# Holomorphic functional calculus

Template:No footnotes In mathematics, holomorphic functional calculus is functional calculus with holomorphic functions. That is to say, given a holomorphic function f of a complex argument z and an operator T, the aim is to construct an operator, f(T), which in a sense extends the function f from complex argument to operator argument.

This article will discuss the case where T is a bounded linear operator on some Banach space. In particular, T can be a square matrix with complex entries, a case which will be used to illustrate functional calculus and provide some heuristic insights for the assumptions involved in the general construction.

## Motivation

### Need for a general functional calculus

In this section T will be assumed to be a n × n matrix with complex entries.

If a given function f is of certain special type, there are natural ways of defining f(T). For instance, if

${\displaystyle p(z)=\sum _{i=0}^{m}a_{i}z^{i}}$

is a complex polynomial, one can simply substitute T for z and define

${\displaystyle p(T)=\sum _{i=0}^{m}a_{i}T^{i}}$

where T0 = I, the identity matrix. This is the polynomial functional calculus. It is a homomorphism from the ring of polynomials to the ring of n × n matrices.

Extending slightly from the polynomials, if f : CC is holomorphic everywhere, i.e. an entire function, with MacLaurin series

${\displaystyle f(z)=\sum _{i=0}^{\infty }a_{i}z^{i},}$

mimicking the polynomial case suggests we define

${\displaystyle f(T)=\sum _{i=0}^{\infty }a_{i}T^{i}.}$

Since the MacLaurin series converges everywhere, the above series will converge, in a chosen operator norm. An example of this is the exponential of a matrix. Replacing z by T in the MacLaurin series of f(z) = ez gives

${\displaystyle f(T)=e^{T}=I+T+{\frac {T^{2}}{2!}}+{\frac {T^{3}}{3!}}+\cdots .}$

The requirement that the MacLaurin series of f converges everywhere can be relaxed somewhat. From above it is evident that all that is really needed is the radius of convergence of the MacLaurin series be greater than ǁTǁ, the operator norm of T. This enlarges somewhat the family of f for which f(T) can be defined using the above approach. However it is not quite satisfactory. For instance, it is a fact from matrix theory that every non-singular T has a logarithm S in the sense that eS = T. It is desirable to have a functional calculus that allows one to define, for a non-singular T, ln(T) such that it coincides with S. This can not be done via power series, for example the logarithmic series

${\displaystyle \ln(z+1)=z-{\frac {z^{2}}{2}}+{\frac {z^{3}}{3}}-\cdots ,}$

converges only on the open unit disk. Substituting T for z in the series fails to give a well-defined expression for ln(T + I) for invertible T + I with ǁTǁ ≥ 1. Thus a more general functional calculus is needed.

### Functional calculus and the spectrum

It is expected that a necessary condition for f(T) to make sense is f be defined on the spectrum of T. For example, the spectral theorem for normal matrices states every normal matrix is unitarily diagonalizable. There leads to a definition of f(T) when T is normal. One encounters difficulties if f(λ) is not defined for some eigenvalue λ of T.

Other indications also reinforce the idea that f(T) can be defined only if f is defined on the spectrum of T. If T is not invertible, then 0 is an eigenvalue. Since the natural logarithm is undefined at 0, one would expect that ln(T) can not be defined naturally. This is indeed the case. As another example, for

${\displaystyle f(z)={\frac {1}{(z-2)(z-5)}}}$

the reasonable way of calculating f(T) would seem to be

${\displaystyle f(T)=(T-2I)^{-1}(T-5I)^{-1}.\,}$

However, this expression is not defined if the inverses on the right-hand side do not exist, that is, if either 2 or 5 are eigenvalues of T.

For a given matrix T, the eigenvalues of T dictate to what extent f(T) can be defined; i.e., f(λ) must be defined for all eigenvalues λ of T. For a general bounded operator this condition translates to "f must be defined on the spectrum of T". This assumption turns out to be an enabling condition such that the functional calculus map, ff(T), has certain desirable properties.

## Functional calculus for a bounded operator

The spectrum σ(T) in light blue and the path γ in red.
The case when the spectrum has multiple connected components and the corresponding path γ.
The case when the spectrum is not simply connected.

Let X be a complex Banach space, and L(X) denote the family of bounded operators on X.

Recall the Cauchy integral formula from classical function theory. Let f : CC be holomorphic on some open set DC, and Γ be a rectifiable Jordan curve in D, that is, a closed curve of finite length without self-intersections. Cauchy's integral formula states

${\displaystyle f(z)={\frac {1}{2\pi i}}\int \nolimits _{\Gamma }{\frac {f(\zeta )}{\zeta -z}}\,d\zeta }$

for any z lying in the inside of Γ, i.e. the winding number of Γ about z is 1.

The idea is to extend this formula to functions taking values in the Banach space L(X). Cauchy's integral formula suggests the following definition (purely formal, for now):

${\displaystyle f(T)={\frac {1}{2\pi i}}\int _{\Gamma }{\frac {f(\zeta )}{\zeta -T}}\,d\zeta ,}$

where (ζ−T)−1 is the resolvent of T at ζ.

Assuming this Banach space-valued integral is appropriately defined, this proposed functional calculus implies the following necessary conditions:

1. As the scalar version of Cauchy's integral formula applies to holomorphic f, we anticipate that is also the case for the Banach space case, where there should be a suitable notion of holomorphy for functions taking values in the Banach space L(X).
2. As the resolvent mapping ζ → (ζ−T)−1 is undefined on the spectrum of T, σ(T), the Jordan curve Γ should not intersect σ(T). Furthermore, the resolvent mapping is holomorphic on the complement of σ(T). So, to obtain a non-trivial functional calculus, Γ must enclose, at least part of, σ(T).
3. The functional calculus should be well-defined in the sense that f(T) has to be independent of Γ.

The full definition of the functional calculus is as follows: For TL(X), define

${\displaystyle f(T)={\frac {1}{2\pi i}}\int \nolimits _{\Gamma }{\frac {f(\zeta )}{\zeta -T}}\,d\zeta ,}$

where f is a holomorphic function defined on an open set DC which contains σ(T), and Γ = {γ1, ..., γm} is a collection of Jordan curves in D such that σ(T) lies in the inside of Γ, and each γi is oriented in the positive sense.

The open set D may vary with f and need not be connected, as shown by the figures on the right.

The following subsections make precise the notions invoked in the definition and show f(T) is indeed well defined under given assumptions.

### Banach space-valued integral

Cf. Bochner integral

For a continuous function g defined in an open neighborhood of Γ and taking values in L(X), the contour integral ∫Γg is defined in the same way as for the scalar case. One can parametrize each γi ∈ Γ by a real interval [a, b], and the integral is the limit of the Riemann sums obtained from ever-finer partitions of [a, b]. The Riemann sums converge in the uniform operator topology. We define

${\displaystyle \int _{\Gamma }g=\sum \nolimits _{i}\int _{\gamma _{i}}g.}$

In the definition of the functional calculus, f is assumed to be holomorphic in an open neighborhood of Γ. It will be shown below that the resolvent mapping is holomorphic on the resolvent set. Therefore the integral

${\displaystyle {\frac {1}{2\pi i}}\int _{\Gamma }{\frac {f(\zeta )}{\zeta -T}}\,d\zeta }$

makes sense.

### The resolvent mapping

The mapping ζ → (ζ−T)−1 is called the resolvent mapping of T. It is defined on the complement of σ(T), called the resolvent set of T and will be denoted by ρ(T).

Much of classical function theory depends on the properties of the integral

${\displaystyle {\frac {1}{2\pi i}}\int _{\Gamma }{\frac {d\zeta }{\zeta -z}}.}$

The holomorphic functional calculus is similar in that the resolvent mapping plays a crucial role in obtaining properties one requires from a nice functional calculus. This subsection outlines properties of the resolvent map that are essential in this context.

#### The 1st resolvent formula

Direct calculation shows, for z1, z2 ∈ ρ(T),

${\displaystyle (z_{1}-T)^{-1}-(z_{2}-T)^{-1}=(z_{1}-T)^{-1}(z_{2}-z_{1})(z_{2}-T)^{-1}.\,}$

Therefore

${\displaystyle (z_{1}-T)^{-1}(z_{2}-T)^{-1}={\frac {(z_{1}-T)^{-1}-(z_{2}-T)^{-1}}{(z_{2}-z_{1})}}.}$

This equation is called the first resolvent formula. The formula shows (z1T)−1 and (z2T)−1 commute, which hints at the fact that the image of the functional calculus will be a commutative algebra. Letting z2z1 shows the resolvent map is (complex-) differentiable at each z1 ∈ ρ(T); so the integral in the expression of functional calculus converges in L(X).

#### Analyticity

Stronger statement than differentiability can be made regarding the resolvent map. The resolvent set ρ(T) is actually an open set on which the resolvent map is analytic. This property will be used in subsequent arguments for the functional calculus. To verify this claim, let z1 ∈ ρ(T) and notice the formal expression

${\displaystyle {\frac {1}{z_{2}-T}}={\frac {1}{z_{1}-T}}\cdot {\frac {1}{1-{\frac {z_{1}-z_{2}}{z_{1}-T}}}}}$

suggests we consider

${\displaystyle (z_{1}-T)^{-1}\sum _{n\geq 0}\left((z_{1}-z_{2})(z_{1}-T)^{-1}\right)^{n}}$

for (z2T)−1. The above series converges in L(X), which implies the existence of (z2T)−1, if

${\displaystyle |z_{1}-z_{2}|<{\frac {1}{\left\|(z_{1}-T)^{-1}\right\|}}.}$

Therefore the resolvent set ρ(T) is open and the power series expression on an open disk centered at z1 ∈ ρ(T) shows the resolvent map is analytic on ρ(T).

#### Neumann series

Another expression for (zT)−1 will also be useful. The formal expression

${\displaystyle {\frac {1}{z-T}}={\frac {1}{z}}\cdot {\frac {1}{1-{\frac {T}{z}}}}}$

${\displaystyle {\frac {1}{z}}\sum _{n\geq 0}\left({\frac {T}{z}}\right)^{n}.}$

This series, the Neumann series, converges to (zT)−1 if

${\displaystyle \left\|{\frac {T}{z}}\right\|<1,\;{\text{i.e.}}\;|z|>\|T\|.}$

#### Compactness of σ(T)

From the last two properties of the resolvent we can deduce that the spectrum σ(T) of a bounded operator T is a compact subset of C. Therefore for any open set D such that σ(T) ⊂ D, there exists a positively-oriented and smooth system of Jordan curves Γ = {γ1, ..., γm} such that σ(T) is in the inside of Γ and the complement of D is contained in the outside of Γ. Hence, for the definition of the functional calculus, indeed a suitable family of Jordan curves can be found for each f that is holomorphic on some D.

### Well-definedness

The previous discussion has shown that the integral makes sense, i.e. a suitable collection Γ of Jordan curves does exist for each f and the integral does converge in the appropriate sense. What has not been shown is that the definition of the functional calculus is unambiguous, i.e. does not depend on the choice of Γ. This issue we now try to resolve.

#### A preliminary fact

For a collection of Jordan curves Γ = {γ1, ..., γm} and a point aC, the winding number of Γ with respect to a is the sum of the winding numbers of its elements. If we define:

${\displaystyle n(\Gamma ,a)=\sum \nolimits _{i}n(\gamma _{i},a),}$

the following theorem is by Cauchy:

Theorem. Let GC be an open set and Γ ⊂ G. If g : CC be holomorphic on G, and for all a in the complement of G, n(Γ, a) = 0, then the contour integral of g on Γ is zero.

We will need the vector-valued analog of this result when g takes values in L(X). To this end, let g : GL(X) be holomorphic, with the same assumptions on Γ. The idea is use the dual space L(X)* of L(X), and pass to Cauchy's theorem for the scalar case.

Consider the integral

${\displaystyle \int _{\Gamma }g\in L(X),}$

if we can show that all φ ∈ L(X)* vanish on this integral then the integral itself has to be zero. Since φ is bounded and the integral converges in norm, we have:

${\displaystyle \phi \left(\int _{\Gamma }g\right)=\int _{\Gamma }\phi (g).}$

But g is holomorphic, hence the composition φ(g): GCC is holomorphic and therefore by Cauchy's theorem

${\displaystyle \int _{\Gamma }\phi (g)=0.}$

#### Main argument

The well-definedness of functional calculus now follows as an easy consequence. Let D be an open set containing σ(T). Suppose Γ = {γi} and Ω = {ωj} be two (finite) collections of Jordan curves satisfying the assumption given for the functional calculus. We wish to show

${\displaystyle \int _{\Gamma }{\frac {f(\zeta )}{\zeta -T}}\,d\zeta =\int _{\Omega }{\frac {f(\zeta )}{\zeta -T}}\,d\zeta .}$

Let Ω′ be obtained from Ω by reversing the orientation of each ωj, then

${\displaystyle \int _{\Omega }{\frac {f(\zeta )}{\zeta -T}}\,d\zeta =-\int _{\Omega '}{\frac {f(\zeta )}{\zeta -T}}\,d\zeta .}$

Consider the union of the two collections Γ ∪ Ω′. Both Γ ∪ Ω′ and σ(T) are compact. So there is some open set U containing Γ ∪ Ω′ such that σ(T) lies in the complement of U. Any a in the complement of U has winding number n(Γ ∪ Ω′, a) = 0 and the function

${\displaystyle \zeta \rightarrow {\frac {f(\zeta )}{\zeta -T}}}$

is holomorphic on U. So the vector-valued version of Cauchy's theorem gives

${\displaystyle \int _{\Gamma \cup \Omega '}{\frac {f(\zeta )}{\zeta -T}}\,d\zeta =0}$

i.e.

${\displaystyle \int _{\Gamma }{\frac {f(\zeta )}{\zeta -T}}\,d\zeta +\int _{\Omega '}{\frac {f(\zeta )}{\zeta -T}}\,d\zeta =\int _{\Gamma }{\frac {f(\zeta )}{\zeta -T}}\,d\zeta -\int _{\Omega }{\frac {f(\zeta )}{\zeta -T}}\,d\zeta =0.}$

Hence the functional calculus is well-defined.

Consequently, if f1 and f2 are two holomorphic functions defined on corresponding neighborhoods D1 and D2 of σ(T) and they are equal on an open set containing σ(T), then f1(T) = f2(T). Moreover, even though the D1 may not be D2, the operator (f1 + f2) (T) is well-defined. Same holds for the definition of (f1·f2)(T).

### On the assumption that f be holomorphic over an open neighborhood of σ(T)

It should perhaps be noted that so far the full strength of this assumption has not been utilized. For convergence of the integral, only continuity was used. For well-definedness, we only needed f be holomorphic on some open set U containing the contours Γ ∪ Ω′ but not σ(T). The assumption will be applied in its entirety for showing the homomorphism property of the functional calculus.

## Properties

### Polynomial case

The linearity of the map ff(T) follows from the convergence of the integral and that linear operations on a Banach space are continuous.

We recover the polynomial functional calculus when f(z) = ∑0 ≤ im ai zi is a polynomial. To prove this, it is sufficient to show, for k ≥ 0 and f(z) = zk, it is true that f(T) = Tk, i.e.

${\displaystyle {\frac {1}{2\pi i}}\int _{\Gamma }{\frac {\zeta ^{k}}{\zeta -T}}d\zeta =T^{k}}$

for any suitable Γ enclosing σ(T). Choose Γ to be a circle of radius greater than the operator norm of T. As stated above, on such Γ, the resolvent map admits a power series representation

${\displaystyle (z-T)^{-1}={\frac {1}{z}}\sum _{n\geq 0}\left({\frac {T}{z}}\right)^{n}.}$

Substituting gives

${\displaystyle f(T)={\frac {1}{2\pi i}}\int _{\Gamma }\left(\sum _{n\geq 0}{\frac {T^{n}}{\zeta ^{n+1-k}}}\right)d\zeta }$

which is

${\displaystyle \sum _{n\geq 0}T^{n}\cdot {\frac {1}{2\pi i}}\left(\int _{\Gamma }{\frac {d\zeta }{\zeta ^{n+1-k}}}\right)=\sum _{n\geq 0}T^{n}\cdot \delta _{nk}=T^{k}.}$

The δ is the Kronecker delta symbol.

### The homomorphism property

For any f1 and f2 satisfying the appropriate assumptions, the homomorphism property states

${\displaystyle f_{1}(T)f_{2}(T)=(f_{1}\cdot f_{2})(T).\,}$

We sketch an argument which invokes the first resolvent formula and the assumptions placed on f. First we choose the Jordan curves such that Γ1 lies in the inside of Γ2. The reason for this will become clear below. Start by calculating directly

{\displaystyle {\begin{aligned}f_{1}(T)f_{2}(T)&={\frac {1}{2\pi i}}\int _{\Gamma _{1}}{\frac {f_{1}(\zeta )}{\zeta -T}}\left[{\frac {1}{2\pi i}}\int _{\Gamma _{2}}{\frac {f_{2}(\omega )}{\omega -\zeta }}d\omega \right]d\zeta \\&={\frac {1}{2\pi i}}\int _{\Gamma _{1}}{\frac {f_{1}(\zeta )}{\zeta -T}}\left[f_{2}(\zeta )\right]d\zeta &&{\text{Cauchy's Integral Formula}}\\&={\frac {1}{2\pi i}}\int _{\Gamma _{1}}{\frac {f_{1}(\zeta )f_{2}(\zeta )}{\zeta -T}}d\zeta \\&=(f_{1}\cdot f_{2})(T)\end{aligned}}}