A}$. Recall that $\Gamma$ was defined by $\Gamma(f):=G^{-1}(f\circ S)$. It is clear by the definition that $\Gamma$ is unital. Also, $G\circ\Gamma(f)=f\circ S$ for every $f\in C(\bigtriangleup)$. Taking the identity function $\mathrm{id}$ we obtain that for every $\phi\in\bigtriangleup$

 $\displaystyle G\circ\Gamma(\mathrm{id})(\phi)$ $\displaystyle=$ $\displaystyle\mathrm{id}(S(\phi))$ $\displaystyle=$ $\displaystyle\phi(x)$ $\displaystyle=$ $\displaystyle G(x)(\phi)$

Since the Gelfand transform is a *-isomorphism, we must have $\Gamma(\mathrm{id})=x$.

Now, let $p:\mathbb{C}\to\mathbb{C}$ be a polynomial of the form $p(\lambda):=\sum c_{n,m}\,\lambda^{n}\overline{\lambda}^{m}$. Notice that $p=\sum c_{n,m}\,\mathrm{id}^{n}\overline{\mathrm{id}}^{m}$. If $F$ is any unital *-homomorphism such that $F(\mathrm{id})=x$, then one must have $F(p)=\sum c_{n,m}\,x^{n}(x^{*})^{m}$. 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 $C(\sigma(x))$. Thus, all such unital *-homomorphisms coincide in $C(\sigma(x))$, and uniqueness is proven. $\square$

## 3 Properties

• The spectral mapping theorem assures that for $f\in C(\sigma(x))$

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

 Title continuous functional calculus Canonical name ContinuousFunctionalCalculus Date of creation 2013-03-22 17:30:02 Last modified on 2013-03-22 17:30:02 Owner asteroid (17536) Last modified by asteroid (17536) Numerical id 11 Author asteroid (17536) Entry type Feature Classification msc 47A60 Classification msc 46L05 Classification msc 46H30 Related topic FunctionalCalculus Related topic PolynomialFunctionalCalculus Related topic BorelFunctionalCalculus Defines continuous functions of normal operators