Advanced Analysis, Notes 18: The holomorphic functional calculus I (motivation, definition, line integrals of holomorphic Banach-space valued functions)
by Orr Shalit
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 . By “unital Banach algebra” we mean that
is a Banach algebra with normalised unit
. For a complex number
we write
for
; in particular
. The spectrum
of an element
is the set
The resolvent set of ,
, is defined to be the complement of the spectrum,
.
1. Motivation
Recall the proof of the key theorem in spectral theory that says that the spectrum of any element is non-empty. For the proof (which is due to Gelfand) we assumed that the spectrum
is empty and we defined for a fixed bounded functional
, the complex valued function
(defined on the complement of the spectrum of
) given by :
We then showed that is an entire function (analytic on the whole plane) and that
as
. It then follows from Louiville’s Theorem that
, so in particular
. But this holds for all
, so it follows from the Hahn-Banach theorem that
– 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 , and denote by
the algebra of all functions that are analytic in some neighbourhood of
. Then there exists a continuous homomorphism
, such that
and
.
Here we denoted by the function
. I will explain below what is meant by “continuous”. Denoting
the theorem says in particular that we will be able to give meaning to “evaluating the function
at
“, provided that the function
is in
, that is, it is analytic in some neighbourhood of
. Moreover, the theorem says that the constant function
is mapped to the identity of
and that the “coordinate” function
is mapped to
; adding to this the homomorphism and continuity properties one sees that this “functional calculus” gives the expression
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
.
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 (where
are complex numbers), then there is no problem to define
. It is easy to check that when holding
fixed the mapping
is a homomorphism from the algebra of polynomials into
.
Though there are no issues in defining for a polynomial
, we note that every polynomial is an entire function, and in particular analytic in a neighbourhood of
, for any
.
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 . If the mapping
is to be a homomorphism, then the only reasonable definition for
is
, and this only makes sense if
is invertible. But this is not an entirely new issue, a problem already arises when we try to evaluate
at
.
More generally, consider the function . We would like to define
, but
is invertible only if
. So if
then there is no reasonable way to define
, but if
then there is an obvious way to define
, and that is
.
This leads us to define the algebra of functions , which contains all rational functions with poles off
. There are two natural ways to define
for
.
First way: We already know now how to define when
is either a polynomial or is of the form
from above. There is also a natural way to define
when
(for
, of course) which is
. But then using partial fraction decomposition we have all the ingredients to define
for all
. This is not the most natural definition, especially since it is not obvious that the mapping
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 be a polynomial. Then
, that is, the spectrum of
is equal to the set of all points
where
.
Exercise A: Prove Lemma 2.
Now if is in
(where
and
are polynomials such that
has no zeros on
) then we define
. This makes sense, because
is invertible. Indeed, the lemma says that
is not in the spectrum of
, since
has no zeros on
.
We point out before moving forward that is precisely the set of rational functions which are holomorphic in some neighbourhood of
; 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 in the function theory related to
can be traced to the Runge approximation theorem.
4. Power series
Not all holomorphic functions are rational. How do we proceed to define 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 that is analytic in a disc
of radius
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
has a power series expansion
Since we have a formula for , is natural to hope to define
by plugging
into the power series. The question is: for what
does this make sense? The answer is very simple. Let us assume that
is precisely the radius of convergence of the series (otherwise, consider a bigger disc). Plugging
in the formula we obtain
. A sufficient condition for this series to converge is that the series
converge. Applying the root test we find that a sufficient condition for this series to converge is that
.
But , while on the other hand, $\lim \|a_n\|^{1/n}$ is equal to the spectral radius of
. We conclude that if the spectral radius of
is strictly less than
, than the series
converges in the norm. In other words,
can be defined as
when
.
5. General analytic functions
Define to be the algebra of all functions which are analytic in some neighbourhood of
. Our goal is to define
for
. Working in this generality requires considerably more effort than the above cases of polynomials, rational functions with poles off
, and power series.
For example, suppose that is equal to the circle
. Let
. There are many functions analytic in
which are neither rational nor represented by power series. Let
be such a function. Since
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
around every point in
, that converges to
in a small neighbourhood of
. Should we plug
in one of these power series? Which one? It turns out that it is senseless to plug
in one of these power series (think about this). What we need is some formula into which we can plug
. 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 be an open subset of
such that the boundary of
is a finite union of
curves. If
is a function holmorphic in a neighborhood of
, then
- For all
.
- For all
,
In this formula the argument appears in avery simple way, without invoking the function
.
If is an element of a unital Banach algebra and
is a function holomorphic in a neighborhood
of
, we find an open set
containing
such that
. Then we simply plug in
instead of
in the above formula to obtain the desired definition of
. More precisely:
(*) .
We know what means when
is in
but not in
, 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
is another open set containing
and
is another set taking the role of
, and
is the boundary of
, we would like to show that defining
by some integral over
gives the same element of
). 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 and taking values in a Banach space (the function appearing in (*) is
). 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
. We will then turn to studying various properties of the formula (*).
Definition 3: Let be a Banach space, and let
be a continuous function. We define the integral of
on the interval
as
where the limit is as taken over all partitions
of
.
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 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: .
2) Additivity: .
3) Triangle inequality: (the integral on the right hand side is a usual Riemann integral, and it is well defined since the assumption on
implies that
is (uniformly) continuous).
4) Linearity + continuity: for every , we have
Again, the right hand makes sense since is continuous.
Returning to formula (*) above,
(*) ,
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 by
then you will get the definition of complex line integral from the theory of complex variables.
Definition 5: Let be a
curve in the complex plane. By this, we mean that there is an injective piecewise
function
such that
. Let
be uniformly continuous function from
into a Banach space
. The line integral of
on
is defined to be
Note that the right hand side is something that we already know how to compute.
Exercise C: Show that
where the limit is taken over the partitions of . In particular, the line integral does not depend on the choice of parametrisation
.
The definition of line integrals extends easily to paths that are not necessarily nor connected:
Definition 6: Let be a finite union of
curves. Then we define \
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 be complex Banach space.
Definition 7: Let be an open set in the complex plane, and let
be a function.
is said to be weakly analytic if for all
, the function
is analytic.
is said to be strongly analytic if for every
, the following limit exists in the norm of
:
.
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 be an open subset of the complex plane and let
be a weakly analytic. Then the following hold:
is continuous.
- If
is bounded and open and
, then
and
.
is strongly analytic.
Remark: Note that the first assertion implies that is uniformly continuous on
, hence the integrals in the second assertion make sense.
Proof: For (1), we assume without loss of generality that and prove that
is continuous at
. Let
be small enough that the closed disc
centered at
is contained in
. Fix
, and consider the function
.
The fact that is weakly analytic implies that
, and hence
, is analytic in a neighborhood of
(the function
has at least one zero at the origin and this zero can be factored out). In particular,
is bounded in
. Since this was true for every
, we conclude that the set
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 for some constant
, which readily implies that
is continuous at
.
From continuity of it follows that the integrals in assertion (2) exist. To prove (2), we use property (4) from Proposition 4 above, saying that for all
,
,
by the classical Cauchy formula, since is weakly analytic. By Hahn-Banach
. The second part of (2) is proved in a similar manner.
Finally, we prove (3). Without loss of generality we assume and prove that
. For
small enough so that
and
we write, using Cauchy’s formula (assertion (2))
which can be reorganized as
.
Writing
we obtain
Since the second integral on the right hand side remains bounded when lies in
, the left hand side converges to the first integral on the right hand side as
.
Example 9: In the proof that the spectrum of an element is non-empty, we showed that for every
, the function
is analytic in . That is just saying that the
-valued function
is weakly analytic. By the above theorem,
is strongly analytic on
.
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.
[…] In this post we continue our discussion of the holomorphic functional calculus for elements of a Banach algebra (or operators). The beginning of this discussion can be found in Notes 18. […]
Why is it senseless to plug in
in one of these power series?
Suppose that
is a diagonal
matrix, with
and
on the diagonal. The spectrum of
is
. Let
. Then
is holomorphic in a neighborhood of
, but has no power series that works everywhere in a nieghborhood of
.
has a power series
around
, and a different power series
around
. If you plug
into either of these series you get a non convergent sum.
Now, if you look at the matrix
, you know what to do. But this is because
is a nice diagonal matrix, given to us in some basis. In general, it is (at least) not clear how to use power series to define
. In this example, we can apply the rational functional calculus to
. But it shows why using power series does not work.
Hi Orr Shalit. Do you have any reference on where to find a proof of the spectral radius formula, mentioned in the motivation, using holomorphic functional calculus?
In Rudin’s book “Functional Analysis” the formula is proved using the holomorphic functional calculus.
The proof given in Arveson’s book that I mention at the top of the post (and similar proofs appear in many books) uses complex function theory, but avoids the holomorphic functional calculus. One might argue that the holomorphic functional calculus lurks in the background of such proofs.