Holomorphic functional calculus

For $x$ in a unital Banach algebra and $f$ holomorphic near $\sigma (x)$, the holomorphic functional calculus defineswhere $\Gamma$ winds once around the spectrum. It is a continuous unital algebra homomorphism and satisfies the spectral mapping theorem $\sigma (f(x))=f(\sigma (x))$.

If a compact spectrum of an element avoids zero and zero belongs to its unbounded complementary component, choose a simple arc from zero to infinity avoiding the Banach algebra spectrum. Its complement is a simply connected neighborhood of the Banach algebra spectrum, admitting a holomorphic logarithm. Apply holomorphic functional calculus to that logarithm and the composition rule for holomorphic functional calculus to the exponential to obtain the displayed identity. An invertible matrix over the complex numbers always satisfies the hypothesis because its finite Banach algebra spectrum cannot separate zero from infinity.

If a spectrum of an element splits into disjoint nonempty compact pieces $X,Y$, choose disjoint open neighborhoods and a holomorphic function equal to one on the first and zero on the second. The holomorphic functional calculus sends it to an idempotent with the corresponding zero/one values under every algebra character. It is neither zero nor identity: if one piece were killed, a reciprocal function on the other piece would make an originally spectral coordinate difference invertible.

The exponential series converges absolutely in a unital Banach algebra. Multiplication of absolutely convergent series gives $e^x{e}^{-x}=1$, and gives $e^{x+y}=e^x{e}^y$ when $x,y$ commute. Termwise differentiation gives $e^{tx}=1+tx+O(t^2)$ near zero. In a C-star algebra, if $h=h^{\ast }$ and $t$ is real, taking adjoints termwise gives $(e^{ith}{)}^{\ast }=e^{-ith}$, so this exponential is a Unitary element of a C-star algebra and has norm one. These facts let a norm bound on a functional recover its real values on Hermitian elements.

If the spectrum of an element $x$ avoids $(-\infty ,0]$, its holomorphic functional calculus value for the principal cube root is the unique $y$ with $y^3=x$ and spectrum in $\{z\ne 0:|\arg z|<\pi /3\}$. For uniqueness, any other such $v$ defines a continuous unital algebra homomorphism $f\mapsto {\Theta }_v(f\circ (z\mapsto z^3))$ from the slit-plane holomorphic functions to the algebra. It sends the coordinate to $x$, so continuity and uniqueness of holomorphic functional calculus identify it with ${\Theta }_x$. Applying it to the principal cube root gives $v=y$. This argument does not assume commutativity of the ambient algebra or separation by algebra characters.

If the spectrum of an element $x$ lies in the open right half-plane, holomorphic functional calculus gives the displayed square root with spectrum in that half-plane. It is unique there. Two such roots $y,v$ have character values equal to the unique scalar square root with positive real part. Every character of an algebra is therefore nonzero on $y+v$, so the character description of the spectrum makes $y+v$ invertible. From $(y-v)(y+v)=0$ one obtains $y=v$. Equal character values alone do not prove equality in a possibly nonsemisimple algebra.

The closed unital algebra generated by a bounded operator and all its resolvents is commutative, because the generators commute. It has exactly the original operator spectrum at that operator: all required inverses outside that spectrum were included, and an inverse in the smaller algebra is also an operator inverse. This permits commutative-algebra linear functional calculus without replacing the spectrum by the possibly larger spectrum of a polynomial-generated algebra.

Equip holomorphic functions on an open spectral neighborhood with the compact-open topology. A fixed admissible contour gives a bound on $\Vert {\Theta }_x(f)\Vert$ by a constant times the supremum of $|f|$ on the contour, proving continuity. The Runge theorem makes rational functions with poles outside the open set dense. A unital homomorphism sending the coordinate function to $x$ must send each inverse coordinate difference to the corresponding resolvent, so its rational values and then all its holomorphic values are forced.

The displayed series converges absolutely in a Banach algebra. It is the holomorphic functional calculus value of the analytic branch $z\mapsto \log (1-z)$ on the unit disk. The composition rule for holomorphic functional calculus and $\exp (\log (1-z))=1-z$ give $\exp (\log (1-x))=1-x$. Thus every sufficiently small norm perturbation of the identity is an exponential.

Choose a cycle $\Gamma$ surrounding $\sigma (x)$ and a cycle $\Delta$ surrounding $f(\sigma (x))$, with $f(\Gamma )$ inside the region of winding number one for $\Delta$. The algebra-homomorphism property givesSubstitute this into the contour formula for $g(f(x))$, interchange integrals, and apply the Cauchy integral formula to the inner $\zeta$ integral. This proves composition as an equality of algebra elements, not merely an equality on characters, which need not separate a nonsemisimple algebra.

Let $x_1,x_2$ belong to a commutative unital Banach algebra, have the same Gelfand transform, and have common spectrum $K$. If $f$ is holomorphic near $K$ and $f^{\prime }$ has no zero on $K$, then $f(x_1)=f(x_2)$ implies $x_1=x_2$. Apply the two-variable holomorphic functional calculus to the divided difference of $f$; its Gelfand transform never vanishes, so it is invertible.