Holomorphic 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 : C → C 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) = e^z 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 e^S = 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.

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. This 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 (recalling that T is an n x n matrix) 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, f → f(T), has certain desirable properties.

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 γ_i ∈ 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 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.

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 (z₁−T)⁻¹ and (z₂−T)⁻¹ commute, which hints at the fact that the image of the functional calculus will be a commutative algebra. Letting z₂ → z₁ 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).

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 (z₂−T)⁻¹. The above series converges in L(X), which implies the existence of (z₂−T)⁻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).

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

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

leads one to consider

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

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

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

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, ..., γ_m} 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.

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.

For a collection of Jordan curves Γ = {γ₁, ..., γ_m} and a point a ∈ C, 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 G ⊂ C be an open set and Γ ⊂ G. If g : G → C is holomorphic, 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 : G → L(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): G ⊂ C → C is holomorphic and therefore by Cauchy's theorem

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

The well-definedness of functional calculus now follows as an easy consequence. Let D be an open set containing σ(T). Suppose Γ = {γ_i} and Ω = {ω_j} are 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 ω_j, 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 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).

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

The linearity of the map f ↦ f(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 ≤ i ≤ m} a_i zⁱ is a polynomial. To prove this, it is sufficient to show, for k ≥ 0 and f(z) = z^k, it is true that f(T) = T^k, 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.

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

${\displaystyle {\begin{aligned}f_{1}(T)f_{2}(T)&=\left({\frac {1}{2\pi i}}\int _{\Gamma _{1}}{\frac {f_{1}(\zeta )}{\zeta -T}}d\zeta \right)\left({\frac {1}{2\pi i}}\int _{\Gamma _{2}}{\frac {f_{2}(\omega )}{\omega -T}}\,d\omega \right)\\&={\frac {1}{(2\pi i)^{2}}}\int _{\Gamma _{1}}\int _{\Gamma _{2}}{\frac {f_{1}(\zeta )f_{2}(\omega )}{(\zeta -T)(\omega -T)}}\;d\omega \,d\zeta \\&={\frac {1}{(2\pi i)^{2}}}\int _{\Gamma _{1}}\int _{\Gamma _{2}}f_{1}(\zeta )f_{2}(\omega )\left({\frac {(\zeta -T)^{-1}-(\omega -T)^{-1}}{\omega -\zeta }}\right)d\omega \,d\zeta &&{\text{First Resolvent Formula}}\\&={\frac {1}{(2\pi i)^{2}}}\left\{\left(\int _{\Gamma _{1}}{\frac {f_{1}(\zeta )}{\zeta -T}}\left[\int _{\Gamma _{2}}{\frac {f_{2}(\omega )}{\omega -\zeta }}d\omega \right]d\zeta \right)-\left(\int _{\Gamma _{2}}{\frac {f_{2}(\omega )}{\omega -T}}\left[\int _{\Gamma _{1}}{\frac {f_{1}(\zeta )}{\omega -\zeta }}d\zeta \right]d\omega \right)\right\}\\&={\frac {1}{(2\pi i)^{2}}}\int _{\Gamma _{1}}{\frac {f_{1}(\zeta )}{\zeta -T}}\left[\int _{\Gamma _{2}}{\frac {f_{2}(\omega )}{\omega -\zeta }}d\omega \right]d\zeta \end{aligned}}}$

The last line follows from the fact that ω ∈ Γ₂ lies outside of Γ₁ and f₁ is holomorphic on some open neighborhood of σ(T) and therefore the second term vanishes. Therefore, we have:

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

Let G ⊂ C be open with σ(T) ⊂ G. Suppose a sequence {f_k} of holomorphic functions on G converges uniformly on compact subsets of G (this is sometimes called compact convergence). Then {f_k(T)} is convergent in L(X):

Assume for simplicity that Γ consists of only one Jordan curve. We estimate

${\displaystyle {\begin{aligned}\left\|f_{k}(T)-f_{l}(T)\right\|&={\frac {1}{2\pi }}\left\|\int _{\Gamma }{\frac {(f_{k}-f_{l})(\zeta )}{\zeta -T}}d\zeta \right\|\\&\leq {\frac {1}{2\pi }}\int _{\Gamma }\left|(f_{k}-f_{l})(\zeta )\right|\cdot \left\|(\zeta -T)^{-1}\right\|d\zeta \end{aligned}}}$

By combining the uniform convergence assumption and various continuity considerations, we see that the above tends to 0 as k, l → ∞. So {f_k(T)} is Cauchy, therefore convergent.

To summarize, we have shown the holomorphic functional calculus, f → f(T), has the following properties:

1. It extends the polynomial functional calculus.
2. It is an algebra homomorphism from the algebra of holomorphic functions defined on a neighborhood of σ(T) to L(X)
3. It preserves uniform convergence on compact sets.

It can be proved that a calculus satisfying the above properties is unique.

We note that, everything discussed so far holds verbatim if the family of bounded operators L(X) is replaced by a Banach algebra A. The functional calculus can be defined in exactly the same way for an element in A.

It is known that the spectral mapping theorem holds for the polynomial functional calculus: for any polynomial p, σ(p(T)) = p(σ(T)). This can be extended to the holomorphic calculus. To show f(σ(T)) ⊂ σ(f(T)), let μ be any complex number. By a result from complex analysis, there exists a function g holomorphic on a neighborhood of σ(T) such that

${\displaystyle f(z)-f(\mu )=(z-\mu )g(z).\,}$

According to the homomorphism property, f(T) − f(μ) = (T − μ)g(T). Therefore, μ ∈ σ(T) implies f(μ) ∈ σ(f(T)).

For the other inclusion, if μ is not in f(σ(T)), then the functional calculus is applicable to

${\displaystyle g(z)={\frac {1}{f(z)-\mu }}.}$

So g(T)(f(T) − μ) = I. Therefore, μ does not lie in σ(f(T)).

The underlying idea is as follows. Suppose that K is a subset of σ(T) and U,V are disjoint neighbourhoods of K and σ(T) \ K respectively. Define e(z) = 1 if z ∈ U and e(z) = 0 if z ∈ V. Then e is a holomorphic function with [e(z)]² = e(z) and so, for a suitable contour Γ which lies in U ∪ V and which encloses σ(T), the linear operator

${\displaystyle e(T)={\frac {1}{2\pi i}}\int _{\Gamma }{\frac {e(z)}{z-T}}\,dz}$

will be a bounded projection that commutes with T and provides a great deal of useful information.

It transpires that this scenario is possible if and only if K is both open and closed in the subspace topology on σ(T). Moreover, the set V can be safely ignored since e is zero on it and therefore makes no contribution to the integral. The projection e(T) is called the spectral projection of T at K and is denoted by P(K;T). Thus every subset K of σ(T) that is both open and closed in the subspace topology has an associated spectral projection given by

${\displaystyle P(K;T)={\frac {1}{2\pi i}}\int \nolimits _{\Gamma }{\frac {dz}{z-T}}}$

where Γ is a contour that encloses K but no other points of σ(T).

Since P = P(K;T) is bounded and commutes with T it enables T to be expressed in the form U ⊕ V where U = T|_PX and V = T|_(1−P)X. Both PX and (1 − P)X are invariant subspaces of T moreover σ(U) = K and σ(V) = σ(T) \ K. A key property is mutual orthogonality. If L is another open and closed set in the subspace topology on σ(T) then P(K;T)P(L;T) = P(L;T)P(K;T) = P(K ∩ L;T) which is zero whenever K and L are disjoint.

Spectral projections have numerous applications. Any isolated point of σ(T) is both open and closed in the subspace topology and therefore has an associated spectral projection. When X has finite dimension σ(T) consists of isolated points and the resultant spectral projections lead to a variant of Jordan normal form wherein all the Jordan blocks corresponding to the same eigenvalue are consolidated. In other words there is precisely one block per distinct eigenvalue. The next section considers this decomposition in more detail.

Sometimes spectral projections inherit properties from their parent operators. For example if T is a positive matrix with spectral radius r then the Perron–Frobenius theorem asserts that r ∈ σ(T). The associated spectral projection P = P(r;T) is also positive and by mutual orthogonality no other spectral projection can have a positive row or column. In fact TP = rP and (T/r)ⁿ → P as n → ∞ so this projection P (which is called the Perron projection) approximates (T/r)ⁿ as n increases, and each of its columns is an eigenvector of T.

More generally if T is a compact operator then all non-zero points in σ(T) are isolated and so any finite subset of them can be used to decompose T. The associated spectral projection always has finite rank. Those operators in L(X) with similar spectral characteristics are known as Riesz operators. Many classes of Riesz operators (including the compact operators) are ideals in L(X) and provide a rich field for research. However if X is a Hilbert space there is exactly one closed ideal sandwiched between the Riesz operators and those of finite rank.

Much of the foregoing discussion can be set in the more general context of a complex Banach algebra. Here spectral projections are referred to as spectral idempotents since there may no longer be a space for them to project onto.

If the spectrum σ(T) is not connected, X can be decomposed into invariant subspaces of T using the functional calculus. Let σ(T) be a disjoint union

${\displaystyle \sigma (T)=\bigcup _{i=1}^{m}F_{i}.}$

Define e_i to be 1 on some neighborhood that contains only the component F_i and 0 elsewhere. By the homomorphism property, e_i(T) is a projection for all i. In fact it is just the spectral projection P(F_i;T) described above. The relation e_i(T) T = T e_i(T) means the range of each e_i(T), denoted by X_i, is an invariant subspace of T. Since

${\displaystyle \sum _{i}e_{i}(T)=I,\,}$

X can be expressed in terms of these complementary subspaces:

${\displaystyle X=\sum _{i}X_{i}.\,}$

Similarly, if T_i is T restricted to X_i, then

${\displaystyle T=\sum _{i}T_{i}.\,}$

Consider the direct sum

${\displaystyle X'=\bigoplus _{i}X_{i}.}$

With the norm

${\displaystyle \left\|\bigoplus _{i}x_{i}\right\|=\sum _{i}\|x_{i}\|,}$

X' is a Banach space. The mapping R: X' → X defined by

${\displaystyle R\left(\bigoplus _{i}x_{i}\right)=\sum _{i}x_{i}}$

is a Banach space isomorphism, and we see that

${\displaystyle RTR^{-1}=\bigoplus _{i}T_{i}.}$

This can be viewed as a block diagonalization of T.

When X is finite-dimensional, σ(T) = {λ_i} is a finite set of points in the complex plane. Choose e_i to be 1 on an open disc containing only λ_i from the spectrum. The corresponding block-diagonal matrix

${\displaystyle \bigoplus _{i}T_{i}}$

is the Jordan canonical form of T.

A holomorphic functional calculus can be defined in a similar fashion for unbounded closed operators with non-empty resolvent set.