Simple Connectivity and Linear Chaos

Joel H. Shapiro

February 4, 1998

Abstract. The goal of this paper is to show that, for domains G of the complex plane, simple connectivity can be characterized by the dynamical properties of certain linear differential operators acting on the space of functions holomorphic on G. The operator-theoretic issues that arise here lead to interesting problems, some of them apparently open. The paper is written so as to be accessible to anyone whose background includes the basics of graduate level complex and functional analysis.

Prologue

The notion of simple connectivity for plane domains stands somewhere between analysis and topology. In beginning complex analysis we learn that every nonvanishing function on such a domain has a holomorphic logarithm, and later on we encounter the Riemann Mapping Theorem, which tells us that simply connected domains are just the ones that are biholomorphically equivalent to the open unit disc. However the concept is most often defined topologically: simply connected domains are the ones in which every closed curve is null-homotopic. In fact if you open Rudin’s classic analysis textbook [18] to page 247, you’ll find the statements I just mentioned embedded in Theorem 13.11—a list of ten equivalent properties, each of which can be taken as the definition of simple connectivity.

The purpose of this paper is to add to these equivalences some further ones that come from the dynamical properties of linear operators. Now you may be surprised to hear that linear operators can have interesting dynamical properties, but I hope to convince you before we’re done that this is indeed the case, and that such phenomena create interesting connections between operator theory, dynamics, and analytic function theory.

I’ve tried to make the exposition accessible to anyone with a graduate-level background in complex and functional analysis. I think of what follows as a continuation of another of my favorite books: [15] by Luecking and Rubel. If you’ve had the pleasure of studying from this little gem, then you should be well prepared for what lies in the pages ahead.

Research supported in part by the National Science Foundation

1 Introduction

This work is set entirely in H(G), the vector space of all functions holomorphic on a plane domain G. In its natural topology—the topology of uniform convergence on compact subsets of G—this space is a complete, metrizable, locally convex linear topological space, or for short, a Fréchet space. A central role will be played by the quintessential linear operator on H(G), the operator D of (complex) differentiation. My goal here is examine how simple connectivity for G can be characterized in terms of certain dynamical properties of both D and those operators that commute with D. In order to state the results economically, I need to introduce some definitions.

1.1 Notation and terminology. Throughout this paper the symbols C, N, and Z denote, respectively, the complex plane, the natural numbers (i.e., the positive integers), and all the integers. The word “operator” and the phrase “linear operator” both mean “continuous linear transformation.” An operator T on a Fréchet space X will be called:

• Nonscalar if it is not a constant multiple of the identity operator;

• Cyclic if there is a vector x ∈ X (called a cyclic vector) whose T -orbit:

Orb(T, x) = {T^{n}x : n = 0, 1, 2, …}

has dense linear span in X;

• Hypercyclic if there is a vector x ∈ X (called a hypercyclic vector) for which Orb(T, x) itself is dense in X; and

• Chaotic [7, page 50, Defn. 8.5]^{∗} if it is hypercyclic and has a dense set of periodic points.

1.2 Notions of cyclicity. Clearly chaotic operators are hypercyclic, and hypercyclic operators are cyclic. Cyclicity is important when you study invariant subspaces (closed subspaces that get taken into themselves by the operator): an operator has a nontrivial invariant subspace precisely when it has a noncyclic vector. Hypercyclicity bears the same relationship to the existence of invariant subsets. You can think of hypercyclicity as a kind of topological randomness. The additional requirement that chaotic operators have dense sets of periodic points superimposes a kind of topological orderliness over all the hypercyclic randomness.

At this point you are entitled to wonder if there are any hypercyclic or chaotic operators at all! After all you’ve never seen such operators in linear algebra courses— for the simple reason that finite dimensional spaces do not support them (you’ll will see why complex spaces don’t support hypercyclic operators in §2.5). In infinitely many dimensions, however, the situation is much different, in fact some of our favorite operators turn out to be chaotic!

∗However, see §1.4 before you check out this reference.

Consider, for example, the first result on hypercyclicity, obtained in 1929 by G.D. Birkhoff [2], who discovered a hypercyclic operator on the space H(C) of entire functions. Now Birkhoff’s example is not something pathological, it is the beloved operator of “translation by a”

T_{a}(f)(z) =^{def} f(z + a) (f ∈ H(C), z ∈ C), (1)

where a can be any nonzero complex number. About twenty years later G.R. MacLane [16] showed that the differentiation operator D has the same property on H(C), and more recently Gilles Godefroy and I showed in [9] that every nonscalar operator on H(C) that commutes with D is chaotic. Thus, for example, every nonscalar constant coefficient differential operator is chaotic on H(C), as is every nontrivial linear combination of translation operators.

You may also be wondering if every hypercyclic operator has to be chaotic. The answer is “no”. Elementary examples of this phenomenon occur in [9, Theorem 6.3] (see also [10, Proposition 4.7(v)] and [12] for a different construction). In [6] Kit Chan and I found that Birkhoff’s original translation operators, when restricted to certain “small” Hilbert spaces of entire functions, are also hypercyclic but not chaotic.

1.3 Chaos and simple connectivity. The work in this paper was motivated by a conversation I had about eight years ago with Carl Prather of Virginia Tech. Prather asked about the fate of the results I just described when the space of entire functions is replaced by H(G) for G an arbitrary plane domain. In what follows I will answer Prather’s question by showing that: If G is a plane domain and P a nonconstant (holomorphic) polynomial, then the following are equivalent:

• P (D) is hypercyclic on H(G).

• P (D) is chaotic on H(G).

• G is simply connected.

The proof of this result will occupy the next section. It turns out that the proof still works when the polynomials in D are replaced by a more extensive class of operators that commute with D. However the result does not extend to all operators that commute with D: In Section 3 I’ll show you an infinitely connected plane domain G whose H(G) supports a chaotic operator that commutes with D. However if you are willing to restrict your attention solely to finitely connected domains, then it will turn out that the above characterization of simple connectivity goes through with polynomials in D replaced by arbitrary nonscalar operators that commute with D.

A recurrent theme in this paper is the issue of “which operators commute with differentiation?” It leads to interesting questions, some of which I discuss in (the mostly expository) Section 4.

The paper closes with a section that takes up the question of whether or not cyclicity can replace hypercyclicity in characterizing simple connectivity. The methods of Section 3 easily eliminate from consideration domains of genus greater than 2, but domains of genus 2—topological annuli—remain problematic. For this I present, in the hope that others may do better, a partial result that suggests an affirmative answer.

1.4 Chaotic footnote. If you look up the reference cited for the definition of “chaotic” (page 50 of Devaney’s book [7]), you won’t see the definition I have given in §1.1. Instead of hypercyclicity you’ll see a requirement of “topological transitivity” that is phrased in terms of open sets rather than orbits. Additionally you will find a further requirement of “sensitive dependence on initial conditions.” There is, however, no problem: the two definitions of “chaotic” are in fact equivalent. Topological transitivity is equivalent, for complete metric spaces, to the existence of a dense orbit (see [9, Section 1]), and in our Fréchet space setting, Devaney’s notion of “sensitive dependence” turns out to be a consequence of hypercyclicity (see [9, Section 6]).

Acknowledgments. As I mentioned above, Carl Prather provided the initial impetus for this work. Paul Bourdon made valuable suggestions on an early version, and encouraged me to think about publishing it. I thank both of these colleagues for their contributions. I also want to thank Pietro Aiena of the University of Palermo, who organized the wonderful conference in whose proceedings this paper appears.

2 Chaotic characterizations of simple connectivity

Much of the work that has been done on hypercyclic operators depends on the following sufficient condition, first discovered by Carol Kitai in her 1982 doctoral dissertation [13], but never published. Robert Gethner and I later rediscovered the result [8], and variations of it have figured prominently in much subsequent work (e.g., [4, 9, 11, 12, 20]). The proof, which is a simple Baire Category argument, occurs in many of these references, so I will omit it here, and instead refer you to Theorem 1.2 of my paper [9] with Godefroy, or to Section 7.1 of my book [20].

2.1 Theorem (“The Hypercyclicity Criterion”). Suppose T is an operator on a Fréchet space X. Suppose further that there are dense subsets X_{0} and Y_{0} of X, and a mapping S : Y_{0} → Y_{0}, such that:

(a) T^{n} → 0 pointwise on X_{0},

(b) S^{n} → 0 pointwise on Y_{0},

(c) T S is the identity map on Y_{0}.

Then T is hypercyclic on X.

2.2 Remark. If you have never seen this result before, don’t let its apparent complexity discourage you. As you will see before too long, it is often very easy to apply. In fact, you can illustrate this for yourself right now by using the Hypercyclicity Criterion to prove Rolewicz’s Theorem [17]—the first hypercyclic result for Hilbert space operators:

Let B denote the backward shift operator on the sequence space ℓ^{2}. Then for any complex number λ of modulus > 1 the operator λB is hypercyclic.

Here B is the operator that takes an ℓ^{2}-sequence (a_{0}, a_{1}, …) to its left-shifted sequence (a_{1}, a_{2}, …). To prove Rolewicz’s Theorem, just take X_{0} to be all sequences with only finitely many nonzero terms, take Y_{0} equal to ℓ^{2}, and set

S = \frac{1}{λ} × the forward shift

where the “forward shift” is the operator that takes (a_{0}, a_{1}, …) to its right-shifted sequence (0, a_{0}, a_{1}, …). Try it!

2.3 Non-hypercyclicity. We also need a convenient condition sufficient for an operator to not be hypercyclic. For this we use a result first observed by Kitai [13], which makes use of the adjoint of an operator on a Fréchet space. Suppose X is a Fréchet space and T is an operator on X. Let X^{∗} denote the dual space of X—the space of continuous linear functionals on X. Then the adjoint of T is the linear transformation defined on X^{∗} by the equation:

T^{∗}ϕ(x) = ϕ(T x) (ϕ ∈ X^{∗} and x ∈ X).

Although it is possible to give X^{∗} a topology in which T^{∗} becomes continuous, I won’t do this here: for us T^{∗} will always be a purely algebraic object.

2.4 A “non-hypercyclicity criterion.” Suppose T is an operator on a Fréchet space X, and that T^{∗} has an eigenvalue. Then T is not hypercyclic.

Proof. The hypothesis states that there is a complex number λ and a continuous linear function ϕ on X, not identically zero, such that T^{∗}ϕ = λϕ. Suppose x ∈ X. The goal is to show that x is not a hypercyclic vector for T, i.e. that the T -orbit of x is not dense in X. If x were hypercyclic for T then the set of values that ϕ takes on Orb(T, x) would be dense in the complex plane. But instead we have for each positive integer n:

ϕ(T^{n}x) =^{def} (T^{n∗}ϕ)(x) = (T^{∗n}ϕ)(x) = λ^{n}(T^{∗}ϕ)(x) =^{def} λ^{n}ϕ(T x),

which you can easily see shows that the set of values {ϕ(T^{n}x)} is not dense in X. Thus x is not hypercyclic for T, as promised. /// 2.5 Remark. I mentioned earlier that no operator on a finite dimensional space is hypercyclic. The above result shows why this is true, at least for complex vector spaces. For if T is an operator on a complex Fréchet space X of dimension 0 < n < ∞ then, because X^{∗} also has dimension n, T^{∗} must have an eigenvalue. Thus T cannot be hypercyclic. Real finite dimensional Fréchet spaces require a somewhat more complicated proof, but this too is possible: it was done by Paul Bourdon [3].

2.6 Complex exponentials. For λ ∈ C let e_{λ} denote the exponential function

e_{λ}(z) = e^{λz}.

Interest in these exponential functions stems from their role as eigenfunctions of the differentiation operator: De_{λ} = λe_{λ}, and so if P is a holomorphic polynomial, then P (D)e_{λ} = P (λ)e_{λ} for each λ ∈ C. Particularly important are the subspaces spanned by various collections of these eigenfunctions. If A is a subset of the complex plane, let

E(A) =^{def} span {e_{λ} : λ ∈ A}.

The following result is well known: I include it here in the interest of completeness.

2.7 Density Theorem. If G is simply connected then E(A) is dense in H(G) whenever A has a limit point in C.

Proof. We use the Hahn-Banach theorem. Suppose ϕ is a continuous linear functional on H(G) that annihilates each exponential function e_{λ} for λ ∈ A. By Hahn-Banach we will be done if we can show that ϕ = 0 on H(G). Since ϕ is continuous, the inverse image of the unit disc contains a basic neighborhood of zero in H(G). Now these basic neighborhoods of zero have the form

N(K, ε) =^{def} {f ∈ H(G) : ||f||_{K} < ε}

where K runs through compact subsets of G, ε through positive reals, and ||f||_{K} is the maximum of |f(z)| as z runs through K.

Thus there is a compact subset K of G and a positive number ε such that |ϕ(f)| ≤ 1 whenever ||f||_{K} < ε, i.e.,

|ϕ(f)| ≤ \frac{1}{ε}||f||_{K} (all f ∈ H(G)) (2)

Now we may, without loss of generality, suppose that K has nonempty interior (since the above inequality just gets better if K is larger), so that H(G) can be injectively regarded, via restriction to K, as a (nonclosed) subspace of C(K). Inequality (2) asserts that, for the C(K)-norm, ϕ is continuous on this subspace, so by the Hahn-Banach Theorem it extends to a continuous linear functional \overline{ϕ} on C(K) itself. The Riesz Representation Theorem provides a complex Borel measure μ on K such that

\overline{ϕ}(f) = ∫_{K} f dμ (f ∈ C(K)).

In particular,

ϕ(e_{λ}) = ∫_{K} e^{λz} dμ(z)

for each λ ∈ C. Now the last equation shows that the function Φ defined on the complex plane by

Φ(λ) =^{def} ϕ(e_{λ}) (λ ∈ C)

is entire, and—upon differentiating repeatedly under the integral sign—that for each non-negative integer n:

(3) Φ^{(n)}(0) = ∫_{K} λ^{n} dμ = ϕ(the monomial function z → z^{n}).

But our hypothesis that the functional ϕ annihilates e_{λ} for each λ ∈ A says that the entire function Φ vanishes at each point of A. Since A has a finite limit point, Φ must vanish on the whole plane, hence the same is true of each of its derivatives. Therefore, by equation (3) the functional ϕ annihilates all monomials, hence all holomorphic polynomials. By Runge’s Theorem and the simple connectivity of G, these polynomials are dense in H(G), hence ϕ is the zero-functional. This completes the proof that E(A) is dense in H(G). ///

With these useful criteria for hypercyclicity, non-hypercyclicity, and density in hand, we can proceed immediately to the central result of this paper:

2.8 Theorem (Chaotic characterization of simple connectivity). Suppose G is a plane domain, and P a nonconstant holomorphic polynomial. Then the following statements are equivalent:

(a) P (D) is hypercyclic on H(G).

(b) P (D) is chaotic on H(G).

(c) G is simply connected.

Proof. (c) ⇒ (b): Suppose G is simply connected. We want to show that P (D) is chaotic. This can be inferred, via Runge’s Theorem and the continuity of the inclusion map H(C) → H(G), from the corresponding result for the case G = C, which was proved in [9, Theorem 6.2]. However, in order to illustrate the ideas involved, I will redo the argument of [9] in the context of H(G).

First we have to show that P (D) is hypercyclic. With an eye towards finding the dense subsets X_{0} and Y_{0} required by the Hypercyclicity Criterion §2.1, we begin by splitting the plane into two sets. Let U denote the open unit disc, and \overline{U} the closed unit disc, and set

A = P^{−1}(U) and B = P^{−1}(C\\overline{U}).

Because P is a polynomial, it has zeros, hence A is not empty. Since P is nonconstant, it is unbounded, hence B is nonempty, and since P is continuous, both A and B are open subsets of the plane. So both A and B have finite limit points, hence the Density Theorem insures that both

X_{0} =^{def} E(A) and Y_{0} =^{def} E(B)

are dense in H(G).

Now λ ∈ A means that |P (λ)| < 1, so for n a positive integer,

P (D)^{n}e_{λ} = P (λ)^{n}e_{λ} → 0 (n → ∞),

hence P (D)^{n} → 0 pointwise on X_{0}.

As for the mapping S, define it on the exponential basis for Y_{0} by:

S(e_{λ}) = \frac{1}{P (λ)}e_{λ} (λ ∈ B),

and extend linearly to Y_{0}. Since T S is the identity on this exponential basis, it is the identity map on Y_{0}. Now λ ∈ B means that |P (λ)| > 1, hence

S^{n}e_{λ} = \frac{1}{P (λ)^{n}} e_{λ} → 0,

so S^{n} → 0 pointwise on Y_{0}. Thus all the conditions needed for the Hypercyclicity Criterion are satisfied, and therefore P (D) is hypercyclic.

To prove that P (D) is chaotic we need to find a dense set of periodic points. Since the range of P is an unbounded open set that contains the origin, its intersection with the unit circle contains an arc. Let R denote the collection of roots of unity that lie in this arc—they form a dense subset of that arc. Then C =^{def} P^{−1}(R) is a subset of the plane that has a limit point (it has infinitely many elements, and it is bounded), so by the Density Theorem, the subspace E(C) is dense in H(G). Now if a ∈ A then P (a) is a root of unity, i.e., P (a)^{n} = 1 for some positive integer n. It follows that P (D)e_{a} = P (a)^{n}e_{a} = e_{a}, so e_{a} is a periodic point of P (D) with period n. Now any linear combination of such periodic vectors is again one (with period equal to the least common multiple of the periods of the component eigenvectors), so every vector in E(C) is a periodic point of P (D). Thus E(C) is the desired dense subset of periodic points, hence P (D) is chaotic.

(b) ⇒ (a): Trivial from the definition of “chaotic”.

(a) ⇒ (c): We will prove the contrapositive statement. Suppose G is not simply connected. We will show that P (D)^{∗} has an eigenvalue, and therefore by the “nonhypercyclicity criterion” of §2.4 we will know that P (D) is not hypercyclic.

Since G is not simply connected there is a smooth Jordan curve γ in G that surrounds some point a ∉ G. This curve induces a continuous linear functional ϕ on H(G) by means of the equation:

ϕ(f) =^{def} ∫_{γ} f(z) dz (f ∈ H(G)).

I claim that ϕ is an eigenvector of P (D)^{∗}. Note that ϕ ≠ 0, since it takes the value 2πi at the function (z − a)^{−1} (this is the crucial use of non-simple connectivity). On the other hand, the null space of ϕ contains every derivative: ϕ(Df) = 0 for every f ∈ H(G). Thus, upon writing P (z) = P (0) + zQ(z), where Q is also a holomorphic polynomial, we see that for each f ∈ H(G):

(P (D)^{∗}ϕ)(f) =^{def} ϕ(P (D)f)
= ϕ(P (0)f + DQ(D)f)
= ϕ(P (0)f) + ϕ(a derivative)
= P (0)ϕ(f),

that is, P (D)^{∗}ϕ = P (0)ϕ. Thus ϕ is an eigenvector of P (D)^{∗} (for the eigenvalue P (0)), and so P (D) is not hypercyclic. ///

2.9 Infinite order differential operators. Suppose F (z) = ∑_{k=0}^{∞} a_{k}z^{k} is a nonconstant entire function for which the series F (D) = ∑_{k} a_{k}D^{k} converges pointwise on H(G). By the Closed Graph Theorem the linear transformation F (D) so defined is continuous on H(G), and the proof above works as well for F (D) as it did for P (D). It turns out that the desired convergence always happens if F is an entire function of “exponential type zero” (we will discuss this matter in Section 4), so the above characterization of simple connectivity can be improved by replacing the polynomial P by any nonconstant entire function of exponential type zero.

3 Beyond P (D)

Does Theorem 2.8 remain true if the constant coefficient differential operators P (D) are replaced by arbitrary nonscalar operators that commute with D? The answer is “yes and no.” I’ll begin with the “no” part: There is a non-simply connected domain G for which H(G) supports a chaotic operator that commutes with D. More precisely:

3.1 Theorem. Let G be the complex plane with the integers removed. Then the operator T of “translation by one” is chaotic on H(G).

Proof. Recall that T is the operator defined by:

T f(z) = f(z + 1) (f ∈ H(G) and z ∈ C).

The first order of business is to show that T is hypercyclic, i.e. to find the subspaces X_{0} and Y_{0} and the inverting operator S required by the Hypercyclicity Criterion (§2.1).

Let

F_{+} = {e_{λ} : Re λ > 0} and F_{−} = {e_{λ} : Re λ < 0},

let

R = \{\frac{1}{(z − n)^{α}} : n ∈ Z and α ∈ N\} , (4)

and set

X_{0} = span {R ∪ F_{−}} and Y_{0} = span {R ∪ F_{+}}.

By the Density Theorem (§2.7), each of the sets F_{+} and F_{−} spans a dense subspace of H(C), so the closures in H(G) of both X_{0} and Y_{0} contain the monomials {z^{n} : n ≥ 0}. Thus the linear span of the union of R with these monomials lies in the closures of both X_{0} and Y_{0}, so by Runge’s theorem, both X_{0} and Y_{0} are dense in H(G).

Clearly T^{n} → 0 pointwise on R. Since T e_{λ} = e^{λ}e_{λ}, the same is true on F_{−}, and therefore on all of X_{0}. Let S to be the operator of “translation by −1”, i.e. S = T^{−1}. Then arguing as above, S → 0 pointwise on Y_{0}. Thus the requirements of the Hypercyclicity criterion are satisfied, so T is hypercyclic on H(G).

To show that T is chaotic requires a bit more work, since the obvious periodic points e_{λ} for λ = 2πiq, where q is a (real) rational number, no longer span a dense subspace of H(G). Fortunately there is another supply of eigenvectors available: I claim that for each point λ of the unit circle and each positive integer α, the series

∑_{n∈Z} \frac{λ^{n}}{(z − n)^{α}} (5)

converges in H(G) to an eigenvector f_{λ,α} of T corresponding to the eigenvalue λ. Here “convergence” means that the sequence of symmetric partial sums

S_{N}(z) =^{def} ∑_{n=−N}^{N} \frac{λ^{n}}{(z − n)^{α}} (6)

converges in H(G).

If α ≥ 2 then the desired convergence is elementary. For if K is a compact subset of G then, uniformly over K,

|z − n|^{−α} = O(|n|^{−α}) as |n| → ∞,

hence the “absolute series”

∑_{n∈Z} |\frac{1}{(z − n)^{α}} |

converges uniformly on K, because α > 1.

However we will also need the case α = 1, and this requires more work. If λ = 1 there is no difficulty: just group corresponding terms of positive and negative index to obtain:

S_{N}(z) = \frac{1}{z} + ∑_{n=1}^{N} \frac{2z}{z^{2} − n^{2}} ,

from which the desired convergence is evident.

If λ ≠ 1 then Dedekind’s Test [14, Theorem 4, page 137] saves the day. For numerical series this states that ∑ a_{n}b_{n} converges whenever the partial sums of ∑ a_{n} form a bounded sequence, and the sequence (b_{n}) converges to zero and has bounded variation, i.e. ∑ |b_{n} − b_{n+1}| < ∞. The result is proved by summation-by-parts, and the proof works equally well for two-sided series. Furthermore, the proof shows that if (b_{n}) is a sequence of functions which, on some set S, converges uniformly to zero and has uniformly bounded variation, then the series ∑ a_{n}b_{n} converges uniformly on S.

Apply this last observation with a_{n} = λ^{n} and b_{n}(z) = (z − n)^{−1}. Because |λ| = 1 and λ ≠ 1, the sequence of numerical sums ∑_{−N}^{N} λ^{n} is bounded, so we need only show that the “multiplying sequence” ((z − n)^{−1} : n ∈ Z), which clearly converges to zero in H(G), also has variation bounded uniformly on each compact subset of G. This, in turn, follows from the fact that the magnitude of the difference between terms of index n and n + 1 is |(z − n)(z − n − 1)|^{−1}, which, on each compact subset of G, is uniformly O(|n|^{−2}) as |n| → ∞ (the “big-oh constant” depending, of course, on the particular compact set).

Summarizing: for each α ∈ N and each complex number λ of modulus one, the series (5) converges in H(G) to a function f_{λ,α}, which is an eigenvector of T with eigenvalue λ. Let F_{Σ} denote the collection of all these eigenvectors where λ is a root of unity, so F_{Σ} is a set of periodic points of T. Let

F_{0} = {e_{λ} : λ is a root of unity},

another collection of periodic points for T. Thus the linear span of F_{0} ∪ F_{Σ} also consists entirely of periodic points, and I claim that this subspace is dense in H(G). This will complete the proof that T is chaotic on H(G).

Recall that the Density Theorem insures that F_{0} spans H(C), so in particular its closed span picks up all the monomials z^{n} for n a non-negative integer. So by Runge’s Theorem it is enough to show that the closure of the span of F_{Σ} contains the set R defined by (4) above.

This is another job for the Hahn-Banach Theorem. Suppose ϕ ∈ H(G)^{∗} annihilates every function in F_{Σ}. By Hahn-Banach it is enough to prove that ϕ also annihilates every function in R. In plain English, we are assuming that ϕ(f_{λ,α}) = 0 for all |λ| = 1 and α ∈ N, and we want to prove that ϕ((z − n)^{−α}) = 0 for α ∈ N and n ∈ Z.

Fix the positive integer α, and for convenience of notation, let’s define the function r_{n} ∈ H(G) by r_{n}(z) = (z − n)^{−α}, and write

a_{n} = ϕ(r_{n}) (n ∈ Z)

for the quantities that we hope to prove are zero. From the continuity of ϕ and the convergence in H(G) of the series (5) to f_{λ,α} it follows that

∑_{n∈Z} a_{n}λ^{n} = ϕ(f_{λ,α}) = 0 (all |λ| = 1). (7)

The desired result now follows from Riemann’s Theorem which asserts that if a trigonometric series converges to zero at every point of the unit circle, then the coefficients must all be zero [21, Theorem IX.3.1, page 326]. But it’s not necessary to be so fancy: I claim that the left-hand side of (7) has square-summable coefficients, so the more familiar uniqueness theorem from the L^{2} theory of Fourier series will also do the job.

For this, note that—as previously observed—the continuity of ϕ means that there is a compact subset K of G and a positive constant C such that

|ϕ(f)| ≤ C||f||_{K} (f ∈ H(G)).

In particular, for each n ∈ Z:

|a_{n}| = |ϕ(r_{n})| ≤ C ||r_{n}||_{K} = C max_{z∈K} |\frac{1}{z − n}|^{α} ≤ \frac{C′}{(|n| + 1)^{α}} ,

where C′ is a finite positive number whose actual value depends on C and K (the finiteness of C′ arises from the fact that K, being compact, lies a positive distance from Z). Since α ≥ 1 this proves that ∑_{−∞}^{∞} |a_{n}|^{2} < ∞, which puts the right-hand side of (7) into L^{2} of the unit circle. Thus by (7) all the coefficients a_{n} must be zero. This completes the proof that T is chaotic. ///

It is no accident that the domain G of the last theorem is infinitely connected. The next result—the “yes” answer to the question that led off this section—shows that for finitely connected domains any nonscalar operator in the commutant of D can be used to characterize simple connectivity.

3.2 Theorem. Suppose G is a finitely connected domain and L a non-scalar operator on H(G) that commutes with D. Then the following are equivalent:

(a) L is hypercyclic on H(G).

(b) L is chaotic on H(G).

(c) G is simply connected.

Proof. The goal is to show that L is chaotic when G is simply connected, and that it fails to be hypercyclic when G is not simply connected.

So first suppose that G is simply connected. The idea behind proving L chaotic is to represent L in terms of D and try to adjust the proof of Theorem 2.8 so that it works in the new situation. Since LD = DL on H(G), for each λ ∈ C the equation De_{λ} = λe_{λ} implies

DLe_{λ} = LDe_{λ} = L(λe_{λ}) = λLe_{λ},

i.e., the function y = Le_{λ} satisfies the complex differential equation y′ = λy on C. Thus for each λ ∈ C there is a complex number F (λ) such that

Le_{λ} = F (λ)e_{λ}. (8)

I claim that F is an entire function. Indeed, fix a point z_{0} ∈ G, and note that for each fixed λ ∈ C the right-hand side of the series representation

e_{λ}(z) =^{def} e^{λz} = ∑_{n=0}^{∞} \frac{z^{n}}{n!} λ^{n}

converges in H(G) to e_{λ}. By the definition of F and the continuity of both the operator L and the linear functional of evaluation at z_{0},

F (λ)e^{λz_{0}} = (Le_{λ})(z_{0}) = ∑_{n=0}^{∞} \frac{(L(z^{n}))(z_{0})}{n!}λ^{n} ,

where on the right you see a convergent numerical series. Thus F is represented by a power series that converges at each point of the plane, hence it is an entire function.

Since G is simply connected we know from the Density Theorem (§2.7) that the complex exponential functions span a dense subspace of H(G). Now in Theorem 2.8 the proof that P (D) is chaotic depended only on the action of that operator on the linear span of the complex exponentials. Once we know (8) and that F is entire, the same proof works word for word, with F in place of P, to prove that L is chaotic.

Now suppose that G is not simply connected. We are supposed to show that L is not hypercyclic. Since G is finitely connected there is a positive integer N such that the complement of G has exactly N components K_{1}, …, K_{N}, where K_{N} is the unbounded component. Let γ_{1}, …, γ_{N} be disjoint rectifiable Jordan curves in G, with γ_{j} surrounding K_{j} but no other K_{i}, and γ_{N} surrounding all the other γ_{j}’s. In other words, let γ_{1}, …, γ_{N} be a homology basis for G in the sense of Ahlfors’s book [1, page 146] (for Rudin’s version of this see [18, Theorem 13.5, page 268]).

Associated with each curve γ_{j} there is a linear functional ϕ_{j} on H(G) defined by:

ϕ_{j}(f) = ∫_{γ_{j}} f(z) dz (f ∈ H(G)).

As in the proof that (a) ⇒ (c) of Theorem 2.8, the functional ϕ_{j} annihilates every derivative in H(G). Moreover it follows from Cauchy’s theorem that every member of H(G) that is annihilated by each the functionals ϕ_{j} for 1 ≤ j ≤ N − 1 is a derivative (see [1, page 146]). Let ran D denote the range of the operator D, i.e. the subspace of H(G) consisting of derivatives. The previous remarks say that the annihilator of ran D, defined by

(ran D)^{⊥} =^{def} {ϕ ∈ H(G)^{∗} : ϕ(f) = 0 for every f ∈ ran D},

has {ϕ_{1}, …, ϕ_{N−1}} as a basis (the functionals are clearly linearly independent, and by Cauchy’s theorem ϕ_{N} is a linear combination of ϕ_{1} … ϕ_{N−1})), and so has dimension N − 1 > 0.

But L commutes on H(G) with the operator D, so L(ran D) ⊂ ran D. From this it follows easily (just as for Hilbert or Banach space operators) that L^{∗}((ran D)^{⊥}) ⊂ (ran D)^{⊥}. Thus L^{∗} has a nontrivial finite dimensional invariant subspace, so it has an eigenvalue, and therefore L satisfies the “non-hypercyclicity criterion” (Theorem 2.4). /// 3.3 Open Question. I don’t know if, for every infinitely connected domain G, there is a chaotic (or even just a hypercyclic) operator on H(G) that commutes with differentiation.

4 Operators that commute with differentiation

Lurking behind the work of Sections 2 and 3 is the problem of finding concrete representations for operators that commute with differentiation. I take this up in more detail now. I’m sure most, if not all, of the results of this section are already known, so what follows should be treated as mostly expository.

4.1 Notation. For a plane domain G, not necessarily simply connected, Com_{D}(G) will denote the collection of operators that commute with D on H(G)—the commutant of D on H(G).

4.2 The characteristic function. The fundamental tool for studying the commutant already resides in the proof of Theorem 3.2, wherein it developed that each L ∈ Com_{D}(G) gives rise to an entire function F defined by the equation

Le_{λ} = F (λ)e_{λ} (λ ∈ C), (9)

which, you will recall, comes from the fact that the function y = Le_{λ} satisfies the differential equation y′ = λy. In order to emphasize the connection between F and L, in this section we’ll write F = F_{L}, and refer to it as the “characteristic function” of L.

4.3 Entire functions of differentiation. Let us say that an entire function F (z) = ∑_{n=0}^{∞} a_{n}z^{n} G-operates on D if the operator series ∑_{n=0}^{∞} a_{n}D^{n} converges pointwise on H(G). In this case we will write

F (D) = ∑_{n=0}^{∞} a_{n}D^{n} (10)

When this happens the Closed Graph Theorem insures that F (D), which is clearly a linear mapping on H(G), is also continuous, and you can check easily that the characteristic function of F (D) is just F.

Two classes of entire functions play a pivotal role in what is to follow:

4.4 Definition. An entire function F is said to be of exponential type if there exist positive constants A and B such that

|F (z)| ≤ A e^{B|z|} (all z ∈ C). (11)

If, in addition, given any positive B there is a constant A such that (11) holds, then F is said to be of exponential type zero.

The next two results show the importance of the notion of “exponential type.”

4.5 Lemma. If L ∈ Com_{D}(G), then its characteristic function F_{L} is of exponential type.

Proof. Recall the basic neighborhood of zero N(K, ε) that we encountered in the proof of Theorem 2.7. Since L is continuous, for every compact subset K of G and ε > 0, the set L^{−1}(N(K, ε)) is also a neighborhood of zero. Thus there exists another compact subset J, which we may assume contains K, and a positive number δ such that L(N(J, δ)) ⊂ N(K, ε). Take ε = 1 and set A = 1/δ. Then

||Lf||_{K} ≤ A||f||_{J} (all f ∈ H(G)). (12)

We may without loss of generality suppose that G contains the origin. Then inequality (12), with K taken to be the singleton containing the origin, asserts that

|F (λ)| = |F (λ))e_{λ}(0)| = |Le_{λ}(0)| ≤ A||e_{λ}||_{J} ≤ Ae^{B|λ|},

where B = max{|z| : z ∈ G}. ///

The issue now is whether or not every entire function of exponential type Goperates on D. The answer depends on the nature of G. First, the good news:

4.6 Theorem. Every entire function of exponential type C-operates on D.

Proof (cf. [9, §5.3], for example). Fix F (z) = ∑_{n=0}^{∞} a_{n}z^{n}, an entire function of exponential type. Our goal is to show that the series ∑_{n=0}^{∞} a_{n}D^{n} converges pointwise on H(G). The restriction on F is that there exist positive constants A and B such that F obeys (11) for all z ∈ C. An elementary orthogonality argument shows that for every r > 0:

a_{n}r^{n} = \frac{1}{2π} ∫_{0}^{2π} F (re^{iθ})e^{−inθ} dθ,

which, in view of the growth restriction on F, implies that

|a_{n}| ≤ Ae^{Br}r^{−n} = AB^{n}e^{Br}(Br)^{−n} (all r > 0).

In particular, the choice r = n/B (which, incidentally, minimizes the right-hand side of the last equation), yields the crucial inequality:

|a_{n}| ≤ A (\frac{Be}{n})^{n} (n = 1, 2, …). (13)

The next order of business is to obtain an estimate on derivatives of entire functions. Suppose f ∈ H(C), fix a compact subset K of C, and set ρ = max{|z| : z ∈ K}. Let r = ρ + 2Be, so, in particular, K lies in the disc {|z| < r}. Finally, set M = max{|f(ζ)| : ζ = r}.

Then the Cauchy formula for derivatives provides the following estimate, valid for every non-negative integer n and every z ∈ K:

|D^{n}f(z)| = \frac{n!}{2π} |∫_{|ζ|=r} \frac{f(ζ)}{(ζ − z)^{n+1}} dζ|

≤ \frac{n! r M}{(r − ρ)^{n+1}}

= C \frac{n!}{(2Be)^{n}} ,

where C = Mr/(r − ρ), a constant independent of n ∈ N and z ∈ K.

This inequality and (13) yield, for each z ∈ K and n ∈ N:

|a_{n}||D^{n}f(z)| ≤ AC (\frac{Be}{n})^{n} \frac{n!}{(2Be)^{n}} = AC \frac{n!}{n^{n}} \frac{AC}{2^{n}} ≤ \frac{AC}{2^{n}} ,

which shows that the series ∑_{n=0}^{∞} a_{n}D^{n}f(z) converges uniformly on K. Since K is an arbitrary compact subset of G, the series ∑_{n=0}^{∞} a_{n}D^{n}f does indeed converge in H(C), as we wished to show. ///

4.7 Corollary. Com_{D}(C) = {F (D) : F is entire, of exponential type}.

Proof. By the previous theorem, if F is entire of exponential type, then F C-operates on D, hence the resulting operator clearly F (D) commutes with D. Conversely if L is an operator that commutes with D, then its characteristic function F_{L} is of exponential type by Lemma 4.5, so the operator F_{L}(D) exists. It is easy to check that for each λ ∈ C, F_{L}(D)e_{λ} = F (λ)e_{λ}, and this, along with (9), guarantees that L = F_{L}(D) on E(C), the linear span of the exponential functions e_{λ} for λ ∈ C. Since E(C) is dense in H(C) (Theorem 2.7), it follows that L = F_{L}(D) on H(C), as desired. ///

In particular, the translation operator T_{a} defined in Section 1 by equation (1) can be represented as an entire function of D. I leave to you to prove the following:

4.8 Corollary. On H(C), T_{a} = e^{aD} for each a ∈ C.

Here’s an example showing that Theorem 4.6 doesn’t hold if C is replaced by an arbitrary simply connected domain G.

4.9 Example. Let G be the domain formed by gluing to the top edge of a horizontal strip squares of unit length, each one unit apart. So you get a “sawtooth” domain that is invariant under the map z → z + 1, and therefore the translation operator T_{1} acts (continuously) on H(G). I claim, however, that the characteristic function e^{z} of T_{1} does not G-operate D, i.e. that the series ∑_{n=0}^{∞} D^{n}/n! does not converge pointwise on H(G).

Indeed, suppose for the sake of contradiction that the series does converge. Then its individual terms converge to zero in H(G), from which it follows easily that the series ∑_{n=0}^{∞} D^{n}/n!2^{n} converges to some operator L on H(G). But on the linear span of the exponential functions, L is easily seen to coincide with T_{1/2}. Since G is simply connected, this linear span is dense in G (Theorem 2.7), hence L is the restriction to H(G) of T_{1/2}, which does not take H(G) into itself. This contradiction shows that the original series could not have converged. ///

The notion of exponential type zero emerges when we try to determine which entire functions act on D for G ≠ C.

4.10 Theorem. Suppose G is any bounded domain, not necessarily simply connected. Then every function of exponential type zero G-operates on D.

Proof. Suppose F is an entire function of exponential type zero. To show it Goperates on D we need to redo the Cauchy integral estimate that occurred in the proof of Theorem 4.6. For this, fix a function f ∈ H(G), a compact subset K of G, and a cycle γ in G\K that “surrounds” K in the sense that the Cauchy integral formula holds over γ for every f ∈ H(G) and every z ∈ K (see [18, Theorem 13.5, page 268]). Let d = dist (γ, K) and let ℓ denote the length of γ. Then the Cauchy formula for the n-th derivative supplies this estimate, valid for every z ∈ K and every non-negative integer n;

|D^{n}f(z)| = \frac{n!}{2π} |∫_{γ} \frac{f(ζ)}{(ζ − z)^{n+1}} dζ|

≤ (\frac{||f||_{γ}}{2πℓd} ) \frac{n!}{d^{n}} ,

so that

||D^{n}f||_{K} ≤ C||f||_{γ} \frac{n!}{d^{n}} (n = 0, 1, 2, …) (14)

where C = (2πℓd)^{−1} is a constant independent of n and f.

Now since F is of exponential type zero, there is a constant A > 0 such that F satisfies estimate (11) with B = d/2e. From this, the coefficient estimate (13) yields

|a_{n}| ≤ A (\frac{d}{2n})^{n} (n = 1, 2, …).

Therefore

|a_{n}| ||D^{n}f||_{K} ≤ AC ||f||_{γ} \frac{n!}{d^{n}} (\frac{d}{2n})^{n}

= AC ||f||_{γ} (\frac{1}{2})^{n} \frac{n!}{n^{n}}

≤ AC ||f||_{γ} (\frac{1}{2})^{n} .

This shows that the series ∑_{n} a_{n}D^{n}f converges uniformly on K, and since K is an arbitrary compact subset of G and f an arbitrary function in H(G), it follows that the operator series ∑_{n} a_{n}D^{n} converges pointwise on H(G) (note also that our estimate explicitly reveals the continuity of the operator represented by the sum). Thus the entire function, F (z) = ∑_{n} a_{n}z^{n}, G-operates on D. ///

Now that Theorem 4.10 is proved, the following consequence, noted in §2.9, is official:

4.11 Corollary. Theorem 2.8 remains true if the phrase “P is a nonconstant holomorphic polynomial” is replaced by “P is a nonconstant entire function of exponential type zero.”

The proof of Theorem 4.10 actually provides more than was promised. Suppose F is entire, of exponential type zero, and set L = F (D). Suppose the compact subset K of G is given. For any 0 < ε < dist (K, ∂G) let K_{ε} denote the “ε-envelope” of K: those points of G that lie within ε of some point of K. The restriction on ε insures that K_{ε} lies entirely in G. Now the cycle γ in the proof of Theorem 4.10 can be chosen to lie in K_{ε}, from which the original argument then yields a constant A = A(K, ε) such that ||Lf||_{K} ≤ A ||f||_{K_{ε}} (all f ∈ H(G)). (15)

By contrast, recall that L is continuous precisely when, given any compact K ⊂ G, there is a constant A and a compact subset J of G (which may, without loss of generality, be assumed to contain K), such that ||Lf||_{K} ≤ A ||f||_{J} for each f ∈ H(G). Thus inequality (15) expresses a property of L that is stronger than continuity, and which I’d like to formalize in the following definition.

4.12 Definition. A linear transformation L on H(G) is supercontinuous if: for every 0 < ε < dist (K, ∂G), and every compact K ⊂ G, there is a constant A = A(K, ε) > 0 such that inequality (15) holds.

Thus the the proof of Theorem 4.10 actually provides this:

For any plane domain G, every entire function F of exponential type zero G-operates on D, and the resulting operator F (D) is supercontinuous on H(G).

For an example of an operator that is not supercontinuous, consider T_{a}, the operator of translation by a nonzero complex number a, acting on H(C). It is easy to check that T_{a} is not supercontinuous on H(C). Note that by Theorem 4.6 and Corollary 4.8, the exponential function e_{a} C-operates on H(C), and T_{a} = e_{a}(D). This motivates the next result, which shows that, for simply connected domains, supercontinuity for an operator in the commutant of D is equivalent to exponential type zero for its characteristic function. The result also shows that, for such operators, supercontinuity is equivalent to two apparently weaker properties that involve the linear functionals of evaluation at points of G.

4.13 Definition. For z ∈ G let ϕ_{z} denote the “evaluation functional” defined on H(G) by:

ϕ_{z}(f) = f(z) (f ∈ H(G)).

It’s an easy exercise to show that each of these evaluation functionals is supercontinuous on H(G).

4.14 Theorem. Suppose G is simply connected and L ∈ Com_{D}(G). Then following statements about L are equivalent:

(a) The characteristic function F of L is of exponential type zero (so L = F (D)).

(b) L is supercontinuous.

(c) L^{∗}ϕ_{z} is supercontinuous on H(G) for every z ∈ G.

(d) L^{∗}ϕ_{z} is supercontinuous on H(G) for some z ∈ G.

Proof. The implication (a) ⇒ (b) was noted above.

(b) ⇒ (c): Since L ∗ ϕ_{z} = ϕ_{z} ∘ L, this follows from the fact that a composition of supercontinuous functions is supercontinuous. In the special case being considered here, the supercontinuity asserted for L^{∗}ϕ_{z} translates into the assertion that for every 0 < ε < dist (z, ∂G) there exists a constant A > 0 such that

|Lf(z)| ≤ A ||f||_{B(ε,z)} (f ∈ H(G)), (16)

where B(ε, z) is the closed ball of radius ε centered at z. But (16) is just what you get by taking K to be the singleton containing z in the definition of supercontinuity for L.

(c) ⇒ (d) is trivial.

(d) ⇒ (a): The hypothesis is that L^{∗}ϕ_{z} = ϕ_{z} ∘ L is supercontinuous at a point z ∈ G, and the goal is to prove that the characteristic function F of L is of exponential type zero. Recall that “F is the characteristic function of L” means that Le_{λ} = F (λ)e_{λ} for each λ ∈ C. Fix 0 < ε < dist (z, ∂G) and λ ∈ C. Then the hypothesis on L^{∗}ϕ_{z} provides a positive constant A = A(ε, z) > 0 such that:

|F (λ)| exp(Re (λz)) = |F (λ)e_{λ}(z)|
= |(Le_{λ})(z)|
= |ϕ_{z}(Le_{λ})|
=^{def} |L^{∗}(ϕ_{z})(e_{λ})|
≤ A ||e_{λ}||_{B(z,ε)}
= A exp ( max_{|ζ−z|≤ε} Re (λζ)) .

Thus

|F (λ)| ≤ A exp ( max_{|ζ−z|≤ε} Re [λ(ζ − z)]) ≤ A e^{ε|λ|}

Since λ is any point of the complex plane, and the constant A does not depend on λ, this proves that F is of exponential type zero. ///

The next result uses the idea of the previous proof to show that supercontinuity for L and exponential type zero for F_{L} are equivalent for bounded simply connected domains.

4.15 Theorem. If G is a bounded, simply connected domain, and F an entire function that G-operates on D, then F is of exponential type zero.

Proof. Choose an increasing sequence {G_{n}} of open subsets of G whose union is G, and such that the closure of G_{n} (compact because G is bounded) lies in G_{n+1}. For f ∈ H(G) let ||f||_{n} = sup{|f(z)| : z ∈ G_{n}}.

Let L = F (D), so that L is continuous on H(G) and Le_{λ} = F (λ)e_{λ}. The continuity of L implies that for each n there is an index m ≥ n and a constant A_{n} > 0 such that

||Lf||_{n} ≤ A_{n}||f||_{m} (all f ∈ H(G)). (17)

Fix a complex number ω of modulus one, and a positive real number r. Then—as in the previous proof—for every z ∈ G_{n},

|F (rω)| exp(r Re (ωz)) = |F (rω)e^{rωz}|
= |L(e_{rω}(z)|
≤ ||L(e_{rω})||_{n}

≤ A_{n} ||e_{rω}||_{m}
= A_{n} sup_{z∈G_{m}} exp(Re (rωz)).

Now let p(ω) = sup_{z∈G} Re (ωz), and write p_{n}(ω) for the corresponding supremum over G_{n}. Then the last chain of inequalities yields:

|F (rω)| exp(rRe (ωz)) ≤ A_{n} exp(r p_{m}(ω)) ≤ A_{n} exp(r p(ω)),

from which it follows, upon taking the supremum of the left-hand side over z ∈ G_{n}, that

|F (rω)| exp(r p_{n}(ω)) ≤ A_{n} exp(r p(ω)),

Summarizing: for each index n, each r > 0 and each ω in the unit circle:

|F (rω)| ≤ A_{n} exp(r[p(ω) − p_{n}(ω)]). (18)

Now the functions p_{n} and p are continuous on the unit circle—it is an elementary exercise to show that for each pair of points ω_{1}, ω_{2} on the unit circle,

|p(ω_{1}) − p(ω_{2})| ≤ (max_{z∈G} |z|) |ω_{1} − ω_{2}|

(I thank Paul Bourdon for pointing this out to me). Clearly the sequence of functions {p_{n}} increases pointwise to p, so by Dini’s theorem, p_{n} → p uniformly on the unit circle.

Suppose ε > 0 is given. Use the uniform convergence mentioned above to choose n so that p(ω) − p_{n}(ω) < ε for each ω in the unit circle. Then (18) shows that for each r > 0 and ω in the unit circle,

|F (rω)| ≤ A_{n} e^{rε},

in other words: F is of exponential type zero. ///

4.16 Corollary. If G is a bounded, simply connected domain, then Com_{D}(G) consists of all operators F (D), where F is an entire function of exponential type zero.

Proof. The proof given above actually showed that if L commutes with D then the characteristic function F_{L} is of exponential type zero, so in particular, L = F_{L}(D). The rest follows as in the proof of Corollary 4.7. ///

4.17 Remark. The function p that occurred in the the proof of Theorem 4.15 occurs frequently in convex analysis and in the theory of entire functions: it is called the support function of the complex conjugate of G.

5 Simple connectivity and cyclicity

This final section investigates the possibility of replacing “hypercyclic” with “cyclic” in Theorem 2.8. In other words:

If G is a plane domain that is not simply connected, does P (D) fail to be cyclic on H(G) for every nonconstant holomorphic polynomial P?

Now if D itself fails to be cyclic, then the same will be true of P (D) for each polynomial P, so the discussion will focus on the differentiation operator itself. As you’ll see next, the methods of Section 2 quickly rule out domains of connectivity larger than two.

5.1 Proposition. If G has connectivity larger than two, then D is not cyclic on H(G).

Proof. The assumption on G is that its complement has (at least) two distinct bounded components K_{1} and K_{2}. Let γ_{1} be a simple, closed, rectifiable curve in G that surrounds K_{1} but not K_{2}, and similarly let γ_{2} surround K_{2} but not K_{1}. Then for j = 1, 2 define the linear functional ϕ_{j} on H(G) by:

ϕ_{j}(f) = ∫_{γ_{j}} f(z) dz (f ∈ H(G)).

As in the proof of Theorem 2.1, neither of these is the zero-functional, and both annihilate all derivatives. In other words, the range of the differentiation operator lies in the intersection of the null spaces of these two functionals. Now the functionals themselves are linearly independent (for example, if z_{1} ∈ K_{1} and f(z) = 1/(z − z_{1}), then ϕ_{1}(f) ≠ 0, but ϕ_{2}(f) = 0, so ϕ_{1} and ϕ_{2} cannot be constant multiplies of each other), so this intersection has codimension at least two. Thus the same is true of the closure of the range of D. But it is easy to see that the closure of the range of a cyclic operator can have codimension no more than one. Thus D is not cyclic on H(G). ///

5.2 Remark. We have previously observed (§2.4) that if T is an operator on a Fréchet space, and T^{∗} has an eigenvalue, then T is not hypercyclic. The argument above shows that if T^{∗} has an eigenvalue of multiplicity larger than 1, then T is not even cyclic.

The case of connectivity two is more difficult, and for this I do not have a complete result. However the following partial result, which requires that G have some additional symmetry about some point of its complement, provides evidence for noncyclicity in this case, too. It will become clear from the proof that the symmetry hypothesis given here can be somewhat relaxed.

5.3 Theorem. Suppose G is a doubly connected domain, and let K denote the bounded component of C\G. Suppose further that G contains an annulus that surrounds K. Then D is not cyclic on H(G).

Proof. Let Ω denote the annulus. By Runge’s Theorem, the restrictions of functions in H(G) to H(Ω) are dense in H(Ω), so H(G) can be regarded as a dense subspace of H(Ω). Thus if D were cyclic on H(G) then it would also be cyclic on H(Ω). So it is enough to prove that D is not cyclic on H(Ω).

The cyclicity or non-cyclicity of D is not altered by translation, so we may without loss of generality assume that Ω has its center at the origin, say Ω = {z ∈ C : r < |z| < R}. Let Ω_{i} = {|z| < R} and Ω_{o} = {|z| > r}.

By Laurent series, each analytic function h on Ω can be written uniquely as f + g where f is analytic in Ω_{i} and g is analytic in Ω_{o} and vanishes at ∞. In other words, H(G) decomposes into a direct sum of closed subspaces H_{i} and H_{o}, where H_{i} = H(Ω_{i}) and H_{o} consists of those functions in H(Ω_{o}) that vanish at ∞.

Now each subspace H_{i} and H_{o} is invariant under differentiation, so D, acting on H(G), decomposes into a direct sum D ⊕ D acting on H_{i} ⊕ H_{o}. Fix a pair (f, g) in H_{i} ⊕ H_{o}. We will show that (f, g) is not cyclic for D ⊕ D. If f is the zero-function this is trivial since then the orbit of (f, g) lies in {0} ⊕ H_{o}, so its span has no chance of being dense in H_{i} ⊕ H_{o}. So we may as well suppose that f is not the zero-function.

The rest of the argument uses an unpublished idea due to James Deddens: we will find a nontrivial continuous linear functional on H_{i} ⊕ H_{o} that annihilates the entire D ⊕ D orbit of (f, g), hence this orbit cannot have dense linear span in H_{i} ⊕ H_{o}.

Fix a circle γ = {|z| = ρ} where r < ρ < R, and for (F, G) ∈ H_{i} ⊕ H_{o} define

Λ(F, G) = ∫_{γ}[g(−z)F (z) + f(−z)G(z)] dz .

The first order of business is to show that Λ is not the zero-functional. Since f is not the zero-function, one of its derivatives, say the N-th one, does not vanish at the origin. Let G(z) = z^{−(N+1)}, a function in H_{o}. Then

Λ(0, G) = \frac{1}{n!} ∫_{γ} \frac{f(−z)}{z^{N+1}} dz = \frac{(−1)^{N}}{2πi} f^{(N)}(0) ≠ 0,

so Λ is not trivial.

Now fix a non-negative integer n. I claim that Λ(D^{n}f, D^{n}g) = 0, which will complete our proof. It follows upon integrating by parts n times that:

Λ(D^{n}f, D^{n}g) =^{def} ∫_{γ}[g(−z)D^{n}f(z) + f(−z)D^{n}g(z)] dz

= ∫_{γ}[f(z)D^{n}g(−z) + f(−z)D^{n}g(z)] dz

= ∫_{γ}[H(z) + H(−z)] dz ,

where H(z) = f(z)D^{n}g(−z) for z ∈ Ω. Parameterize γ, say by γ(t) = ρ exp(2πit) for 0 ≤ t ≤ 1, to obtain:

∫_{γ} H(−z) dz = ∫_{0}^{1} H(−γ(t))γ′(t) dt

= − ∫_{0}^{1} H(−γ(t))(−γ)′(t) dt
= − ∫_{γ} H(z) dz ,

where the last line follows from the fact that the parameterization t → −ρ exp(2πit) (with 0 ≤ t ≤ 1) also represents γ, with the same orientation. Thus Λ(D^{n}f, D^{n}g) = 0, as promised. ///

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Department of Mathematics Michigan State University East Lansing, MI 48824 USA

Email: shapiro@math.msu.edu