### Advanced Analysis, Notes 18: The holomorphic functional calculus I (motivation, definition, line integrals of holomorphic Banach-space valued functions)

This course, Advanced Analysis, contains some lectures which I have not written up as posts. For the topic of Banach algebras and C*-algebras the lectures I give in class follow pretty closely Arveson’s presentation from “A Short Course in Spectral Theory” (except that we do more examples in class). But there is one topic  – the holomorphic functional calculus -for which I decided to take a slightly different route, and for the students’ reference I am writing up my point of view.

Throughout this lecture we fix a unital Banach algebra $A$. By “unital Banach algebra” we mean that $A$ is a Banach algebra with normalised unit $1_A$.  For a complex number $t \in \mathbb{C}$ we write $t$ for $t \cdot 1_A$; in particular $1 = 1_A$.  The spectrum $\sigma(a)$ of an element $a \in A$ is the set

$\sigma(a) = \{t \in \mathbb{C} : a- t \textrm{ is not invertible in } A\}.$

The resolvent set of $a$, $\rho(a)$, is defined to be the complement of the spectrum,

$\rho(a) = \mathbb{C} \setminus \sigma(a)$.

#### 1. Motivation

Recall the proof of the key theorem in spectral theory that says that the spectrum of any element $a \in A$ is non-empty. For the proof (which is due to Gelfand) we assumed that the spectrum $\sigma(a)$ is empty and we defined for a fixed bounded functional $f \in A^*$, the complex valued function $u_f$ (defined on the complement of the spectrum of $a$) given by :

$u_f(z) = f((a-z)^{-1}).$

We then showed that $u_f$ is an entire function (analytic on the whole plane) and that $u_f(z) \rightarrow 0$ as $z \rightarrow \infty$. It then follows from Louiville’s Theorem that $u_f \equiv 0$, so in particular $f(a^{-1}) = 0$. But this holds for all $f \in A^*$, so it follows from the Hahn-Banach theorem that $a^{-1} = 0$ – but this is absurd since this element is supposed to be invertible.

The proof is simple enough once one knows the idea, but it takes a genius like Gelfand to come up with an idea like this. The beautiful and profound idea making this proof work is that when studying Banach algebras over the complex numbers it should be useful to exploit the powerful theory of analytic functions in a complex variable; with hindsight that does indeed seem like a natural thing to do.

The theory of analytic functions is used also in a non-trivial way in obtaining the spectral radius formula, and also (in a rather trivial way) in the construction of the Neumann series. Our goal in these lectures is to develop the so-called holomorphic functional calculus ( or analytic functional calculus – we will use the words holomorphic and analytic interchangeably). This functional calculus gives a systematic way of applying analytic functions to Banach algebra elements. Ultimately, we will obtain the following theorem.

Theorem 1(holomorphic functional calculus): Let $a \in A$, and denote by $\mathcal{O}(a)$ the algebra of all functions that are analytic in some neighbourhood of $\sigma(a)$. Then there exists a continuous homomorphism $\Psi : \mathcal{O}(a) \rightarrow A$, such that $\Psi(1) = 1$ and $\Psi(z) = a$.

Here we denoted by $z$ the function $z \mapsto z$. I will explain below what is meant by “continuous”. Denoting $f(a) = \Psi(f)$ the theorem says in particular that we will be able to give meaning to “evaluating the function $f$ at $a$“, provided that the function $f$ is in $\mathcal{O}(a)$, that is, it is analytic in some neighbourhood of $\sigma(a)$. Moreover, the theorem says that the constant function $1$ is mapped to the identity of $A$ and that the “coordinate” function $z$ is mapped to $a$; adding to this the homomorphism and continuity properties one sees that this “functional calculus” gives the expression $f(a)$ a reasonable meaning whenever we can think of one. Before we prove the theorem, we take our time to look at simpler cases of giving value to the expression $f(a)$.

#### 2. Polynomials

The simplest functions which we can always apply to elements of a unital algebra (not necessarily a Banach algebra) are, of course, polynomials. If $p(z) = c_0 + c_1 z + \ldots + c_n z^n$ (where $c_0, c_1, \ldots, c_n$ are complex numbers), then there is no problem to define $p(a) = c_0 + c_1 a + \ldots + c_n a^n$. It is easy to check that when holding $a$ fixed the mapping $p \mapsto p(a)$ is a homomorphism from the algebra of polynomials into $A$.

Though there are no issues in defining $p(a)$ for a polynomial $p$, we note that every polynomial is an entire function, and in particular analytic in a neighbourhood of $\sigma(a)$, for any $a \in A$.

#### 3. Rational functions

What are the next simplest functions after polynomials? It depends who you ask, but a good case can be made for rational functions, being a natural extension of polynomials that is defined purely in arithmetical terms.

Consider the function $g(z) = z^{-1}$. If the mapping $f \mapsto f(a)$ is to be a homomorphism, then the only reasonable definition for $g(a)$ is $a^{-1}$, and this only makes sense if $a$ is invertible. But this is not an entirely new issue, a problem already arises when we try to evaluate $g$ at $0 \in \mathbb{C}$.

More generally, consider the function $r_\lambda(z) = (z-\lambda)^{-1}$. We would like to define $r_\lambda(a) = (a-\lambda)^{-1}$, but $a-\lambda$ is invertible only if $\lambda \in \rho(a) = \mathbb{C} \setminus \sigma(a)$. So if $\lambda \in \sigma(a)$ then there is no reasonable way to define $r_\lambda(a)$, but if $\lambda \in \rho(a)$ then there is an obvious way to define $r_\lambda(a)$, and that is $r_\lambda(a) = (a - \lambda)^{-1}$.

This leads us to define the algebra of functions $Rat(a)$, which contains all rational functions with poles off $\sigma(a)$. There are two natural ways to define $r(a)$ for $r \in Rat(a)$.

First way: We already know now how to define $r(a)$ when $r$ is either a polynomial or is of the form $r(z) = r_\lambda(z) = (z-\lambda)$ from above. There is also a natural way to define $r(a)$ when $r = r_\lambda^k$ (for $\lambda \notin \sigma(a)$, of course) which is $r(a) = ((a-\lambda)^{-1})^k$. But then using partial fraction decomposition we have all the ingredients to define $r(a)$ for all $r \in Rat(a)$. This is not the most natural definition, especially since it is not obvious that the mapping $r \mapsto r(a)$ defined like this is a homomorphism.

Second way: Here is a more natural definition. It requires the following lemma whose simple proof is omitted (we will prove something more general later).

Lemma 2: Let $q$ be a polynomial. Then $\sigma(q(a)) = q(\sigma(a))$, that is, the spectrum of $q(a)$ is equal to the set of all points $q( \lambda)$ where $\lambda \in \sigma(a)$.

Exercise A: Prove Lemma 2.

Now if $r = p/q$ is in $Rat(a)$ (where $p$ and $q$ are polynomials such that $q$ has no zeros on $\sigma(a)$) then we define $r(a) = p(a) q(a)^{-1}$. This makes sense, because $q(a)$ is invertible. Indeed, the lemma says that $0$ is not in the spectrum of $q(a)$, since $q$ has no zeros on $\sigma(a)$.

We point out before moving forward that $Rat(a)$ is precisely the set of rational functions which are holomorphic in some neighbourhood of $\sigma(a)$; it is here that we first see that the notions of functional calculus and spectrum are connected.

The rational functions play a role that is more important than may seem at first. Without going into details, let me mention that (in my opinion) the role of $Rat(a)$ in the function theory related to $a$ can be traced to the Runge approximation theorem.

#### 4. Power series

Not all holomorphic functions are rational. How do we proceed to define $f(a)$ for functions which are not rational? We now finally make use of the fact that we are in a Banach algebra, and consider power series. This is a very natural class of functions to consider, but we will later argue that this is not the way to go.

Let us consider a function $f$ that is analytic in a disc $D_r$ of radius $r$ around the origin (it is not hard, and is left as an exercise for the reader, to figure out how what follows should be modified to work for functions analytic in a disc that is not centred at the origin). We know from function theory that $f$ has a power series expansion

$f(z) = \sum_{n=0}^\infty c_n z^n .$

Since we have a formula for $f$, is natural to hope to define $f(a)$ by plugging $a$ into the power series. The question is: for what $a$ does this make sense? The answer is very simple. Let us assume that $r$ is precisely the radius of convergence of the series (otherwise, consider a bigger disc). Plugging $a$ in the formula we obtain $\sum_{n=0}^\infty c_n a^n$. A sufficient condition for this series to converge is that the series $\sum_n \|c_n a^n\|$ converge. Applying the root test we find that a sufficient condition for this series to converge is that

$\limsup |c_n|^{1/n} \|a^n\|^{1/n} < 1$.

But $\limsup |c_n|^{1/n} = 1/r$, while on the other hand, $\lim \|a_n\|^{1/n}$ is equal to the spectral radius of $a$. We conclude that if the spectral radius of $a$ is strictly less than $r$, than the series $\sum c_n a^n$ converges in the norm. In other words, $f(a)$ can be defined as $f(a) = \sum c_n a^n$ when $\sigma(a) \subset D_r$.

#### 5. General analytic functions

Define $\mathcal{O}(a)$ to be the algebra of all functions which are analytic in some neighbourhood of $\sigma(a)$. Our goal is to define  $f(a)$ for $f \in \mathcal{O}(a)$. Working in this generality requires considerably more effort than the above cases of polynomials, rational functions with poles off $\sigma(a)$, and power series.

For example, suppose that $\sigma(a)$ is equal to the circle $\{z: |z|=1\}$. Let $U = \{z: 3/4<|z|<5/4\}$. There are many functions analytic in $U$ which are neither rational nor represented by power series. Let $f$ be such a function. Since $f$ is not a rational function we do not have the natural algebraic formulas given in the rational case. We know that there is a power series $\sum c_n(z_0) (z-z_0)^n$ around every point in $z_0 \in U$, that converges to $f$ in a small neighbourhood of $z_0$. Should we plug $a$ in one of these power series? Which one? It turns out that it is senseless to plug $a$ in one of these power series (think about this).  What we need is some formula into which we can plug $a$. What formulas do we have at our disposal?

If you remember enough from the course on function theory, you will recall that the decisive tool was not power series, but the Cauchy integral formula. Recall the statement:

Cauchy’s theorem and integral formula: Let $V$ be an open subset of $\mathbb{C}$ such that the boundary of $V$ is a finite union of $C^1$ curves. If $f$ is a function holmorphic in a neighborhood of $\overline{V}$, then

1. For all $\int_{\partial V} f(z) dz = 0$
2. For all $w \in U$

$f(w) = \frac{1}{2 \pi i}\int_{\partial V} \frac{f(z) dz}{z - w}.$

In this formula the argument $w$ appears in avery simple way, without invoking the function $f$.

If $a$ is an element of a unital Banach algebra and $f$ is a function holomorphic in a neighborhood $U$ of $a$, we find an open set $V$ containing $\sigma(a)$ such that $\overline{V} \subset U$. Then we simply plug in $a$ instead of $w$ in the above formula to obtain the desired definition of $f(a)$. More precisely:

(*) $f(a) := \frac{1}{2 \pi i} \int_{\partial V} f(z) (z-a)^{-1} dz$.

We know what $f(z) (z-a)^{-1}$ means when $z$ is in $U$ but not in $K = \sigma(a)$, but it takes some work to make clear the meaning of the integral, and a little more work to see that the definition is independent of the choices made (if $U'$ is another open set containing $\sigma(a)$ and $V'$ is another set taking the role of $V$, and $C'$ is the boundary of $V'$, we would like to show that defining $f(a)$ by some integral over $C'$ gives the same element of $A$). The rest of this post and the next one is devoted to making sense of the above formula and studying its properties.

#### 6. Integrating functions with values in a Banach space

To make sense of the formula (*) above, we need to explain how to integrate a function defined on a curve $C \subset \mathbb{C}$ and taking values in a Banach space (the function appearing in (*) is $z \mapsto f(z)(z-a)^{-1}$). Then we will show that for certain Banach-space-valued functions, those that are analytic, the integral (*) is independent of the choices made, in particular it is independent of the path of integration $C$. We will then turn to studying various properties of the formula (*).

Definition 3: Let $X$ be a Banach space, and let $f:[a,b] \rightarrow X$ be a continuous function. We define the integral of $f$ on the interval $[a,b]$ as

$\int_a^b f(t) dt = \lim \sum_{j=0}^{n-1} f(t_j)(t_{j+1}-t_j) ,$

where the limit is as $\max \{t_{j+1}-t_j : j=0,1, \ldots, n-1\} \rightarrow 0$ taken over all partitions $a = t_0 < t_1 < \ldots < t_n = b$ of $[a,b]$

Exercise B: Prove that the above limit does indeed exist (it is very similar to the proof that the Riemann sums of a real valued continuous function converge, but be careful since $X$ is Banach-space valued and there are no upper sums or lower sums).

Note that a continuous function from one metric space to another is always uniformly continuous on any compact set; this observation is crucial to the solution of the above exercise. We will have occasion to integrate only (uniformly) continuous functions.

Proposition 4 (properties of the integral): The above integral enjoys the following properties:

1) Linearity: $\int_a^b (f(t) + c g(t) ) dt = \int_a^b f(t) dt + c \int_a^b g(t) dt$.

2) Additivity: $\int_a^c f(t) dt = \int_a^b f(t) dt + \int _b^c f(t) dt$.

3) Triangle inequality: $\| \int_a^b f(t) dt\| \leq \int_a^b \|f(t)\| dt$ (the integral on the right hand side is a usual Riemann integral, and it is well defined since the assumption on $f$ implies that $t \mapsto \|f(t)\|$ is (uniformly) continuous).

4) Linearity + continuity: for every $\rho \in X^*$, we have

$\rho \left(\int_a^b f(t) dt \right) = \int_a^b \rho(f(t)) dt .$

Again, the right hand makes sense since $t \mapsto \rho(f(t))$ is continuous.

Returning to formula (*) above,

(*) $f(a) = \int_C f(z) (z-a)^{-1} dz$,

we see that we are going to need to know how to compute complex line integrals of functions with values in a Banach space. In the following definition, if you replace the Banach space $X$ by $\mathbb{C}$ then you will get the definition of complex line integral from the theory of complex variables.

Definition 5: Let $\Gamma$ be a $C^1$ curve in the complex plane. By this, we mean that there is an injective piecewise $C^1$ function $\gamma : [a,b]\rightarrow \mathbb{C}$ such that $\Gamma = \gamma([a,b])$. Let $f : \Gamma \rightarrow X$ be uniformly continuous function from $\Gamma$ into a Banach space $X$. The line integral of $f$ on $\Gamma$ is defined to be

$\int_\Gamma f(z) dz = \int_a^b f(\gamma(t)) \gamma'(t) dt.$

Note that the right hand side is something that we already know how to compute.

Exercise C: Show that

$\int_\Gamma f(z) dz = \lim \sum_{j=0}^{n-1} f(\gamma(t_j)) (\gamma(t_{j+1})-\gamma(t_j)) ,$

where the limit is taken over the partitions of $[a,b]$. In particular, the line integral does not depend on the choice of parametrisation $\gamma$.

The definition of line integrals extends easily to paths that are not necessarily $C^1$ nor connected:

Definition 6: Let $\Gamma = \Gamma_1 \cup \ldots \cup \Gamma_n$ be a finite union of $C^1$ curves. Then we define \

$\int_\Gamma f(z) dz = \sum_{k=1}^n \int_{\Gamma_k} f(z) dz .$

In function theory, complex line integrals were particularly useful when applied to analytic functions. To make progress we require a notion of a analytic functions with values in a Banach space.

#### 7. Analytic Banach-space valued functions

Throughout this section, let $X$ be complex Banach space.

Definition 7: Let $U$ be an open set in the complex plane, and let $f : U \rightarrow X$ be a function.

1. $f$ is said to be weakly analytic if for all $\rho \in X^*$, the function $\rho \circ f : U \rightarrow \mathbb{C}$ is analytic.
2. $f$ is said to be strongly analytic if for every $z_0 \in U$, the following limit exists in the norm of $X$:

$\lim_{z\rightarrow z_0 } \frac{f(z) - f(z_0)}{z-z_0}$.

It is clear that a strongly analytic function is weakly analytic and continuous. We shall see below that the converse also holds.

Theorem 8: Let $U$ be an open subset of the complex plane and let $f : U \rightarrow X$ be a weakly analytic. Then the following hold:

1. $f$ is continuous.
2. If $V$ is bounded and open and $\overline{V} \subset U$, then $\int_{\partial V} f(z) dz = 0$ and $f(w) = \frac{1}{2\pi i} \int_{\partial V} \frac{f(z)dz}{z-w}$
3. $f$ is strongly analytic.

Remark: Note that the first assertion implies that $f$ is uniformly continuous on $\partial V$, hence the integrals in the second assertion make sense.

Proof: For (1), we assume without loss of generality that $0 \in U$ and prove that $f$ is continuous at $0$. Let $r>0$ be small enough that the closed disc $D_{r}$ centered at $0$ is contained in $U$. Fix $\rho \in X^*$, and consider the function

$g(z) = \frac{\rho(f(z)) - \rho(f(0))}{z}$.

The fact that $f$ is weakly analytic implies that $\rho\circ f$, and hence $g$, is analytic in a neighborhood of $D_{r}$ (the function $\rho \circ f (z) - \rho \circ f(0)$ has at least one zero at the origin and this zero can be factored out). In particular, $g$ is bounded in $D_r$. Since this was true for every $\rho \in X^*$, we conclude that the set

$\{z^{-1}(f(z) - f(0)) : 0<|z|\leq r\}$

is weakly bounded. By a corollary to the uniform boundedness principle (adapt the proof of Theorem 6 in Lecture 11), this set is norm bounded. It follows that $\|f(z) - f(0)\| \leq Mz$ for some constant $M$, which readily implies that $f$ is continuous at $0$.

From continuity of $f$ it follows that the integrals in assertion (2) exist. To prove (2), we use property (4) from Proposition 4 above, saying that for all $\rho \in X^*$,

$\rho (\int_{\partial V} f(z) dz ) = \int_{\partial V}\rho \circ f(z) dz = 0$,

by the classical Cauchy formula, since $f$ is weakly analytic. By Hahn-Banach $\int_{\partial V} f(z) dz = 0$. The second part of (2) is proved in a similar manner.

Finally, we prove (3). Without loss of generality we assume $0 \in U$ and prove that $\lim_{w\rightarrow 0} \frac{f(w) - f(0)}{w} = 0$. For $r$ small enough so that $D_r \subset U$ and $|w| we write, using Cauchy’s formula (assertion (2))

$2 \pi i \frac{f(w)-f(0)}{w} = \int_{|z|=r}\frac{f(z) dz}{w(z-w)} - \int_{|z|=r}\frac{f(z) dz}{wz}$

which can be reorganized as

$2 \pi i \frac{f(w)-f(0)}{w} = \int_{|z|=r} \frac{f(z) dz}{z(z-w)}$.

Writing

$\frac{f(z)}{z(z-w)} = \frac{f(z)}{z^2} + \left( \frac{wf(z)}{z^2(z-w)}\right)$

we obtain

$\frac{f(w) - f(0)}{w} = \frac{1}{2 \pi i} \int_{|z|=1} \frac{f(z) dz}{z^2} + \frac{w}{2 \pi i} \int_{|z|=r} \frac{f(z)dz}{z^2(z-w)} .$

Since the second integral on the right hand side remains bounded when $w$ lies in $D_{r/2}$, the left hand side converges to the first integral on the right hand side as $w \rightarrow 0$.

Example 9: In the proof that the spectrum of an element $a \in A$ is non-empty, we showed that for every $f \in X^*$, the function

$u(z) = f((a-z)^{-1})$

is analytic in $\mathbb{C} \setminus \sigma(a)$. That is just saying that the $A$-valued function $U(z) = (a-z)^{-1}$ is weakly analytic. By the above theorem, $U$ is strongly analytic on $\mathbb{C} \setminus \sigma(a)$.

Remark: From now on we will only use the adjective “analytic”, since the two notions are the same.

We will continue discussing the holomorphic functional calculus in the next post.