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Holomorphic functional calculus is contour independent

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $A$ be a unital complex Banach algebra, $a \in A$ , and let $f$ be holomorphic on an open set $U$ containing $σ_{A} (a)$ . Then the value $f (a)$ of Holomorphic functional calculus is independent of the admissible cycle used to compute it. Moreover, if $V \subseteq U$ is another open set containing $σ_{A} (a)$ and $g$ is holomorphic on $V$ with $g = f$ on a neighbourhood of $σ_{A} (a)$ , then $f (a) = g (a)$ with $g$ computed from $V$ .

Facts & Assumptions

Given: An assumed Axiom of Choice, a unital complex Banach algebra $A$ , an element $a \in A$ with nonempty compact spectrum $σ_{A} (a)$ , an open $U \supseteq σ_{A} (a)$ , a holomorphic $f : U \to C$ , and two admissible cycles $Γ_{1}, Γ_{2}$ in $U ∖ σ_{A} (a)$ .

[L1]

The definition chooses an admissible cycle $Γ_{0}$ and sets $f (a) = \frac{1}{2 π i} \int_{Γ_{0}} f (z) R (z, a) d z$ ; for every admissible cycle the displayed integral is defined because the integrand is continuous on its trace (Holomorphic functional calculus).

[L2]

A cycle $Γ$ is null-homologous in an open set $Ω$ exactly when $n (Γ, p) = 0$ for every $p \notin Ω$ ; equivalently $n (Γ, p)$ vanishes at every point outside $Ω$ (Null-homologous cycles and homologous cycles in an open set).

[L3]

If $F$ is continuous and weakly holomorphic on an open $Ω$ and $Γ$ is a cycle with trace in $Ω$ that is null-homologous in $Ω$ , then $\int_{Γ} F d z = 0$ (Banach-valued Cauchy integral vanishes).

[L4]

For fixed $a$ the map $z \mapsto R (z, a)$ is holomorphic on $ρ_{A} (a)$ and the product of a scalar holomorphic function with it is weakly holomorphic: for every bounded linear functional $φ$ on $A$ the map $z \mapsto φ (f (z) R (z, a))$ is holomorphic on $U \cap ρ_{A} (a)$ (Resolvent is Banach-valued holomorphic, Spectrum and resolvent set in a Banach algebra).

[L6]

For every compact $K$ contained in an open set $U$ there is a finite polygonal cycle with index $1$ on $K$ and index $0$ outside $U$ (Admissible cycle around a compact plane set).

Proof

technique · direct

1.1

Put $Ω : = U ∖ σ_{A} (a)$ , an open set containing both traces $Γ_{1}^{*}, Γ_{2}^{*}$ ; the difference $Γ_{1} - Γ_{2}$ is a cycle with trace in $Ω$ whose index at $p$ is $n (Γ_{1}, p) - n (Γ_{2}, p)$ .

L2L4algebra

1.2

The map $F (z) : = f (z) R (z, a)$ is continuous on $Ω$ and weakly holomorphic there: for a bounded linear functional $φ$ the composition $φ (F (z)) = f (z) φ (R (z, a))$ is a product of the holomorphic scalar function $f$ and the holomorphic scalar function $φ (R (\cdot, a))$ , hence holomorphic on the open subset $Ω$ of $ρ_{A} (a)$ .

L4algebra

1.3

Germ independence: let $V \supseteq σ_{A} (a)$ be open and let $g$ be holomorphic on $V$ with $f = g$ on a neighbourhood $W$ of $σ_{A} (a)$ . Apply [L6] to the compact set $σ_{A} (a)$ and the open set $U \cap V \cap W$ : this gives an admissible cycle for both $(f, U)$ and $(g, V)$ with trace in $(U \cap V \cap W) ∖ σ_{A} (a)$ , hence lying in $W$ , where $f = g$ .

L6algebra

2.1

For $p \in σ_{A} (a)$ both indices equal $1$ , and for $p \notin U$ both equal $0$ , by admissibility; hence $n (Γ_{1} - Γ_{2}, p) = 0$ for every $p \notin U ∖ σ_{A} (a)$ , that is, $Γ_{1} - Γ_{2}$ is null-homologous in $Ω : = U ∖ σ_{A} (a)$ .

step 1.1L2algebra

3.1

By [L3] applied to $F$ and the cycle $Γ_{1} - Γ_{2}$ , null-homologous in $Ω$ by [step 2.1], one has $\int_{Γ_{1} - Γ_{2}} F d z = 0$ ; by additivity of the chain integral this gives $\int_{Γ_{1}} f (z) R (z, a) d z = \int_{Γ_{2}} f (z) R (z, a) d z$ , hence $f (a)$ does not depend on the admissible cycle.

step 1.2step 2.1L1

4.1

On the trace of the cycle of [step 1.3] the two integrands coincide, $f (z) R (z, a) = g (z) R (z, a)$ , so the two integrals agree; by [step 3.1] applied to each function separately, $f (a) = g (a)$ .

step 1.3step 3.1L1

5.1

Both assertions of the statement are proved: cycle independence by [step 3.1] and germ independence by [step 4.1].

step 3.1step 4.1∎

Depends on

Used by

Holomorphic functional calculus homomorphism Theorem

Dependency tree · two levels

42 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Theo Bühler and Dietmar A. Salamon, Functional Analysis — Theorem 5.25(i), printed pp. 228–229 (standard reference, not scraped)
Vahid Shirbisheh, Lectures on C-star Algebras, v2 — the contour-independence discussion after Definition 2.5.1, printed p. 47 (standard reference, not scraped)