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Let $A$ be a unital Banach algebra over complex numbers and call elements of $A$ "vectors". Let $\Omega$ be an open set in $\mathbb{C}$ and $H(\Omega)$ the space of analytic functions on $\Omega$. There are two kinds of holomorphic calculus we can do on $A$.

Vector-valued Analytic Functions

A vector-valued function $ F: \Omega \to A $ is called analytic if the limit

$$ F'(z) = \lim_{h \to 0} \frac{F(z+h)-F(z)}{h} $$

exists for all $ z \in \Omega $, where the limit is taken w.r.t. the norm on $A$.

A basic result says that $F$ is analytic if and only if $l \circ F: \Omega \to \mathbb{C} $ is analytic in the classical complex analytic sense for every bounded functional $l \in A^*$.

Many theorems in classical complex analysis can be extended to an vector-valued analytic function $F$. For example, we have the Cauchy integral formula:

$$ F(z) = \frac{1}{2 \pi i} \int_\Gamma \frac{F(\zeta)}{\zeta - z} d\zeta $$

where $\Gamma$ is a simple closed contour in $\Omega$ that encloses $z$. The integral is the Bochner integral, which is defined as the norm limit of Riemann partial sums.

Holomorphic Functional Calculus

Let $a \in A$ be a vector with spectrum $\sigma(a)$ contained in $\Omega$. Any classical analytic function $ f: \Omega \to \mathbb{C} $ gives rise to a vector $f(a) \in A$, which is defined to be the Bochner integral:

$$ f(a) = \frac{1}{2 \pi i} \int_\Gamma \frac{f(\zeta)}{\zeta - a} d\zeta $$

For simplicity let's assume the resolvent set $\Omega - \sigma(a)$ is connected, in which case $\Gamma$ can be any simple closed contour in $\Omega$ that encloses all of $\sigma(a)$ (otherwise $\Gamma$ should be a disjoint union of curves that are located and oriented appropriately).

The holomorphic functional calculus defines a continuous algebra homomorphism $ \Phi_a: H(\Omega) \to A $ by $ \Phi_a(f) = f(a) $ for $f \in H(\Omega)$, which satisfies the spectral mapping theorem:

$$ \sigma(f(a)) = f(\sigma(a)) $$

Some observations

  1. Both kinds of calculus output vectors. The difference is, in the first case, we first fix a vector-valued function $ F: \Omega \to A $, then each complex number $ z \in \Omega $ gives a vector $F(z) \in A$. In the second case, we first fix a vector $a \in A$, then each analytic function $ f \in H(\Omega) $ gives another vector $f(a) \in A$.

  2. Both kinds of calculus have Cauchy integral formulae. The difference is, in the integrand of the first integral, the numerator $F(\zeta)$ is vector while the denominator $\zeta - z$ is number. In the integrand of the second integral, the numerator $f(\zeta)$ is number while the denominator $\zeta - a$ is vector.

  3. Holomorphic functional calculus has the spectral mapping theorem $\sigma(f(a)) = f(\sigma(a))$, but vector-valued analytic functions don't have a similar result for the spectrum $\sigma(F(z))$.

Questions

  1. Based on observations 1 and 2, I feel that there is sort of duality going on between the two kinds of calculus. Are there some established results on this could-be duality?
  2. As said in observation 3, there is no spectral mapping theorem for a vector-valued analytic function $ F: \Omega \to A $. Let $\sigma(F)$ be the union of all spectra $ \sigma(F) = \bigcup_{z \in \Omega} \sigma(F(z)) $. What do we know about $\sigma(F)$? Is it the analytic image of a closed set? Is it even closed?
  3. More about the duality. Let $\Sigma \subseteq \mathbb{C}$ be an open set that contains the closure of $\sigma(F)$ (in case $\sigma(F)$ is not closed). For each $z \in \Omega$ and each analytic function $f \in H(\Sigma)$, we get a vector $f(F(z)) \in A$. Thus, we have a family of maps

$$ \begin{align*} F_f: \Omega &\to A \\ z &\mapsto f(F(z)) \end{align*} \quad \text{for each} \quad f \in H(\Sigma) $$

Is each $F_f$ analytic? How is it related to the orignal function $ F: \Omega \to A $? Do we have some version of Montel's theorem for the family $ \{ F_f \}_{f \in H(\Sigma)} $?

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    $\begingroup$ In spirit, this is $\sum_{n \geq 0} A_n z^n$ vs. $\sum_{n \geq 0} a_n Z^n$, where small letters are complex scalars and capital letters are elements of the algebra. In the finite-dimensional setting, algebraically, you have, say, $M_n(\mathbb{C})[[z]] \cong M_n (\mathbb{C}[[z]])$ for the first type of series. For the second type of series pick for concreteness a basis, e.g. $Z = \sum_{1\leq i,j\leq n} E_{ij} z_{ij}$. Then the second series is a matrix of formal power series of $n^2$ indeterminates that obey a system of generalized first order partial differential equations with const. coeff. $\endgroup$
    – M.G.
    Commented Apr 1, 2021 at 10:01
  • $\begingroup$ On a side note I've always wondered if there is holomorphic functional calculus for vector-valued holomorphic functions. There should be, but I never found a reference. $\endgroup$
    – M.G.
    Commented Apr 1, 2021 at 10:05
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    $\begingroup$ On a more general note, I would argue that there are a two red herings here: (1) Vector-valued holomorphic functions are typically defined in the more general setting where the target space is just a Banach space and thus does not carry any multiplicatice structure. The holomorphic functional calculus, on the other hand, requires a Banach algebra, i.e., both concepts are actually defined in different frameworks. $\endgroup$ Commented Apr 2, 2021 at 11:53
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    $\begingroup$ @M.G.: "I've always wondered if there is holomorphic functional calculus for vector-valued holomorphic functions." Very interesting idea! The thought had never crossed my mind before, but now as you mention it, it seems quite a natural thing to do. So I would be really interested in learning about such a theory, too. $\endgroup$ Commented Apr 2, 2021 at 12:02
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    $\begingroup$ Concerning holomorphic functional calculus for vector-valued holomorphic functions: While there are several useful notions of a spectrum for nonlinear operators (see e.g. J. Appell, E. De Pascale, A. Vignoli, Nonlinear spectral theory), there is no hope to obtain a corresponding functional calculus due to lack of commutativity and distributivity of nonlinear operators: Already the 3 representations of the same quadratic polynomial x^2+x=x(x+1)=(x+1)x give 2 different results even for simple nonlinear operators/maps. $\endgroup$ Commented Apr 7, 2021 at 4:12

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