In mathematics, particularly in operator theory and C*-algebra theory, the continuous functional calculus is a functional calculus which allows the application of a continuous function to normal elements of a C*-algebra.

In advanced theory, the applications of this functional calculus are so natural that they are often not even mentioned. It is no overstatement to say that the continuous functional calculus makes the difference between C*-algebras and general Banach algebras, in which only a holomorphic functional calculus exists.

If one wants to extend the natural functional calculus for polynomials on the spectrum ${\displaystyle \sigma (a)}$ of an element ${\displaystyle a}$ of a Banach algebra ${\displaystyle {\mathcal {A}}}$ to a functional calculus for continuous functions ${\displaystyle C(\sigma (a))}$ on the spectrum, it seems obvious to approximate a continuous function by polynomials according to the Stone-Weierstrass theorem, to insert the element into these polynomials and to show that this sequence of elements converges to ${\displaystyle {\mathcal {A}}}$. The continuous functions on ${\displaystyle \sigma (a)\subset \mathbb {C} }$ are approximated by polynomials in ${\displaystyle z}$ and ${\displaystyle {\overline {z}}}$, i.e. by polynomials of the form ${\textstyle p(z,{\overline {z}})=\sum _{k,l=0}^{N}c_{k,l}z^{k}{\overline {z}}^{l}\;\left(c_{k,l}\in \mathbb {C} \right)}$. Here, ${\displaystyle {\overline {z}}}$ denotes the complex conjugation, which is an involution on the complex numbers. To be able to insert ${\displaystyle a}$ in place of ${\displaystyle z}$ in this kind of polynomial, Banach *-algebras are considered, i.e. Banach algebras that also have an involution *, and ${\displaystyle a^{*}}$ is inserted in place of ${\displaystyle {\overline {z}}}$. In order to obtain a homomorphism ${\displaystyle {\mathbb {C} }[z,{\overline {z}}]\rightarrow {\mathcal {A}}}$, a restriction to normal elements, i.e. elements with ${\displaystyle a^{*}a=aa^{*}}$, is necessary, as the polynomial ring ${\displaystyle \mathbb {C} [z,{\overline {z}}]}$ is commutative. If ${\displaystyle (p_{n}(z,{\overline {z}}))_{n}}$ is a sequence of polynomials that converges uniformly on ${\displaystyle \sigma (a)}$ to a continuous function ${\displaystyle f}$, the convergence of the sequence ${\displaystyle (p_{n}(a,a^{*}))_{n}}$ in ${\displaystyle {\mathcal {A}}}$ to an element ${\displaystyle f(a)}$ must be ensured. A detailed analysis of this convergence problem shows that it is necessary to resort to C*-algebras. These considerations lead to the so-called continuous functional calculus.

continuous functional calculus—Let ${\displaystyle a}$ be a normal element of the C*-algebra ${\displaystyle {\mathcal {A}}}$ with unit element ${\displaystyle e}$ and let ${\displaystyle C(\sigma (a))}$ be the commutative C*-algebra of continuous functions on ${\displaystyle \sigma (a)}$, the spectrum of ${\displaystyle a}$. Then there exists exactly one *-homomorphism ${\displaystyle \Phi _{a}\colon C(\sigma (a))\rightarrow {\mathcal {A}}}$ with ${\displaystyle \Phi _{a}({\boldsymbol {1}})=e}$ for ${\displaystyle {\boldsymbol {1}}(z)=1}$ and ${\displaystyle \Phi _{a}(\operatorname {Id} _{\sigma (a)})=a}$ for the identity.

The mapping ${\displaystyle \Phi _{a}}$ is called the continuous functional calculus of the normal element ${\displaystyle a}$. Usually it is suggestively set ${\displaystyle f(a):=\Phi _{a}(f)}$.

Due to the *-homomorphism property, the following calculation rules apply to all functions ${\displaystyle f,g\in C(\sigma (a))}$ and scalars ${\displaystyle \lambda ,\mu \in \mathbb {C} }$:

One can therefore imagine actually inserting the normal elements into continuous functions; the obvious algebraic operations behave as expected.

The requirement for a unit element is not a significant restriction. If necessary, a unit element can be adjoined, yielding the enlarged C*-algebra ${\displaystyle {\mathcal {A}}_{1}}$. Then if ${\displaystyle a\in {\mathcal {A}}}$ and ${\displaystyle f\in C(\sigma (a))}$ with ${\displaystyle f(0)=0}$, it follows that ${\displaystyle 0\in \sigma (a)}$ and ${\displaystyle f(a)\in {\mathcal {A}}\subset {\mathcal {A}}_{1}}$.