### K-spectral sets and the holomorphic functional calculus

In two previous posts I discussed the holomorphic functional calculus as part of a standard course in functional analysis (lectures notes 18 and 19). In this post I wish to discuss a slightly different approach, which relies also on the notion of K-spectral sets, and relies a little less on contour integration of Banach-space valued functions.

In my very personal opinion this approach is a little more natural then the standard one, and it would be even more natural if one was able to altogether remove the dependence on Banach-space valued integrals (unfortunately, right now I don’t know how to do this completely).

#### 1. K-spectral sets

Throughout this post $A$ denotes a Banach algebra with a normalized unit, and $a$ is a fixed element of $A$. If $X \subseteq \mathbb{C}$ we denote by $Rat(X)$ the (analytic) rational functions with poles off $X$. If $f$ is a continuous function on $X$ then we denote $\|f\|_X = \sup_{z\in X}|f(z)|$.

Definition 1: Let $X \subset \mathbb{C}$ be a compact set such that $\sigma(a) \subseteq X$ and let $K$ be positive number. $X$ is said to be a K-spectral set for $a$ if for every $f \in Rat(X)$ the following inequality holds:

$\|f(a)\| \leq K \|f\|_X.$

In this definition it is important that we know how to define $f(a)$ when $f \in Rat(X)$. Since $\sigma(a) \subseteq X$, we know how to do this by the functional calculus. However, one does not need the elaborate machinary of the functional calculus, since there is a natural interpretation of $f(a)$ for every $f \in Rat(\sigma(a))$ (see Sections 2 and 3 in Notes #18). One way is to simply plug $a$ in the partial fraction decomposition of $f$. Another, perhaps even more natural way, is to write $f = g/h$, where $h$ has no zeros in $X$, and define $f(a) = g(a) h(a)^{-1}$ (this definition is legal since for every polynomial $h$, it holds that $h(\sigma(a)) = \sigma(h(a))$).

Remark: The interested reader might want to look at this extensive survey on K-spectral sets. If the constant $K$ can be chosen to be $1$, then $X$ is simply called a spectral set for $a$. Spectral sets happen to be one of my research interests, I referred to the notion in a previous post.

Theorem 2: For every bounded open set $W$ containing $\sigma(a)$, its closure $X:= \overline{W}$ is a K-spectral set for some $K$. In other words, the functional calculus $f \mapsto f(a)$ is a bounded map from $(\mathcal{O}(X),\|\cdot\|_X)$ to $(A, \|\cdot \|)$

Remark: In the theorem, $\mathcal{O}(X)$ denotes the algebra of functions analytic in some neighborhood of $X$, and we equip it with the sup norm (on $X$) which makes it into a normed space (which is not complete).

Proof: This follows from item (5) in the main theorem on the holomorphic functional calculus (see Section 1, Notes 19). Indeed, the functional calculus is a linear map, and a linear map between normed spaces is continuous if and only if it is bounded (no harm is caused by the fact $(\mathcal{O}(X),\|\cdot\|_X)$ is not complete).

Thus, we see that every neighborhood of the spectrum is a K-spectral set. Below we will show how to define the analytic functional calculus, using the rational functional calculus and assuming that every neighborhood is a K-spectral set. If one was able to prove that the rational functional calculus is bounded (i.e., that the closure of every bounded open neighborhood of $\sigma(a)$ is a K-spectral set for some $K$) without using Cauchy integrals, then one would have a way to define the analytic functional calculus without using Banach-space valued integrals. I believe (a belief that is based only on a hunch and not on historical documents or something like that) that this (the construction of a natural analytic functional calculus without recourse to integration) was the reason why von Neumann introduced spectral sets in the first place.

#### 2. Definition of the holomorphic functional calculus via K-spectral sets

Lemma 3: The rational functional calculus given for $f \in Rat(\sigma(a))$ by  $f = g/h \mapsto f(a) := g(a) h^{-1}(a)$ is a well defined homomorphism from $Rat(\sigma(a))$ into $A$.

Remark: In writing $f = g/h$ we are assuming that $g$ and $h$ are polynomials such that $h$ has no roots on $\sigma(a)$.

Proof: Suppose that $f = g/h = p/q$. Then $gq = ph$, thus $g(a)q(a) = p(a) h(a)$, since the polynomial functional calculus (see Section 2 in Notes 18) is obviously a well defined homomorphism. But then we get $g(a) h(a)^{-1} = p(a) q(a)^{-1}$, and this means that $f(a)$ gets the same value, no matter how we represented it as a quotient of two polynomials.

Once it has been shown that the mapping is well defined, it follows easily from algebraic manipulations that it is a homomorphism.

To define the analytic functional calculus, we need Runge’s Theorem from the theory of analytic functions of one complex variable (we use a simpler version than the version appearing in the linked Wikipedia article).

Theorem 4 (Runge’s Theorem): Let $X \subset \mathbb{C}$ be a compact set, and suppose that $f \in \mathcal{O}(X)$, that is, $f$ is analytic in some open neighborhood of $X$. Then there is a sequence $\{r_n\}$ of rational functions with poles off $X$ such that

$\|r_n - f\|_X \rightarrow 0$.

Now using Runge’s Theorem and the rational functional calculus, and assuming that every neighborhood of $\sigma(a)$ is a K-spectral set for $a$, we define the analytic functional calculus as follows:

Let $f \in \mathcal{O}(\sigma(a))$. Then there is some open neighborhood $U\supset \sigma(a)$ such that $f \in \mathcal{O}(U)$. One may find another bounded open set $V$ containing $\sigma(a)$ such that $V \subset \overline{V} \subset U$. By assumption, $X:= \overline{V}$ is a K-spectral set for $a$, where $K$ is some positive real constant. By Runge’s theorem, there is a sequence $r_n \in Rat(X)$  that converges uniformly to $f$ on $X$. Therefore, $\{r_n\}$ is a Cauchy sequence in the sup norm. Since $X$ is a K-spectral set, $\|r_m(a) - r_n(a) \| \leq K \|r_m - r_n\|_X$, thus $\{r_n(a)\}$ is a Cauchy sequence in the Banach space $A$. Define $f(a)$ to be the limit of the sequence $\{r_n(a)\}$.

Exercise A: Show that the mapping $f \mapsto f(a)$ is well defined; that is, $f(a)$ does not depend on the choice of sequence $\{r_n\}$ of rational functions converging uniformly to $f$, nor on the open sets $V$ and $U$ appearing in the above paragraph, but only on the function $f$ itself.

#### 3. Some properties of the functional calculus

Under the assumption that every bounded neighborhood of the spectrum is a K-spectral set (for some $K$) the functional calculus constructed above can be shown to have all the nice properties of the functional calculus constructed in the last two lectures (of course it has the same properties, since it is the same functional calculus).

Compare the following proof with the proof of Section 3 in the last lecture.

Proposition 5 (the functional calculus is a homomorphism): If $f,g \in \mathcal{O}(\sigma(a))$, then $fg(a) = f(a) g(a)$

Proof: Let $U\supset \sigma(a)$ be an open set where both $f$ and $g$ are analytic. Let $V$ be a bounded open set containing $\sigma(a)$ such that $\overline{V} \subset U$, and let $q_n$ and $r_n$ be sequences of rational functions with poles off $\overline{V}$ that converge uniformly to $f$ and $g$, respectively. Then $q_n r_n$ converges to $fg$ uniformly on $\overline{V}$. Thus, using the fact that the rational functional calculus is a homomorphism, we find

$fg(a) = \lim (q_n r_n )(a) = \lim q_n(a) r_n(a) = f(a) g(a),$

as required (don’t forget the one also needs to show linearity, but it is a simple matter and we omit it).

We now prove another property of the functional calculus. We did not prove it in the previous lectures (it was given in the homework exercises – see Exercise 7 here). We will assume the Spectral Mapping Theorem (Property 4 of the functional calculus from the previous lecture), which says that if $f \in \mathcal{O}(\sigma(a))$, then

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

Proposition 6: Let $a \in A$. Let $g\in \mathcal{O}(\sigma(a))$ and let $f \in \mathcal{O}(\sigma(g(a)))$. Then $f \circ g \in \mathcal{O}(\sigma(a))$ and

(**) $f \circ g (a) = f(g(a))$

Proof: That $f \circ g \in \mathcal{O}(\sigma(a))$ is elementary, and is left to the reader. The fact that $f \circ g (a) = f(g(a))$ holds for rational functions (with poles in the right places) follows easily from the well definedness of the rational functional calculus.

We now show that formula (**) holds . Some care is needed to do this right, so we will do it wrong. But here is the idea. First we show that (**) holds for rational $f$ and arbitrary $g$.

Suppose that $U$ is an open set where $f$ is analytic. If $x, x_n \in A$, and $x_n \rightarrow x$, and the spectrum of $x, x_n$ are all in $U$, then $f(x), f(x_n)$ are defined and $f(x_n) \rightarrow f(x)$ (this follows by writing the partial fraction decomposition and using continuity of addition, multiplication, and the inverse in a Banach algebra). Now let $r_n$ be a sequence of rational functions converging uniformly to $g$ in a neighborhood of $\sigma(a)$. Then $r_n(a) \rightarrow g(a)$, whence

$f\circ g(a) = \lim f \circ r_n (a) = \lim f(r_n(a)) = f(g(a)) ,$

where the second equality follows from the fact that (**) holds for rational functions. Thus (**) holds for $f$ rational and $g$ analytic.

We now show that (**) for $f$ and $g$ analytic. Let $f,g$ be analytic, and let $q_n$ be a sequence of rationals converge uniformly to $f$ in a neighborhood of $\sigma(g(a))$. Then

$f\circ g (a) = \lim q_n \circ g(a) = q_n(g(a)) = f(g(a)) ,$

from our definition of the functional calculus. This completes the proof.

The usual proof of this fact (which some students handed in) is to do some rather tricky Cauchy integration.

#### 4. Proof that every neighborhood of the spectrum if a K-spectral set

Here I will write a proof of this fact that does not use Banach-space valued Cauchy integrals, when I find such a proof (or when a reader tells me how to do it).