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 naturally extends the function f from complex argument to operator argument. More precisely, the functional calculus defines a continuous algebra homomorphism from the holomorphic functions on a neighbourhood of the spectrum of T to the bounded operators. 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
p ( z ) = ∑ i = 0 m a i z i {\displaystyle p(z)=\sum _{i=0}^{m}a_{i}z^{i}}
is a complex polynomial, one can simply substitute T for z and define
p ( T ) = ∑ i = 0 m a i T i {\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
f ( z ) = ∑ i = 0 ∞ a i z i , {\displaystyle f(z)=\sum _{i=0}^{\infty }a_{i}z^{i},}
mimicking the polynomial case suggests we define
f ( T ) = ∑ i = 0 ∞ a i T i . {\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
f ( T ) = e T = I + T + T 2 2 ! + T 3 3 ! + ⋯ . {\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
ln ( z + 1 ) = z − z 2 2 + z 3 3 − ⋯ , {\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. 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
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