Holomorphic functional calculus

Assume the Axiom of Choice (The Axiom of Choice). Let $A$ be a unital complex Banach algebra, let $a\in A$, and let $f$ be a function holomorphic on an open set $U\subseteq \mathbb{C}$ containing the spectrum ${\sigma }_A(a)$ (Spectrum and resolvent set in a Banach algebra); the spectrum is a nonempty compact subset of the plane (Spectrum is nonempty compact and norm bounded). By Admissible cycle around a compact plane set applied to the compact set ${\sigma }_A(a)\subseteq U$ there is a finite polygonal complex cycle $\mathrm{\Gamma }$ with trace in $U\setminus {\sigma }_A(a)$ such that

$n(\mathrm{\Gamma },z)=1\text{for }z\in {\sigma }_A(a),n(\mathrm{\Gamma },z)=0\text{for }z\mathrm{\notin }U;$

such a cycle is called admissible for $(f,U)$ (or simply admissible). Define

$f(a):=\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}f(z)R(z,a)dz\in A,$

where $R(z,a)=(z1-a{)}^{-1}$ is the resolvent and the integral is that of Banach algebra valued contour integral over the chain $\mathrm{\Gamma }$ (Complex chains, their traces, and cycles). The integrand $z\mapsto f(z)R(z,a)$ is continuous on the trace of $\mathrm{\Gamma }$: $f$ is holomorphic on $U$, and $z\mapsto R(z,a)$ is norm continuous on the resolvent set Resolvent is Banach-valued holomorphic, which contains ${\mathrm{\Gamma }}^{\ast }$. So the integral exists, and $f(a)\in A$.

The construction describes the value attached to the germ of $f$ near ${\sigma }_A(a)$: two holomorphic functions $f_1$ on $U_1$ and $f_2$ on $U_2$ with the same germ at ${\sigma }_A(a)$ — that is, agreeing on some neighbourhood of ${\sigma }_A(a)$ — give the same $f(a)$. The value is also independent of which admissible cycle is used; that is Holomorphic functional calculus is contour independent, and until it is proved the notation $f(a)$ refers to the value computed from any one chosen admissible cycle.

• The hypothesis is nonempty and the cycle exists without choice. The spectrum of $a$ is nonempty and compact, so the admissible cycle of Admissible cycle around a compact plane set always exists; the construction inside that lemma uses only finitely many grid cells.

• Notation for operators. For a nonzero complex Banach space $X$ and $T\in \mathcal{B}(X)$ the definition applies with $A=\mathcal{B}(X)$ and gives $f(T)=\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}f(z)(z1-T{)}^{-1}dz\in \mathcal{B}(X)$, the Dunford integral of the resolvent. The spectrum is taken in $\mathcal{B}(X)$ (ex-bounded-operators-form-a-noncommutative-banach-algebra).

• What is not part of the definition. The definition does not assert that $f\mapsto f(a)$ is multiplicative, that it preserves polynomials, or that $\sigma (f(a))=f(\sigma (a))$; those properties are proved from this definition in Holomorphic functional calculus homomorphism and Holomorphic spectral mapping and composition. In particular the contour independence of the value is a theorem, and the notation is provisional until then.

• Wider or smaller domains of holomorphy. Only the germ at ${\sigma }_A(a)$ matters: enlarging $U$ beyond a neighbourhood of the spectrum does not change the value, and shrinking it is allowed as long as it still contains the spectrum and the cycle lies inside it. Both statements follow from Holomorphic functional calculus is contour independent.

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