The Invariant Subspace Problem by Jonathan A. Noel a thesis submitted in partial fulfillment of the requirements for the degree of Bachelor of Science (Hons.) in the Department of Mathematics & Statistics c !Jonathan A. Noel 2011 We accept this thesis as conforming to the required standards: Robb Fry Dept. of Mathematics & Statistics Thesis Supervisor Examiner 2 Examiner 3 Dated April 23, 2011, Kamloops, British Columbia, Canada THOMPSON RIVERS UNIVERSITY DEPARTMENT OF MATHEMATICS & STATISTICS Permission is herewith granted to Thompson Rivers University to circulate and to have copied for non-commercial purposes, at its discretion, the above title upon request of individuals or institutions. -------------------------------Signature of Author the author reserves other publication rights, and neither the thesis nor extensive extracts from it may be printed or otherwise reproduced without the author’s written permission. the author attests that permission has been obtained for the use of any copy-righted material appearing in this thesis (other than brief excerpts requiring only proper acknowledgement in scholarly righting) and that all such use is clearly acknowledged. ii Abstract The notion of an invariant subspace is fundamental to the subject of operator theory. Given an operator T on a Banach space X, a closed subspace M of X is said to be a non-trivial invariant subspace for T if T (M ) ⊆ M and M #= {0}, X. A famous unsolved problem, called the “invariant subspace problem,” asks whether every bounded linear operator on a Hilbert space (more generally, a Banach space) admits a non-trivial invariant subspace. In this thesis, we discuss the greatest achievements in solving this problem for special classes of linear operators. We include several positive results for linear operators related to compact operators and normal operators, and negative results for certain linear operators on Banach spaces. Our goal is to build up the theory from the basics, and to prove the main results in a way that is accessible to a student who is relatively new to the world of functional analysis. iii Acknowledgements First and foremost, I would like to thank my supervisor Robb Fry for agreeing to work with a self-described graph theorist. I appreciate the time and effort taken to read and respond to several drafts of this thesis, guide me through difficult material, and help me fill many gaps in my knowledge of basic functional analysis. Despite facing some difficult circumstances, Robb has been an excellent mentor and teacher. I was first inspired to study mathematics by learning analysis from Robb; it is only fitting that I am finishing my degree in the same way. I consider myself very fortunate to have studied at Thompson Rivers University, and will always be grateful to the professors in the Department of Mathematics and Statistics who have helped me along the way. Special thanks go out to Rick Brewster for introducing me to the incredible world of mathematical research. I also thank my family and friends, for without them I would not be the person, or the mathematician, that I am today. The completion of this thesis signals a new era in my life, where I will have to leave the ones that I love in order to pursue my goals. It has become clearer than ever that the people closest to you are the most easily taken for granted. I am forever grateful to my parents and siblings for their constant love and support. iv Contents Abstract iii Acknowledgements iv 1 Introduction 1 2 Preliminaries 3 2.1 Eigenvectors and Finite-Dimensional Spaces . . . . . . . . . . . . . . . . 3 2.2 Cyclic Subspaces and Separability . . . . . . . . . . . . . . . . . . . . . . 6 2.2.1 8 Some Additional Facts About Cyclic Vectors . . . . . . . . . . . . 3 Positive Results 3.1 10 Spectral Theory and Compact Operators . . . . . . . . . . . . . . . . . . 10 3.1.1 Compact Operators . . . . . . . . . . . . . . . . . . . . . . . . . . 14 3.2 von Neumann’s Result and Some Extensions . . . . . . . . . . . . . . . . 17 3.3 Lomonosov’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 3.4 Operators Related to Normal Operators . . . . . . . . . . . . . . . . . . 23 4 Counterexamples on Banach Spaces 26 4.1 History and Controversy . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 4.2 A Counterexample on !1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 v 4.3 Constructing (ei )i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 4.4 First Properties of (ei )i . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 4.5 The Linear Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 4.5.1 A Truncated Version Of T . . . . . . . . . . . . . . . . . . . . . . 35 4.5.2 A Norm On The Set Of Polynomials . . . . . . . . . . . . . . . . 36 4.6 A Compactness Argument . . . . . . . . . . . . . . . . . . . . . . . . . . 37 4.7 Tweaking Our Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . 39 4.8 Final Arguments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 4.9 Every Unit Vector Is T -Cyclic . . . . . . . . . . . . . . . . . . . . . . . . 46 4.10 Sharpness of Lomonosov’s Theorem . . . . . . . . . . . . . . . . . . . . . 46 5 Conclusion 49 A Vector Spaces 52 B Norms and Inner Products 55 B.1 Some Norm Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 B.2 Banach Spaces and Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . 59 C Linear Operators 62 D Polynomials 65 E Properties of !1 68 vi Chapter 1 Introduction There are many fundamental problems in mathematics which remain unsolved. Often they can be stated in relatively simple terms, requiring little background knowledge, and yet somehow their solutions continue to elude humankind. The archetypal example of this is the famous ‘twin prime conjecture’ in number theory.1 One of these tantalizing open problems is the so-called ‘invariant subspace problem’ in functional analysis. Given a linear operator T on a vector space X, a subset S of X is said to be T -invariant if T (S) ⊆ S. A vector subspace M of X which is T -invariant is called an invariant subspace for T . All linear operators have invariant subspaces; For example {0} and X are obviously invariant under T and so these are referred to as trivial invariant subspaces. The problem, in a general form, is stated as follows: The Invariant Subspace Problem. If T is a bounded linear operator on a Banach space X, does it follow that T has a non-trivial closed invariant subspace? Note, for a non-zero vector x the linear span of {x, T x, T 2 x, . . . } is T -invariant, usually not equal to X, but may not be closed. Thus, the major difficulty of the invariant 1 The twin prime conjecture states that there are infinitely many prime numbers p such that p + 2 is also prime. 1 2 subspace problem comes from requiring that the T -invariant subspace be simultaneously non-trivial and closed. In Chapter 2 we show that the problem is solved easily in the case that X is either finite-dimensional or non-separable. As it turns out, it is also solved in the negative for certain Banach spaces. The reduction to infinite-dimensional separable Hilbert spaces, however, remains one of the most famous and elusive open problems in functional analysis. The Invariant Subspace Problem (as it stands today). If T is a bounded linear operator on an infinite-dimensional separable Hilbert space H, does it follow that T has a non-trivial closed invariant subspace? For the remainder of the thesis, let us simply say invariant subspace when referring to a closed invariant subspace. In Chapter 3 we present some of the most important achievements in proving that certain operators have non-trivial invariant subspaces. This includes theorems of von Neumann and Lomonosov on compact operators, and Brown’s results for operators related to normal operators. In the final chapter, we exhibit an example from C. J. Read of a bounded operator on the classical Banach space !1 having only the trivial invariant subspaces. In addition, we provide a brief discussion of the history and controversy surrounding the first known counterexamples on Banach spaces. For a list of standard definitions and notation, please refer to the appendices. Chapter 2 Preliminaries In this chapter, we give detailed solutions to the invariant subspace problem for Banach spaces which are either finite-dimensional (too small) or non-separable (too large). Although these reductions are quite straightforward, the solutions raise some important themes which shall be returned to throughout the thesis. 2.1 Eigenvectors and Finite-Dimensional Spaces As with most problems in functional analysis, the invariant subspace problem only remains unsolved for infinite-dimensional spaces. In this section, we provide a solution to the finite-dimensional case by making clever use of the Fundamental Theorem of Algebra. We use freely the fact that finite-dimensional subspaces of normed spaces are closed. The proof of this is not difficult, and follows easily from the fact that finite-dimensional normed spaces are complete. Note. In this section, X denotes a real or complex normed space of dimension n ≥ 0 and T is an arbitrary linear operator on X. The solution is built up from an elementary fact regarding n-dimensional spaces. 3 2.1. EIGENVECTORS AND FINITE-DIMENSIONAL SPACES 4 Remark 2.1.1. Given x ∈ X, the vectors of the set Sn (x, T ) = {x, T x, T 2 x, . . . , T n x} are linearly dependent. This is simply because the above set has cardinality n + 1, and therefore cannot be linearly independent. Thus, we can ensure the existence of scalars α0 , α1 , . . . , αn so that: α0 x + α1 T x + α2 T 2 x + · · · + αn T n x = 0. Let us define a polynomial p(t) = !n i i=0 αi t . Applying Corollaries D.8 and D.9, we can rewrite p as p(t) = rm (t)rm−1 (t) . . . r1 (t) for m ≤ n where each polynomial ri , 1 ≤ i ≤ m, has degree 1 or 2. We can assume degree 1 if scalars are taken to be complex. We use this to prove the following result. Proposition 2.1.2. Suppose that n ≥ 1. Then every operator T on X has an invariant subspace M of dimension 1 or 2. If X is complex, then M can be chosen to have dimension 1. Proof. Let x be an arbitrary vector of X\{0} and, as above, define a polynomial p(t) = rm (t)rm−1 (t) . . . r1 (t) so that p(T )x = 0. Let us choose j to be the minimum index so that rj (T )rj−1 (T ) . . . r1 (T )x = 0 and define u = rj−1 (T ) . . . r1 (T )x = 0 (if j = 1, simply let u = x). By minimality of j, we have that u #= 0 and rj (T )u = 0. Recall that rj has degree 1 or 2. First suppose that rj (t) = αt + β for α #= 0. Then, we have rj (T )u = (αT + βI)u = 0. In other words, T u = −α−1 βu. We define M = ({u}). The space M is easily seen to be T -invariant, and M is 1- 2.1. EIGENVECTORS AND FINITE-DIMENSIONAL SPACES 5 dimensional since u #= 0. Note that this is always possible when X is complex, since we may assume deg(rj ) = 1. On the other hand, we may have rj (t) = αt2 + βt + λ where α #= 0. In this case we obtain: T 2 u = −α−1 βT u − α−1 λu We simply choose M = ({u, T u}), which is seen to be T -invariant and has dimension either 1 or 2. The ideas in the above proof are closely related to the well-known notions of eigenvalues and eigenvectors. A scalar λ is called an eigenvalue for T if there exists a non-zero vector x such that T x = λx, or equivalently, (T − λI)x = 0; In this case, x is called an eigenvector for T corresponding to λ, or simply an associated eigenvector when T and λ are clear from context. In proving Proposition 2.1.2, we used the fact that eigenvalues are equivalent to 1-dimensional invariant subspaces. Indeed, if λ is an eigenvalue for T with associated eigenvector x, then ({x}) is a 1-dimensional T -invariant subspace. On the other hand, if M is 1-dimensional and T -invariant, then any non-zero vector x of M is an eigenvector for T since the vectors of {x, T x} ⊆ M are linearly dependent. Let us now characterize operators on finite-dimensional spaces with non-trivial invariant subspaces. Theorem 2.1.3. Let X be an n-dimensional Banach space and T : X → X a linear operator. Then T has a non-trivial invariant subspace if, and only if, either n ≥ 3 or n = 2 and T has an eigenvalue. Proof. First, if n = 0 or 1, then the only subspaces of X are {0} and X and so T cannot have a non-trivial invariant subspace. In the case that n = 2, the only nontrivial subspaces of X are 1-dimensional. As noted in the paragraph before the theorem, the existence of a 1-dimensional invariant subspace for T is equivalent to T having an 2.2. CYCLIC SUBSPACES AND SEPARABILITY 6 eigenvalue. Finally if n ≥ 3, then T has an invariant subspace M of dimension 1 or 2 by Proposition 2.1.2. Since the dimension of M differs from that of X and {0}, we must have that M is non-trivial. The result follows. Any linear operator on a 2-dimensional complex space satisfies the hypotheses of Theorem 2.1.3 by Proposition 2.1.2. Consider the following example showing that linear operators on 2-dimensional real spaces may not have eigenvalues. Example 2.1.4. Define an operator Tθ which rotates each vector in R2 by θ radians counter clockwise about the origin. The explicit definition of this operator on a vector x = (x1 , x2 ) ∈ R2 is given as follows: Tθ x = (x1 cos(θ) − x2 sin(θ), x1 sin(θ) + x2 cos(θ)) Provided that R2 is equipped with the Euclidean norm, the operator Tθ is norm preserving; That is, +Tθ x+ = +x+ for every vector x ∈ R2 (by the identity cos2 (θ) + sin2 (θ) = 1). It follows that the only possible eigenvalues for Tθ are λ = 1 or −1. In either case, the equation Tθ x = λx for a non-zero vector x implies that sin(θ) = 0. Therefore, Tθ has an eigenvalue if, and only if, θ is an integer multiple of π. 2.2 Cyclic Subspaces and Separability The solution for finite-dimensional spaces breaks down in infinitely many dimensions for an obvious reason: There may be (and often is) no n ≥ 0 such that the vectors of Sn (x, T ) are linearly dependent. In fact, the vectors of S(x, T ) = {x, T x, T 2 x, . . . } may even be linearly independent, as the next example shows. 2.2. CYCLIC SUBSPACES AND SEPARABILITY 7 Example 2.2.1. For 1 ≤ p ≤ ∞ consider the unilateral shift operator U+ on !p defined by U+ (x0 , x1 , . . . ) = (0, x0 , x1 , . . . ). If x is non-zero then it is easy to see that the vectors of S(x, T ) are linearly independent. This is because distinct vectors in S(x, T ) begin with a different number of zeros. However, U+ does have a non-trivial invariant subspace (in fact, many). For example, the set of sequences with zero in the first coordinate (the range of U+ ) is closed in !p and U+ -invariant. Note. Throughout this section, let X be a Banach space over a field F = R or C. We let T : X → X denote an arbitrary bounded linear operator on X. As it turns out, a natural object to study in the general case is the notion of a cyclic subspace. Given non-zero x ∈ X, we define the cyclic subspace generated by x, denoted W (x, T ), to be the smallest closed subspace of X containing S(x, T ). As an explicit formula, we can express it as W (x, T ) = [S(x, T )]. (2.1) For a non-zero vector x ∈ X, we say that x is a cyclic vector for T or that x is T -cyclic if W (x, T ) = X. Proposition 2.2.2. The operator T has only the trivial invariant subspaces if, and only if, every non-zero vector of X is a cyclic vector for T . Proof. If M is a non-trivial invariant subspace for T , then for every non-zero vector x ∈ M we have W (x, T ) ⊆ M and so x is not a cyclic vector for T . On the other hand, if x ∈ X is a non-zero non-cyclic vector for T , then W (x, T ) #= X. Also, since x #= 0 and x ∈ W (x, T ), we have W (x, T ) #= {0}. It follows that W (x, T ) is a non-trivial invariant subspace for T . The result follows. 8 2.2. CYCLIC SUBSPACES AND SEPARABILITY It is sometimes useful to define cyclic subspaces in terms of polynomial combinations of T . Since {p(T )x : p(t) ∈ F[t]} = ({T n x : n ≥ 0}) we have that W (x, T ) = {p(T )x : p(t) ∈ F[t]}. (2.2) Using this idea, it is immediate that cyclic subspaces are separable. Indeed, define a set Q by Q = Q if F = R and Q = {a + bi : a, b ∈ Q} if F = C. The set D of polynomials over Q is easily seen to be countable. The fact that {p(T )x : p ∈ D} is dense in W (x, T ) follows from density of Q in R. We omit this argument here. Theorem 2.2.3. If X is non-separable, then every bounded linear operator T : X → X has a non-trivial invariant subspace. Proof. Let T be a bounded linear operator on a non-separable Banach space X, and choose x ∈ X\{0}. Since W (x, T ) is separable and X is not, we have that W (x, T ) #= X. So, x is not T -cyclic. The result follows by Proposition 2.2.2. 2.2.1 Some Additional Facts About Cyclic Vectors The final results of this section highlight important properties of cyclic vectors that are used later in our discussion of Lomonosov’s Theorem (Section 3.3) and Read’s counterexample on !1 (Sections 4.2 - 4.9). Proposition 2.2.4. A vector x ∈ X is T -cyclic if, and only if, for every non-empty open set U of X there is some p(t) ∈ F[t] such that p(T )x ∈ U . Proof. First suppose that x is T -cyclic and let U be an open set of X containing a point y. Then, since y ∈ X = [{p(T )x : p(t) ∈ F[t]}] there must be some polynomial p such that p(T )x ∈ U . On the other hand, if x is not T -cyclic, then U = X − W (x, T ) is a non-empty open set such that p(T )x ∈ / U for any polynomial p. The result follows. 2.2. CYCLIC SUBSPACES AND SEPARABILITY 9 Proposition 2.2.5. Suppose that x0 is T -cyclic. If x is a vector of X such that x0 ∈ W (x, T ), then x is T -cyclic. Proof. Recall, W (x0 , T ) = X is the smallest invariant subspace of T containing x0 . Therefore, since W (x, T ) is T -invariant and contains x0 we must have X ⊆ W (x, T ), the reverse inclusion being trivial. Therefore x is T -cyclic. Proposition 2.2.6. An operator T on X has only the trivial invariant subspaces if, and only if, every unit vector is T -cyclic. Proof. The argument is simple. We have that T has only the trivial invariant subspaces if, and only if, every non-zero vector of X is T -cyclic by Proposition 2.2.2. However, a non-zero vector x is T -cyclic if, and only if, the unit vector +x+−1 x is T -cyclic by Proposition 2.2.5. The result follows. The main result of Chapter 4 applies the following corollary. Corollary 2.2.7. Let T be an operator on X with cyclic vector x0 . The operator T has only the trivial invariant subspaces if, and only if, for every unit vector x and ε > 0 there is a polynomial q so that +q(T )x − x0 + < ε. Chapter 3 Positive Results We begin to investigate the more ‘interesting’ case of the invariant subspace problem: infinite-dimensional separable Banach spaces. Several of the crucial techniques apply only to complex spaces, so they are our central focus. Note. In this chapter, X denotes an arbitrary Banach space and H a Hilbert space. Both spaces are assumed to be complex, infinite-dimensional and separable. This chapter samples some of the important breakthroughs in showing that certain classes of bounded operators do indeed possess non-trivial invariant subspaces. Before presenting these results, we must cover important background on spectral theory and become familiar with the class of compact operators. 3.1 Spectral Theory and Compact Operators Earlier in Section 2.1, we demonstrated the usefulness of eigenvalues in characterizing operators on finite-dimensional spaces with non-trivial invariant subspaces. Spectral theory extends the concept of eigenvalues to infinite-dimensional spaces in a natural way. 10 3.1. SPECTRAL THEORY AND COMPACT OPERATORS 11 Definition 3.1.1. Given a bounded linear operator T on X we define the spectrum of T , denoted by σ(T ), to be the set of scalars α ∈ C such that T − αI is not invertible (bijective). The point spectrum of T , denoted by σp (T ), is the set of all scalars λ ∈ C such that T − λI is not injective. On an n-dimensional space, it is well known that a linear operator T is injective if, and only if, it is surjective and therefore we have σ(T ) = σp (T ) in this case. This is not necessarily true for operators on infinite-dimensional spaces, although σp (T ) ⊆ σ(T ) clearly holds. Lemma 3.1.2. If T : X → X is a bounded linear operator, then σp (T ) is precisely the set of eigenvalues for T . Proof. Suppose that λ ∈ σp (T ). Since T −λI is not injective, there are distinct vectors x and y in X such that (T − λI)(x) = (T − λI)(y). This implies that T (x − y) = λ(x − y). Since x #= y we have x − y #= 0 and so x − y is an eigenvector for T corresponding to λ. Now, suppose that λ is an eigenvalue for T and let x #= 0 be an associated eigenvector. Then we have T x = λx, which implies (T − λI)x = 0. Since T − λI also maps the zero vector to 0, we have that T − λI is not injective. Therefore, λ ∈ σp (T ), as desired. We introduce the well-known concept of an eigenspace. While it may seem that this definition belongs more in Section 2.1, it was actually not necessary for our treatment of finite-dimensional spaces. Eigenspaces are needed briefly, however, for our proof of of Theorem 3.3.2 at the end of the chapter. Definition 3.1.3. If T : X → X is a linear operator with eigenvalue λ, the eigenspace corresponding to λ is defined to be the set Wλ = {x ∈ X : T x = λx}. It is clear that any eigenspace Wλ for T is T -invariant since T simply acts as a 3.1. SPECTRAL THEORY AND COMPACT OPERATORS 12 multiple of the identity on Wλ . It is easily shown that an eigenspace is also a closed subspace, provided that T is bounded. Lemma 3.1.4. If λ is an eigenvalue for a bounded linear operator T : X → X, then Wλ is a non-trivial closed invariant subspace for X. As one can imagine, operators which are closely related to one another tend to share invariant subspaces. Consider, for example, the following definition. Definition 3.1.5. Let T be a linear operator on X. (a) We say that a linear operator A on X commutes with T if AT = T A. (b) A subspace M of X is said to be a hyperinvariant subspace for T if A(M ) ⊆ M for every operator A which commutes with T . Proposition 3.1.6. Let T be an operator on X and suppose that λ is an eigenvalue for T with eigenspace Wλ . Then Wλ is a hyperinvariant subspace for T . Proof. Let A be an operator which commutes with T and fix any x ∈ Wλ . We have T Ax = AT x = Aλx = λAx. Thus, T Ax = λAx and so Ax ∈ Wλ as desired. Understanding the spectral properties of bounded operators is very important to many areas of operator theory. One of the crucial ideas is that the spectrum is bounded, and therefore contained within some closed disc in C. Definition 3.1.7. Suppose that T : X → X is a bounded operator such that σ(T ) is non-empty. The spectral radius of T , denoted by rσ (T ), is defined by rσ (T ) = sup{|α| : α ∈ σ(T )}.1 1 It can also be shown that σ(T ) is compact, and therefore the supremum in the definition of rσ can be replaced by a maximum, see [35, Section 3.3]. 3.1. SPECTRAL THEORY AND COMPACT OPERATORS 13 The following theorem gives us a very useful formula for calculating the spectral radius of a bounded operator. This result is quite well known, but the proof is quite involved and would distract from the main focus of this thesis; See for example [35, Section 3.3]. Theorem 3.1.8 (Gelfand Spectral Radius Formula). If T : X → X is bounded, then +T n +1/n → rσ (T ). Of special importance is the class of operators with spectral radius equal to zero, which are involved in a few of the results discussed later in the chapter. Definition 3.1.9. A bounded linear operator T : X → X such that +T n +1/n → 0 is said to be quasinilpotent.2 Example 3.1.10. We provide two examples of quasinilpotent operators on the Banach space !p for 1 ≤ p ≤ ∞. First, consider the linear operator T : !p → !p defined by T (x0 , x1 , . . . ) = (0, x0 , 0, x2 , 0, . . . ). Clearly, T 2 = 0 and therefore T is quasinilpotent (in fact, T is nilpotent). For our next example, let α = (αn )∞ i=1 be a sequence of positive real numbers such that αn < n−n for all n ≥ 1. We let Tα : !p → !p be the ‘weighted backwards shift’ operator Tα (x0 , x1 , . . . ) = (α1 x1 , α2 x2 , . . . ). It is easy to argue that for any n ≥ 2 and vector x satisfying +x+ ≤ 1, we have +Tαn x+ ≤ αn αn−1 . . . α1 < αn . Thus, +Tαn +1/n ≤ n−1 and so Tα is quasinilpotent. The next lemma highlights a simple property of quasinilpotent operators, which is used to prove a special case of Lomonosov’s Theorem in Section 3.3. Lemma 3.1.11. If T is a quasinilpotent operator on X, then for every scalar c we have +(cT )n + → 0. 2 Quasinilpotent operators generalize nilpotent operators. An operator T : X → X is nilpotent if T = 0 for some n ≥ 1. n 3.1. SPECTRAL THEORY AND COMPACT OPERATORS 14 Proof. Given n ≥ 1 let us define an = +(cT )n +, which is seen to equal |c|n +T n +. The sequence an must approach zero as n → ∞ (in fact, rather quickly) since the sequence 1/n an = |c|+T n +1/n approaches zero. 3.1.1 Compact Operators We give a brief but informative introduction to compact operators. With the exception of Section 3.4, compact operators are involved in all of the invariant subspace theorems in this chapter. The belief is that once the reader obtains a sufficient ‘feel’ for compact operators, these main results should seem somewhat natural and intuitive. Definition 3.1.12. Let Y be a Banach space. A linear operator K : X → Y is said to be compact if K(S) is relatively compact for every bounded subset S of X. The definition of compact operators as it appears above is due to Riesz [46]. The notion of a completely continuous operator, first studied by Hilbert [31], is equivalent to a compact operator on a separable Hilbert space. The theory of compact operators is very rich, especially on Hilbert spaces. This section merely samples some of the most well-known and useful results. Proposition 3.1.13. Every compact linear operator K : X → Y is bounded. Proof. Recall, +K+ = sup{+Kx+ : +x+ ≤ 1}. Since S = {x : +x+ ≤ 1} is bounded and K is compact, we have that K(S) is compact. Of course, compact sets must be bounded. Indeed, if F ⊆ X is unbounded, then the collection {Bn (0) : n ≥ 1} would be an open cover for F having no finite subcover and therefore F cannot be compact. Thus, K(S) ⊆ K(S) is also bounded. The result follows. As it turns out, simple examples of compact operators are not hard to come by. In fact, all bounded finite-rank operators are compact. This comes from the basic result 3.1. SPECTRAL THEORY AND COMPACT OPERATORS 15 from analysis: A subset of a finite-dimensional normed space is relatively compact if, and only if, it is bounded. Hence, the following result. Proposition 3.1.14. Every bounded finite-rank operator from a Banach space into a Banach space is compact. Definition 3.1.15. Let K(X, Y ) denote the set of all compact operators mapping X to Y . We simply write K(X) in the case that Y = X. The set K(X, Y ) is a closed subspace of B(X, Y ) as illustrated by the next proposition. Actually K(X) is a closed ideal in B(X). Recall, a subspace M of B(X) is called an ideal if T A, AT ∈ M for every T ∈ B(X) and A ∈ M .3 Proposition 3.1.16. We have the following: (a) If K1 , K2 ∈ K(X, Y ) are compact and α is a scalar, then K1 + K2 and αK1 are compact. (b) If (Kn )∞ n=0 ⊆ K(X, Y ) converges to a bounded operator K, then K is compact. (c) If T : X → Y and A : Y → Z are bounded and one of T or A is compact, then T A is compact. Using Propositions 3.1.14 and 3.1.16 (b), it follows that the norm limit of a monotone sequence of bounded finite-rank operators is compact. We use this fact to obtain our first example of a compact operator having infinite-dimensional range. Example 3.1.17. We consider an operator on !p for 1 ≤ p ≤ ∞. Let (αi )∞ i=0 be a sequence of positive real numbers such that αi → 0 and let D be the ‘diagonal operator’ on !p defined by D(x0 , x1 , . . . ) = (α0 x0 , α1 x1 , . . . ). Consider the sequence (Di )∞ i=0 of 3 Generally, subset I of a ring R is an ideal if I is closed under addition and for each i ∈ I and r ∈ R we have ir, ri ∈ I. 3.1. SPECTRAL THEORY AND COMPACT OPERATORS 16 bounded finite-rank operators such that Di (x0 , x1 , . . . ) = (α0 x0 , α1 x1 , . . . αi xi , 0, 0, . . . ). Each Di is compact by Proposition 3.1.14. Given a vector x such that +x+ ≤ 1 we have +Dx − Di x+ ≤ αi+1 . Therefore, Di → D and so D is compact. By a similar argument, the operator Tα from Example 3.1.10 is compact as well as being quasinilpotent. As one can imagine, it is very useful to represent compact operators as the norm limit of a sequence of bounded finite-rank operators. The space X is said to have the approximation property if for every Banach space Y the set of bounded finite-rank operators mapping Y to X is dense in K(Y, X). Many important spaces have the approximation property including Hilbert spaces and !p spaces, 1 ≤ p < ∞. A famous and longstanding open problem in functional analysis, called the ‘approximation problem,’ asked whether every Banach space has the approximation property. A counterexample was finally given in a 1973 paper by Per Enflo [18] on a separable Banach space.4 Enflo is also famous for discovering the first counterexample for the invariant subspace problem on a Banach space [20]. See Chapter 4 for more information. From the ideas presented so far, one should recognize that, in some sense, compact operators bridge the gap between bounded finite-rank operators and general bounded operators. Compact operators even have spectral properties that are very similar to operators on finite-dimensional spaces, as illustrated by the next theorem. Theorem 3.1.18 (Riesz [46]). If K is a compact operator on X, then we have σ(K) = σp (K) ∪ {0}. A consequence of Theorem 3.1.18 is that non-zero scalar operators are not compact. This result can also be proved directly using the well-known fact that the unit 4 Enflo’s counterexample also solved a problem of Stanis!law Mazur, posed in 1936. Mazur’s problem was written in the so-called “Scottish Book” of open problems kept by Polish mathematicians who frequented the Scottish Café in Lwów. He offered the reward of a live goose to anyone who could come up with a solution. More than thirty years later, Mazur was true to his word. After lecturing on his solution in Warsaw, Enflo was awarded with a live goose. 3.2. VON NEUMANN’S RESULT AND SOME EXTENSIONS 17 sphere SX is compact if, and only if, X is finite-dimensional. Corollary 3.1.19. If X is infinite-dimensional, then the only compact scalar operator on X is the zero operator. Proof. For some scalar λ #= 0, define a scalar operator T by T x = λx for all x ∈ X. We have that T is bijective and so 0 is not an element of σ(T ). Therefore T is not compact by Theorem 3.1.18. Clearly the zero operator is compact. The result follows. We obtain another immediate corollary by Lemma 3.1.11 and Theorem 3.1.18. Corollary 3.1.20. If K : X → X is compact with no eigenvalues, then K is quasinilpotent. Moreover, for an arbitrary scalar c we have that +(cK)n + → 0. 3.2 von Neumann’s Result and Some Extensions Given a bounded finite-rank operator F : H → H, the range of F is an immediate nontrivial invariant subspace for F . Since Hilbert spaces have the approximation property, it can be seen that each compact operator K on H admits a sequence of ‘approximately K-invariant’ finite-dimensional subspaces. While this argument does not prove that compact operators satisfy the invariant subspace problem, it does seem somewhat reasonable that this may be true. Sometime during the 1930s John von Neumann proved that compact operators have non-trivial invariant subspaces, but did not decide to publish it. The proof was rediscovered and finally published by N. Aronszajn and K. T. Smith [7] in 1954. Theorem 3.2.1 (von Neumann, proved in [7]). Every compact operator on H has a non-trivial invariant subspace. 3.2. VON NEUMANN’S RESULT AND SOME EXTENSIONS 18 While von Neumann’s original proof uses orthogonal projections, and therefore applies only to Hilbert spaces, Aronszajn and Smith also included an alternative proof that extends to general Banach spaces. von Neumann’s Theorem resisted generalization for more than a decade after the Aronszajn and Smith paper, and not for lack of interest. Finally, in 1966 Bernstein and Robinson [13] extended the result to the slightly larger class of polynomially compact operators. A linear operator T on a Banach space is said to be polynomially compact if there is a non-zero polynomial p ∈ C[t] such that p(T ) is compact. Theorem 3.2.2 (Bernstein and Robinson [13]). Every polynomially compact operator on H has a non-trivial invariant subspace. Clearly all compact operators are polynomially compact by considering the polynomial p(t) = t, however, the converse is not true. Consider the following example. Example 3.2.3. Let K be a compact operator on X and let α be any scalar such that α ∈ / σ(K). The operator T = K − αI is bijective, and therefore not compact by Theorem 3.1.18 (ie. 0 ∈ / σ(T − αI)). However, T is polynomially compact with polynomial p(t) = t + α. Also, any nilpotent operator is polynomially compact (and may not be compact) by considering the polynomial p(t) = tn for sufficiently large n. An interesting aspect of Bernstein and Robinson’s proof is that it used the relatively new techniques of non-standard analysis, which builds up the foundations of analysis based on a rigorous definition of ‘infinitesimal’ numbers. Shortly after, the proof was translated into standard analysis by Halmos [26]. The next major generalization was achieved by Arveson and Feldman [8] in 1968. First, consider the following definition. Definition 3.2.4. For a bounded linear operator T on X, the uniformly closed algebra generated by T , denoted by A(T ), is defined to be the subspace [{I, T, T 2 , . . . }] of B(X). 3.3. LOMONOSOV’S THEOREM 19 Alternatively, A(T ) is the smallest closed subspace of B(X) containing T and I which is closed under function composition. If T is a bounded operator, then A(T ) can be thought of as the closure of the set of polynomial combinations of T , or the set of all operators which can be norm approximated by polynomial combinations of T . Theorem 3.2.5 (Arveson and Feldman [8]). If T : H → H is a bounded quasinilpotent operator such that A(T ) contains a non-zero compact operator, then T has a non-trivial invariant subspace. Some further generalizations were also discovered. For example, Arveson and Feldman’s result was extended to Banach spaces [23], and to the following: If T is quasinilpotent and the closure of the set of rational functions of T contains a non-zero compact operator, then T has a non-trivial invariant subspace [36,38]. Also, the Arveson and Feldman’s proof highlighted a new notion of quasitriangular operators, which would be extracted and studied by Halmos [27]. 3.3 Lomonosov’s Theorem While the techniques of von Neumann and subsequent generalizations yielded many interesting and surprising theorems during the 1950s and 60s, their effectiveness was reaching its limit by the 70s. Just as this was occurring, a young mathematician named Victor Lomonosov introduced a new and powerful technique [34]. Recall the definition of a hyperinvariant subspace, Definition 3.1.5 (b). Theorem 3.3.1 (Lomonosov [34]). If A is a non-scalar operator on X which commutes with a non-zero compact operator K, then A has a non-trivial hyperinvariant subspace. 3.3. LOMONOSOV’S THEOREM 20 Lomonosov’s Theorem was a significant breakthrough for several reasons. For one, it applies to Banach spaces and not just Hilbert spaces. Even restricted to Hilbert spaces, however, Lomonosov’s Theorem is still more general than anything that was previously known; see Proposition 3.3.3. Moreover, his technique allowed for a short and elegant proof. Theorem 3.3.1 describes a ‘commuting chain’ of operators K − A − T such that • K is non-zero and compact, • A is nonscalar, which implies that • T has a non-trivial invariant subspace. We provide a simple proof discovered by Hilden [37] of a weak version of the theorem. It contains many of the same ideas as Lomonosov’s original proof, but avoids the technical Schauder Fixed Point Theorem. The tradeoff is that Hilden’s proof only applies to commuting chains K − T of length two. In Section 4.10 we give a delightful argument from [50] showing that Lomonosov’s Theorem cannot be extended to commuting chains of length four. Theorem 3.3.2 (Lomonosov [34]). If T : X → X commutes with a non-zero compact operator K, then T has a non-trivial invariant subspace. Proof (Hilden, proved in [37]). The proof is by contradiction. Suppose to the contrary that T does not have a non-trivial invariant subspace. First, we may assume that +K+ = 1 as the operator +K+−1 K is compact and also commutes with T . We argue that K cannot have any eigenvalues. Indeed, if K had an eigenvalue λ, then we would have Wλ #= X since K is non-scalar (Corollary 3.1.19). By Lemma 3.1.4 and Proposition 3.1.6 the subspace Wλ would be a non-trivial invariant subspace for T , 21 3.3. LOMONOSOV’S THEOREM a contradiction. Therefore, we assume that K has no eigenvalues. By Corollary 3.1.20 we deduce the following: +(cK)n + → 0 for every scalar c ∈ C (3.1) Next, let us choose some x0 ∈ X such that +Kx0 + > +K+ = 1. This implies +x0 + > 1 as well by the definition of the operator norm. Let U denote the set B1 (x0 ). By definition of x0 , neither the closure of U nor K(U ) contains the zero vector (this is the source of our contradiction). For each polynomial p ∈ C[t], let θ(p) = p(T )−1 (U ). Since p(T ) is continuous, we have that θ(p) is open. Since we are assuming that T has only the trivial invariant subspaces, every non-zero vector of X must be a cyclic vector for T (Proposition 2.2.2). It follows that for every non-zero x ∈ X there is some polynomial p ∈ C[t] such that p(T )x ∈ U by Proposition 2.2.4. Thus, the collection {θ(p)}p is an open cover K(U ) (in fact, it covers all of X\{0}). The set U is bounded and so K(U ) is compact since K is a compact operator. It follows that there is a finite subcollection of {θ(p)}p which covers K(U ). Thus, we may let F be a finite set of polynomials such that {θ(p)}p∈F covers K(U ). Let c = max{+p(T )+ : p ∈ F }. The rest of the proof has been appropriately called “Hilden’s ping-pong technique.” Since Kx0 ∈ K(U ) ⊆ K(U ) we have that Kx0 ∈ θ(p1 ) for some p1 ∈ F . Therefore, p1 (T )Kx0 ∈ U . It follows that Kp1 (T )Kx0 ∈ K(U ) and so we may choose p2 ∈ F such that p2 (T )Kp1 (T )Kx0 ∈ U . Continuing this process for any positive integer n gives pn (T )Kpn−1 (T )K . . . p1 (T )Kx0 ∈ U where pi ∈ F for each i. By (3.1), given arbitrary ε > 0 we may choose n large enough so that +(cK)n x0 + < ε. We obtain the following inequality: +pn (T )Kpn−1 (T )K . . . p1 (T )Kx0 + = +pn (T )pn−1 (T ) . . . p1 (T )K n x0 + (as T K = KT ) 3.3. LOMONOSOV’S THEOREM 22 ≤ +pn (T )++pn−1 (T )+ . . . +p1 (T )++K n x0 + ≤ +(cK)n x0 + < ε Thus, we have a sequence of points in U converging to 0, contradicting the fact that 0∈ / U . Therefore, T must have a non-trivial invariant subspace. The result follows. All of the main results from Section 3.2 follow from Theorem 3.3.2, as we shall now demonstrate. Proposition 3.3.3. Let T and A be non-zero bounded linear operators on X. If the uniformly closed algebra generated by T contains A, then T commutes with A. Proof. Suppose that A ∈ A(T ). Given ε > 0, let p be a polynomial such that +p(T ) − A+ < 2$Tε $ . We obtain the following inequality: +AT − T A+ ≤ +AT − p(T )T + + +T p(T ) − T A+ (since p(T )T = T p(T )) ≤ +A − p(T )++T + + +T ++p(T ) − A+ < ε Therefore, we must have +AT − T A+ = 0, which implies AT = T A. The result follows. Corollary 3.3.4. If T : H → H is a bounded operator such that the uniformly closed algebra generated by T contains a non-zero compact operator, then T has a non-trivial invariant subspace. Many operator theorists were curious whether Lomonosov’s Theorem could solve the invariant subspace problem, at least for complex separable Hilbert spaces. That is, it was not clear whether a bounded operator on a Hilbert space could fail to satisfy the hypotheses of Theorem 3.3.1. The first natural candidate was the unilateral shift U+ on !2 ; however, an example of a non-scalar operator commuting with both U+ and a compact operator was discovered by Cowen [16]. Finally in 1980 Hadwin et 3.4. OPERATORS RELATED TO NORMAL OPERATORS 23 al. [24] discovered a class of operators, called ‘quasianalytic shifts,’ which do not satisfy Lomonosov’s hypotheses, ending the seven year search. We mention that Theorem 3.3.1 has been extended to real spaces by Hooker [32]. He proves that a bounded linear operator on a real or complex space which commutes with a compact operator and does not satisfy an irreducible polynomial equation has a non-trivial hyperinvariant subspace. On a complex space, this only rules out scalar operators by the Fundamental Theorem of Algebra, but the case for real spaces is more complicated. Fortunately, he also proves that a bounded linear operator on a real space commuting with a compact operator cannot satisfy any irreducible polynomial equation anyways. Finally, we would like to point out that Lomonosov’s Theorem can often provide us with not only an invariant subspace, but a sequence of nested invariant subspaces. For instance, let K : X → X be compact with invariant subspace M1 . It is easily shown that the restriction K1 of K to M1 is also compact. So, provided that the dimension of M1 is at least 2, we have that K1 has a non-trivial invariant subspace M2 ! M1 , which is seen to be an invariant subspace for K as well. A similar property of normal operators is discussed in the next section. 3.4 Operators Related to Normal Operators We begin with a standard definition from linear algebra. Definition 3.4.1. Let T be a bounded linear operator on H. A bounded operator T ∗ is said to be the adjoint of T if for all x, y ∈ H we have (T x, y) = (x, T ∗ y). The fact that every bounded operator has a unique adjoint is well-known. The concept of an adjoint is extremely important to the study of linear operators on Hilbert spaces. In some sense, adjoints extend the idea of complex conjugation. To illustrate 3.4. OPERATORS RELATED TO NORMAL OPERATORS 24 this, note that the adjoint of a scalar operator λI is λI. Indeed, for x, y ∈ H we have (λx, y) = λ(x, y) = λ(y, x) = λ(y, x) = (λy, x) = (x, λy). The notion of normal operators (and matrices) is very well-studied. We provide this definition here. Definition 3.4.2. A bounded linear operator N on H is said to be normal if N commutes with N ∗ . Normal operators generalize self-adjoint and unitary operators. Recall, an operator T is self-adjoint if T ∗ = T and unitary if T ∗ = T −1 .5 Note, for example, that for a real scalar λ the operator λI is self-adjoint (since λ is self-conjugate). The fact that normal operators have non-trivial invariant subspaces has been known for some time. This follows from a few results including Fuglede’s Theorem and the Spectral Theorem for normal operators. We only mention Fuglede’s Theorem briefly, for a deeper examination see [40, Section 1.5]. Theorem 3.4.3 (Fuglede’s Theorem [22]). Let T and N be bounded operators where N is normal. If T commutes with N , then T commutes with N ∗ . The study of operators related to normal operators has been an interesting and fruitful area of research. A natural generalization is the following, due to Halmos [25]. Definition 3.4.4. Let T be an operator on H. We say that T is subnormal if T is the restriction of a normal operator to an invariant subspace. That is, if there is a Hilbert space H & ⊇ H and a normal operator N on H & that is equal to T on H. Every normal operator is obviously subnormal by letting H & = H, but the converse is not true. In 1950 Halmos [25] asked whether every subnormal operator has 5 Equivalently, an operator T is unitary if it has dense range and preserves the inner product. That is, (T x, T y) = (x, y) for all x, y ∈ H. 3.4. OPERATORS RELATED TO NORMAL OPERATORS 25 a non-trivial invariant subspace. This problem was finally solved by Scott Brown [14] in 1978. His solution made clever use of a deep result of Sarason [48] and introduced powerful new techniques. Theorem 3.4.5 (Brown [14]). Every subnormal operator on H has a non-trivial invariant subspace. Brown’s theorem is equivalent to the following surprising result: If M is an infinite-dimensional invariant subspace for a normal operator N , then M contains a proper subspace other than {0} which is also N -invariant. His work was applied and generalized by many researchers, including Brown himself who extended his result in [15] to hyponormal operators with ‘thick’ spectra. Definition 3.4.6. An operator T on H is said to be hyponormal if +T ∗ x+ ≤ +T x+ for all x ∈ H. As it turns out, subnormal operators are hyponormal but the converse is not true. Brown’s full result on hyponormal operators cannot be stated without developing a lot of necessary background. An interesting special case is given as follows. Theorem 3.4.7 (Brown [15]). If T is a hyponormal operator on H such that σ(T ) has non-empty interior, then T has a non-trivial invariant subspace. This concludes our treatment of operators on Banach and Hilbert spaces known to have non-trivial invariant subspaces. We admit that this merely scratches the surface of invariant subspace theorems, especially on Hilbert spaces. For a more in-depth study of invariant subspaces for operators on Hilbert spaces, see [40]. Chapter 4 Counterexamples on Banach Spaces The theorems of Lomonosov and Brown together constitute the strongest evidence for a positive answer to the invariant subspace problem. The results of this chapter, however, contrast sharply. We discuss the known counterexamples to the problem on Banach spaces, and outline the proof of the simplest such example on !1 . 4.1 History and Controversy A few years after Lomonosov’s Theorem was proved came another monumental breakthrough in the study of the invariant subspace problem. In 1975 Per Enflo discovered the first example of an operator on a Banach space having only the trivial invariant subspaces. He gave an outline of the proof in 1976 [19]. However, his full solution was not submitted until 1981 and did not appear in print until 1987 [20]. The delay was mainly due to the sheer complexity of Enflo’s construction. It was so formidable that few people had the time and energy to work through the fine details. This, combined with the fact that early versions contained several minor errors, made Enflo’s paper a nightmare for referees. 26 4.1. HISTORY AND CONTROVERSY 27 Davie [17]: “Enflo’s successful completion of the task is a remarkable achievement; however the latter part of his paper is so impenetrable that it is destined to be admired rather than read.” Radjavi and Rosenthal [39]: “We’ve heard lots of rumors of the form: soand-so spent two months working very hard on the manuscript, found minor correctable mistakes, got about 1/3 of the way through, then quit in exhaustion.” This is all in spite of the fact that, on the first level, Enflo’s idea is both natural and interesting. He notices that a linear operator T with a cyclic vector x acts as a shift on the set {x, T x, T 2 x, . . . }. He likens this shift behaviour to the action of multiplying a polynomial p ∈ C[t] by the independent variable t. His goal then, is to construct a suitable norm + · + on the set of polynomials that achieves the following: (1) the operator T on the completion X which maps T : p(t) /→ tp(t) is bounded, and (2) every non-zero x ∈ X is T -cyclic. In other words, his idea is to construct the Banach space to suit his operator, not the other way around. As Enflo’s paper crawled through the publication process, C. J. Read developed a counterexample of his own and submitted it for publication [41]. The paper was of similar length and complexity to Enflo’s, however it was published much earlier in 1984. He did not cite Enflo’s work, except to say the following: Read [41]: “This is the only counterexample which the author knows to be valid. P. Enflo has produced two preprints purporting to contain examples of operators without an invariant subspace, obtained by very different methods to our own. However, these preprints have been in existence since 1976 and 1981, and neither has yet been published.” 4.1. HISTORY AND CONTROVERSY 28 The fact that Read would publish his example first, without giving proper credit to Enflo, was seen by many mathematicians to be in bad form. This was made worse by the fact that several facets of Read’s construction were actually based on Enflo’s ideas. Like Enflo, Read begins with a shift operator on the set of finitely non-zero sequences (isomorphic to C[t]) and constructs the space as he goes. Beauzamy [11]: “To give such precision is uncommon, and would be of no interest, if C. Read had not, several times, unelegantly and unsuccessfully, tried to claim priority towards the solution of the problem. This behaviour might be condemned with stronger words, but we remember we are presently writing for posterity.” There is no question among mathematicians that Enflo’s ideas were behind the first counterexample on a Banach space, and for that he should receive full credit. Since then, however, many others have emerged. Beauzamy [10] provides a significant simplification of Enflo’s original example (although, it is still very complicated). Apart from this, most advances have been developed by Read. The first such achievement came in 1985 when he managed to give a counterexample on the classical Banach space !1 [42]. A simplification was given by A. M. Davie, which appears in [11, Chapter XIV]. Another construction by Read of a linear opearator on !1 in [43] is the simplest known counterexample to date. We outline this construction in Section 4.2. Read has also discovered counterexamples having additional interesting properties. For one, in 1997 he showed that his example from [43] could be modified to have the additional property of being quasinilpotent [45]. Even more interesting is an example given by Read in 1988 of an operator T on !1 such that every non-zero vector is hypercyclic [44] (as opposed to just cyclic). We say that a vector x of X is hypercyclic for T if the set {x, T x, T 2 x, . . . }, called the orbit of x, is dense in X. This example 4.2. A COUNTEREXAMPLE ON !1 29 proves the incredible result: There is an operator T on !1 such that no non-trivial closed set S ⊆ !1 is T -invariant. Lastly, we make note of some recent work showing that the invariant subspace problem actually has a positive solution on certain infinite-dimensional separable Banach spaces. Argyros and Haydon [6] have shown that there is such a space X so that every operator T : X → X can be written as T = K + λI where K is compact and λ is a scalar.1 By Lomonosov’s Theorem, such operators are guaranteed to have non-trivial invariant subspaces, solving the problem positively in this case. 4.2 A Counterexample on !1 We guide the reader through the construction of the simplest known counterexample to the invariant subspace problem on the Banach space !1 , due to C. J. Read [43]. See Appendix E for some of the relevent notation and definitions. The example is constructed as a simple shift operator on a basis (ei )∞ i=0 for F , different from the canonical basis (fi )∞ i=0 . The properties of T are therefore directly determined by the construction of (ei )i , which we shall provide shortly. The construction and proof revolve around a somewhat mysterious increasing sequence d = (di )∞ i=1 of natural numbers. Often we require that d increases sufficiently rapidly so that certain deductions can be made. The phrase ‘d increases sufficiently rapidly’ is taken to mean that for some m > 1 the number dm is bounded below by a real-valued function of d1 , d2 , . . . , dm−1 . We can ensure that the sequence d is welldefined, provided that only finitely many such bounds are required. ∞ Note. We define (ai )∞ i=1 and (bi )i=1 to be the sequences of odd and even terms of d: ai = d2i−1 and bi = d2i for i ≥ 1. Let a0 = 1, v0 = 0 and define vn = n(an + bn ) for 1 This solves the so-called ‘scalar-plus-compact’ problem. 4.2. A COUNTEREXAMPLE ON !1 30 n ≥ 1. As previously mentioned, the linear operator T is defined so that for each i ≥ 0 i ∞ we have T ei = ei+1 . Since (ei )∞ i=0 = (T e0 )i=0 spans F , which is dense in !1 , we have that e0 is T -cyclic regardless of how (ei )i is defined. Our main goal of the proof is outlined as follows. Objective. We show that for each unit vector x and ε > 0 there is some polynomial q so that +q(T )x − e0 + < ε. The fact that T has no non-trivial invariant subspaces follows by Corollary 2.2.7. Since d is unbounded it suffices to show that for any k ≥ 1 there is a polynomial q so 3 . In order to achieve this, we focus on an associated vector y that +q(T )x − e0 + < ak−1 based on x such that • y is an element of some special compact set Kk,n for n > k; • We can obtain sufficient control over +q(T )(x − y)+. We then decompose x into x = (x − y) + y. The polynomial q is actually defined via a compactness argument after strategically choosing y ∈ Kk,n . Our conclusion to the proof has the following form: +q(T )x − e0 + ≤ +q(T )(x − y)+ + +q(T )y − e0 + ≤ 1 ak−1 + 2 ak−1 = 3 ak−1 < ε. On the way to this conclusion, we need several bounds regarding polynomial combinations of T on y. Let us move on with the construction of (ei )i . 4.3. CONSTRUCTING (ei )i 4.3 31 Constructing (ei)i The first application of d is to induce a partitioning P of the natural numbers into finite intervals. The purpose of P is to facilitate the construction of (ei )i . For each i ≥ 1, the vector ei is then defined based on which set of P contains i. Definition 4.3.1. We define P as the collection of all intervals in N of the following form, for 1 ≤ r ≤ n. (vn−1 , an ) if r = 1; (4.1.1) (vn−r+1 + (r − 1)an , ran ) if r > 1; (4.1.2) [ran , vn−r + ran ]; (4.2) (nan + (r − 1)bn , r(an + bn )); (4.3) [r(an + bn ), nan + rbn ]. (4.4) Note that we may assume that an > vn−1 + 1 and bn > (n − 1)an + 1 for all n ≥ 1 to ensure that intervals described in (4.1.1), (4.1.2), and (4.3) are non-empty. We are required to show that P is indeed a partitioning of the natural numbers. We do not prove this rigorously, as it should be easy enough to see after going over a few examples. Example 4.3.2. Let us consider the sets of P for fixed n = 2. The interval (v1 , v2 ] can be written as a union of the following (disjoint, non-empty) sets: (v1 , a2 ) ∪ [a2 , v1 + a2 ] ∪ (v1 + a2 , 2a2 ) ∪ [2a2 , 2a2 ]∪ (2a2 , a2 + b2 ) ∪ [a2 + b2 , 2a2 + b2 ] ∪ (2a2 + b2 , v2 ) ∪ [v2 , v2 ] 4.3. CONSTRUCTING (ei )i 32 Each set of P for n = 2 appears in the above union.2 A more general argument shows that the sets of P for fixed n form a partitioning of (vn−1 , vn ]. Since v0 = 0 we have that P partitions N. We suggest that the reader tries to duplicate this example for n = 3 to get a better ‘feel’ for how the partition works in general. Therefore, we may assume that every integer i ≥ 1 is contained in precisely one set of P. We are now in position to define the sequence (ei )∞ i=0 . We warn the reader that the full definition is complex, and is difficult to comprehend out of context. Fortunately, it is not yet necessary to grasp the fine details of the definition, only a few minor facts which are clarified shortly. Definition 4.3.3. Define f0 = e 0 . (4.5) Given I ∈ P and i ∈ I, we define ei in the following way: √ If I is as in (4.1.1) or (4.1.2), let fi = 2(h−i)/ an ei where h = (r − 1/2) an ; If I is as in (4.2), let fi = an−r (ei − ei−ran ); √ (4.6) (4.7) If I is as in (4.3), let fi = 2(h−i)/ bn ei where h = (r − 1/2) bn ; (4.8) If I is as in (4.4), let fi = ei − bn ei−bn . (4.9) Remark 4.3.4. Let 1 ≤ r ≤ n. For some of the arguments given later, it is important to notice the following: (a) fran = an−r (eran − e0 ) by (4.7). 3 √ (b) If i ∈ (nan + bn , 2(an + bn )), then ei = 2(i− 2 bn )/ bn fi by (4.8). 2 Taking the sets in the order that they appear in the union, the sets are described in (4.1.1), (4.2), (4.1.2), (4.2), (4.3), (4.4), (4.3), (4.4). The sequence of r values is 1, 1, 2, 2, 1, 1, 2, 2. 4.4. FIRST PROPERTIES OF (ei )i 33 (c) If i ∈ [an + bn , nan + bn ], then ei = fi + bn ei−bn by (4.9). 4.4 First Properties of (ei)i For each integer m ≥ 0 let Em be the span of (ei )m i=0 . In Definition 4.3.3, each ei is defined by writing fi in terms of e0 , e1 , . . . , ei where ei makes a non-zero contribution. Using this fact, we obtain a simple but important lemma. Lemma 4.4.1. For m ≥ 0 we have Em = Fm . Moreover, ∪m Em = ∪m Fm = F . Proof. Each vector fi for 0 ≤ i ≤ m can be written as fi = !i j=0 λm,j ej where λi,i #= 0. This is easily seen to be invertible. That is, ei can be written as !i j=0 βi ei where βi #= 0. Therefore, we obtain Em = Fm . The fact that ∪m Em = ∪m Fm = F follows immediately. Recall that the construction of each vector ei for i ≥ 1 depends heavily on the sequence d. However, for fixed i the vector ei only depends on certain terms of d. Consider the following. Note 4.4.2. Let i ≥ 1 and n ≥ 1 be given. • If i ≤ nan , then the definition of ei depends only on the choice of (at most) a1 , b1 , . . . , bn−1 , an ; • If i ≤ vn , then the definition of ei depends only on the choice of (at most) a1 , b 1 , . . . , a n , b n . For each integer m ≥ 0 let Jm : Fm → Fm be the linear isomorphism such that Jm (ei ) = fi 34 4.5. THE LINEAR OPERATOR for 0 ≤ i ≤ m. In the case that m = nan for some n ≥ 1, the vectors e0 , e1 , e2 , . . . , em depend only on the choice of a1 , b1 , . . . , bn−1 , an by Note 4.4.2. Thus, Jm (and, in particular, its norm) is affected only by the choice of these terms of d and so we may define a real-valued function Mn such that: −1 Mn (a1 , b1 , . . . , bn−1 , an ) ≥ max{+Jnan +, +Jna +}. n (4.10) Similarly, by Note 4.4.2 there is a function Nn mapping into the real numbers so that: +}. Nn (a1 , b1 , . . . , an , bn ) ≥ max{+Jvn +, +Jv−1 n (4.11) The purpose of functions Mn and Nn is to highlight the fact that, for example, bn can n−1 be chosen large with respect to (ai )ni=0 and (bi )i=0 without affecting the value of +Jnan +. 4.5 The Linear Operator We define the linear operator T which is the central focus of the counterexample. For now, we consider T as a linear operator on F . Later we see that T can be extended uniquely to a bounded operator on !1 . Definition 4.5.1. Given the sequence (ei )∞ i=0 , we define T : F → F to be the unique linear operator such that T (ei ) = ei+1 for each i ≥ 0. In some sense, T is based on the unilateral shift U+ . This makes it somewhat surprising that T would not have any non-trivial invariant subspaces, since there are 35 4.5. THE LINEAR OPERATOR many U+ -invariant subspaces which are quite easy to describe. For example, the range of U+ suffices, see Example 2.2.1. The construction of T has a fundamental difference in the fact that T m (F ) = ({ei : i ≥ m}) is dense in !1 . In particular, T (F ) contains e0 , as we shall now demonstrate. Lemma 4.5.2. Given n > k ≥ 1, we have +e(n−k+1)an − e0 + = 1 ak−1 . Proof. Letting r = n − k + 1 we recognize that fran = an−r (eran − e0 ) by Remark 4.3.4 1 1 (a). So, we have +eran − e0 + = an−r = ak−1 . 4.5.1 A Truncated Version Of T At certain points it is useful to study the action of T on certain finite subspaces. For this, we define a truncated version of T on Fm . Definition 4.5.3. Let Tm : Fm → Fm be the linear operator so that Tm (ei ) =   e for 0 ≤ i ≤ m − 1, i+1  0 for i = m. One useful property of Tm is that it allows us to ‘isolate’ certain vectors ei . To see what we mean, consider the following lemma. Lemma 4.5.4. Let y = !m i=α λi ei where λα #= 0. Then there is some polynomial r so that r(Tm )y = eα and deg(r) ≤ m − α. Proof. Given j ≥ 1 we have (Tm )j y = !m j i=α λi (Tm ) ei = !m i=α+j λi−j ei by definition of 36 4.5. THE LINEAR OPERATOR Tm . So, for j ≥ 1 we have α+j−1 j y − λα+j λ−1 α (Tm ) y = % λi e i + i=α α+j−1 = % λi e i + m % m % (λi − λα+j λ−1 α λi−j )ei i=α+j (λi − λα+j λ−1 α λi−j )ei . i=α+j+1 i=α ! & Essentially, we have taken a vector y = m i=α λi ei and constructed a new vector y = !m i=α βi ei where βα+j = 0. Now it should be clear that there is a polynomial r satisfying the above properties. 4.5.2 A Norm On The Set Of Polynomials For ease of demonstration, it is useful to define a norm function + · +p on the set F[t] of polynomials. We define + · +p in a way that is analogous to the standard 1-norm on F , where polynomials 1, t, t2 , . . . are taken to be the standard unit vectors in F[t]. Definition 4.5.5. Given a polynomial p = !m i=0 αi t i over F, define +p+p = !m i=0 |αi |. By applying a slight variation of Proposition E.1, it is clear that +T +e = 1. Using this fact, we obtain the following lemma. Lemma 4.5.6. If x ∈ F and p is a polynomial, then +p(T )x+e ≤ +p+p +x+e . Proof. Let p(t) = !m i i=0 αi t . We obtain +p(T )x+e ≤ +p(T )+e +x+e ≤ m % i=0 as desired. |αi |+T i +e +x+e ≤ +p+p +x+e , 37 4.6. A COMPACTNESS ARGUMENT 4.6 A Compactness Argument We define an operator to help us construct a sequence of compact sets which is crucial to the proof. Definition 4.6.1. For n > k ≥ 1 let τk,n : Fnan → Fnan be the unique linear operator so that: τk,n (ei ) =   e i  0 for 0 ≤ i < (n − k)an , for (n − k)an ≤ i ≤ nan . The linear operator τk,n is a standard projection Fnan → F(n−k)an −1 , where n > k ≥ 1. Our sequence of compact sets Kk,n is defined as follows. Definition 4.6.2. For n > k ≥ 1 let us define a set Kk,n ⊆ Fnan in the following way: & ' 1 Kk,n = y ∈ Fnan : +y+ ≤ an and +τk,n y+ ≥ . an Lemma 4.6.3. For n > k ≥ 1 the set Kk,n is non-empty and compact. Proof. We have that e0 ∈ Kk,n since +e0 + = +f0 + = 1 ≤ an and +τk,n e0 + = +e0 + = 1 ≥ a1n for all n ≥ 2. Therefore Kk,n is non-empty. Now, to prove compactness it suffices to show that Kk,n is closed and bounded since Fnan is finite-dimensional. The linear operator τk,n is continuous since Fnan is −1 finite-dimensional. It follows that the τk,n (S) is closed for any closed subset S of Fnan . The intersection of two closed sets is closed, and so −1 Kk,n = {y ∈ Fnan : +y+ ≤ an } ∩ τk,n (& 1 y ∈ Fnan : +y+ ≥ an ') is closed. Also, y ∈ Kk,n implies +y+ ≤ an by definition, and so Kk,n is bounded. Let us give our first of many results regarding polynomial combinations of T and Tnan on vectors of Kk,n . 38 4.6. A COMPACTNESS ARGUMENT Lemma 4.6.4. Let n > k ≥ 1 and choose y ∈ Kk,n . Then there is a polynomial p so that tan | p, deg(p) ≤ nan , and p(Tnan )y = e(n−k+1)an . Proof. Let us write y = !nan i=α λi ei where λα #= 0. By Lemma 4.5.4 there is a polynomial r so that r(Tnan )y = eα and deg(r) < nan −α. Since y ∈ Kk,n we have that τk,n y #= 0 and so α < (n − k)an . Therefore j = (n − k + 1)an − α > an and the polynomial p(t) = tj r(t) satisfies the desired properties. Here is how the compactness argument works. By Lemma 4.6.4 for every y ∈ Kk,n there is a polynomial p so that tan | p, (4.12) deg(p) ≤ nan . (4.13) +p(Tm )y − e(n−k+1)an + < 1 , an (4.14) By Lemma 4.6.4 and compactness of Kk,n there is a finite set Pk,n of polynomials satisfying (4.12) and (4.13) so that for every y ∈ Kk,n there is some p ∈ Pk,n satisfying (4.14). Now, notice that the choice polynomials which constitute Pk,n are dependent only on n the definition of vectors (ei )na i=0 . It follows by Note 4.4.2, and the fact that Pk,n is finite, that for each n ≥ 2 there is a real-valued function Ln so that +p+p ≤ Ln (a1 , b1 , . . . , bn−1 , an ) for each k < n and p ∈ Pk,n . Let us summarize the results of this section. Proposition 4.6.5. Let n > 1 be given. There is a real-valued function Ln with the following property. For each k so that 1 ≤ k < n and y ∈ Kk,n there is a polynomial p satisfying tan | p, 39 4.7. TWEAKING OUR POLYNOMIALS deg(p) ≤ nan , +p+p ≤ Ln (a1 , b1 , . . . , bn−1 , an ), +p(Tnan )y − e0 + ≤ 1 1 + . an ak−1 Proof. The polynomial p is chosen from the set Pk,n so that +p(Tnan )y − e(n−k+1)an + < 1 . ak−1 The fact that p satisfies the first three properties is simply by definition. For the fourth property we invoke the triangle inequality. We have +p(Tnan )y − e0 + ≤ +p(Tnan )y − e(n−k+1)an + + +e(n−k+1)an − e0 + ≤ 1 1 + . an ak−1 by Lemma 4.5.2. The result follows. 4.7 Tweaking Our Polynomials In this section, we would like to construct, for each y ∈ Kk,n , a polynomial q satisfying 2 +q(T )y − e0 + < ak−1 in such a way that q has many other desirable properties. As bn it turns out, the polynomial that we want is q(t) = tbn p(t) where p is chosen as in Proposition 4.6.5. It is clear that q satisfies the following: +q+p = tan +bn | q; (4.15) deg(q) ≤ nan + bn ; (4.16) +p+p Ln (a1 , b1 , . . . , bn−1 , an ) ≤ . bn bn (4.17) 4.7. TWEAKING OUR POLYNOMIALS 40 We obtain the following bound via the triangle inequality: * * bn * * T * +q(T )y − e0 + ≤ * *q(T )y − bn p(Tnan )y * + * * bn * *T * * * bn p(Tnan )y − p(Tnan )y * + +p(Tnan )y − e0 + (4.18) 1 Notice that we have already proven +p(Tnan )y − e0 + ≤ a1n + ak−1 in Proposi- tion 4.6.5. Therefore, we may obtain control over +q(T )y − e0 + by observing bounds on * * * * bn * bn * * * *q(T )y − Tbn p(Tnan )y * and * Tbn p(Tnan )y − p(Tnan )y *. This is done in Lemmas 4.7.3 and 4.7.2 respectively. Let us assume that d increases so rapidly that for each n the values an and bn satisfy bn ≥ Ln (a1 , b1 , . . . , bn−1 , an )Mn (a1 , b1 , . . . , bn−1 , an )a2n , (4.19) bn ≥ Ln (a1 , b1 , . . . , bn−1 , an )2(n−1)an +1 an−1 , (4.20) an ≥ an−1 Nn−1 (a1 , b1 , . . . , an−1 , bn−1 ). (4.21) bn ≥ 4nan , (4.22) an ≥ 3an−1 . (4.23) These bounds are applied in the coming results. Lemma 4.7.1. Given n > k ≥ 1 choose y ∈ Kk,n and let p be a polynomial as in Proposition 4.6.5. We have that +p(T )y+e ≤ Ln (a1 , b1 , . . . , bn−1 , an )Mn (a1 , b1 , . . . , bn−1 , an )an . 41 4.7. TWEAKING OUR POLYNOMIALS Proof. By Lemma 4.5.6 and by Lemma E.3 we get +p(T )y+e ≤ +p+p +y+e ≤ +p+p +Jnan ++y+. This is bounded above by Ln (a1 , b1 , . . . , bn−1 , an )Mn (a1 , b1 , . . . , bn−1 , an )an by (4.10), definition of p, and the fact that +y+ ≤ an . Lemma 4.7.2. Given n > k ≥ 1 choose y ∈ Kk,n , and let p be a polynomial as in Proposition 4.6.5. We have * * bn * *T * *≤ 1. p(T )y − p(T )y na na n n * an * bn Proof. Let us write p(Tnan )y = !nan i=an λi ei . Recall that ei = fi + bn ei−bn whenever i ∈ [an + bn , nan + bn ] by Remark 4.3.4 (c). Therefore * + na , na * bn * * n n * bn % % *T * * T * * * * λi e i − λi e i * * bn p(Tnan )y − p(Tnan )y * = * * * bn i=a i=a n n * na +b * * na +b * nan nan n n n n * % * * % * % % λi−bn λi−bn * * * * =* ei − λi e i * = * (fi + bn ei−bn ) − λi e i * * * * * bn bn i=a i=a i=a +b i=a +b n n n n n n * na +b * nan n n * % * λ 1 % * i−bn * =* fi * = |λi |. * * bn b n i=a i=a +b n n n This last expression is equal to b1n +p(Tnan )y+e . Applying Lemma 4.7.1, we get 1 1 +p(Tnan )y+e ≤ Ln (a1 , b1 , . . . , bn−1 , an )Mn (a1 , b1 , . . . , bn−1 , an )+y+. bm bn Recall that y ∈ Kk,n and so +y+ ≤ an . Therefore, by (4.19) this final expression is bounded above by a1n , as desired. The result follows. 42 4.7. TWEAKING OUR POLYNOMIALS Lemma 4.7.3. Given n > k ≥ 1 choose y ∈ Kk,n . Let p be a polynomial as in Proposibn tion 4.6.5, and q(t) = tbn p(t). We have * * bn * * T 1 *q(T )y − p(Tnan )y * ≤ . * * bn an !2nan Proof. Since tan | p, deg(p) ≤ nan and y ∈ Fnan we can write p(T )y = i=an λi ei !nan for scalars λi . Note that p(Tnan )y = i=an λi ei since Tnan is truncated. So, p(T )y − ! n p(Tnan )y = 2na i=nan +1 λi ei . We get T bn T bn T bn p(Tnan )y = (p(T )y − p(Tnan )y) = q(T )y − bn bn bn + 2na %n λi e i i=nan +1 , Therefore * * * * * * 2na 2na n +bn n +bn * * bn % % * * λi−bn * * λ T * * i−b n * * *q(T )y − p(Tnan )y * ei * ≤ *=* * bn e i * * * * bn b n i=na +b +1 i=na +b +1 n ≤ n n n +p(T )y+e max{+ei + : nan + bn < i ≤ 2nan + bn }, bn where +p(T )y+e ≤ Ln (a1 , b1 , . . . , bn−1 , an )Mn (a1 , b1 , . . . , bn−1 , an )an by Lemma 4.7.1. So, given (4.19), we obtain * * bn * * max{+ei + : nan + bn < i ≤ 2nan + bn } T *q(T )y − *≤ p(T )y . na n * * bn an Now it suffices to show that +ei + ≤ 1 whenever i ∈ (nan + bn , 2nan + bn ]. By (4.22) we have bn ≥ 4nan > 2(n − 1)an . This implies that 2nan + bn < 2(an + bn ), and so 3 √ (nan + bn , 2nan + bn ] ⊆ (nan + bn , 2(an + bn )). Thus, we have ei = 2(i− 2 bn )/ bn fi by 43 4.7. TWEAKING OUR POLYNOMIALS √ 3 Remark 4.3.4 (b) and so +ei + = 2(i− 2 bn )/ bn . Dealing with the exponent, we see that bn + 2nan ≥ i and (4.22) imply 3 1 3 i − bn ≤ bn + 2nan − bn = 2nan − bn ≤ 0. 2 2 2 So, we have +ei + ≤ 1. The result follows. Let us summarize the results of this section and the last. Proposition 4.7.4. Let n > k ≥ 1 and choose some y ∈ Kk,n . There is a polynomial q satisfying the following: tan +bn | q; deg(q) ≤ nan + bn ; +q+p ≤ 1 2(n−1)an +1 a +q(T )y − e0 + ≤ , k−1 2 ak−1 . Proof. The polynomial q is chosen as in Lemma 4.7.3. The fact that q satisfies the first two properties is simply by definition. For the third, we have +q+p = 1 Ln (a1 , b1 , . . . , bn−1 , an ) ≤ (n−1)an +1 , bn 2 ak−1 by (4.17) and (4.20). For the fourth property of q, we combine the bounds in (4.18), Proposition 4.6.5, and Lemmas 4.7.2 and 4.7.3 to obtain: 1 1 +q(T )y − e0 + ≤ + + an an ( 1 1 + an ak−1 ) 2 We have that this final expression is bounded above by ak−1 by (4.23). The result 44 4.8. FINAL ARGUMENTS follows. 4.8 Final Arguments The work up to this point has allowed us to obtain a certain amount of control over vectors in Kk,n . However, there are still a few crucial questions which remain unanswered. • Is T bounded on F ? • How should we decompose an arbitrary unit vector x into (x − y) + y for y ∈ Kk,n ? We take the time now to settle these questions, before moving on to prove the main result. The special decomposition of unit vectors makes use of a special finite-rank linear operator, Q0n , which we define now. Definition 4.8.1. For n ≥ 1, let Q0n : F → Fnan be the linear operator such that Q0n (fi ) =    f   i for 0 ≤ i ≤ nan , −am−r ei−ram     0 for i ∈ [ram , vm−r + ram ], where 1 ≤ m − n < r ≤ m, otherwise. Let us obtain an explicit upper bound on +Q0n +. Lemma 4.8.2. We have +Q0n + ≤ an . Proof. Recall that +Q0n + = sup{+Q0n fi +i ≥ 0} by Proposition E.1. We have that Q0n fi = fi or 0 unless j ∈ [ram , ram + vm−r ] for 0 < m − n < r ≤ m. In this case we have * * * +Q0n fi + = am−r +ei−ran + ≤ an−1 *Jv−1 f n−1 i−ran ≤ an−1 Nn−1 (a1 , b1 , . . . , an−1 , bn−1 ) ≤ an 45 4.8. FINAL ARGUMENTS by (4.11) and (4.21). The result follows. The next two lemmas are crucial to the proof. However, we have chosen to omit their proofs as they are overly technical and not particularly enlightening. For proofs, see [43]. Lemma 4.8.3. For any unit vector x and k ≥ 1 there is some n > k so that y = Q0n x ∈ Kk,n . Lemma 4.8.4. Provided that d increases sufficiently rapidly, then for every n ≥ 1 we have: +T + < 2 and +T an +bn (I − Q0n )+ < 2. By Lemmas 4.8.2 and 4.8.4 we have that T and Q0n can be extended uniquely to bounded operators on !1 (by Proposition C.16). For the rest of the proof, the symbols T and Q0n refer to their extensions to !1 . The next result gives us control over the value of +q(T )(I − Q0n )+, which is needed to prove the main result. Proposition 4.8.5. Let y be a vector in Kk,n and choose a polynomial q as in Proposi1 . tion 4.7.4. We have +q(T )(I − Q0n )+ < ak−1 Proof. Let r be the polynomial so that q(t) = tan +bn r(t). Note that deg(r) ≤ (n − 1)an . Also, q and r have the same coefficients and so +q+p = +r+p . We have that +q(T )(I − Q0n )+ = +r(T )T an +bn (I − Q0n )+ ≤ +r(T )++T an +bn (I − Q0n )+ < 2+r(T )+ ≤ 2+r+p +T +deg(r) ≤ +q+p 2(n−1)an +1 ≤ by Lemma 4.8.4 and Proposition 4.7.4. 1 ak−1 , 4.9. EVERY UNIT VECTOR IS T -CYCLIC 4.9 46 Every Unit Vector Is T -Cyclic We are now in position to prove the main result. Theorem 4.9.1. The operator T : !1 → !1 has only the trivial invariant subspaces. Proof. It suffices to show that every unit vector is T -cyclic. So, we let x be any unit 3 < ε. Our goal is vector and let ε > 0 be arbitrary. Choose k large enough so that ak−1 3 to prove that there exists a polynomial q so that +q(T )x − e0 + ≤ ak−1 . By Lemma 4.8.3 we may let n > k so that y = Q0n ∈ Kk,n . Let us choose a polynomial q as in Proposition 4.7.4. We have +q(T )x − e0 + ≤ +q(T )(x − y)+ + +q(T )y − e0 + ≤ +q(T )(I − Q0n )++x+ + +q(T )y − e0 + < 1 ak−1 + 2 ak−1 = 3 ak−1 <ε by Proposition 4.8.5, definition of q, and the fact that +x+ = 1. The result follows. Surprisingly, Read [43] also shows that under additional conditions on d we can ensure that either: (1) T k has no non-trivial invariant subspaces for any integer k ≥ 1, or alternatively (2) T k has non-trivial invariant subspaces for all k ≥ 1. 4.10 Sharpness of Lomonosov’s Theorem Recall that Lomonosov’s Theorem applies to a ‘commuting chain’ of operators K −A−T where A is non-scalar and K is compact. Intuitively, one may wonder whether the result can be generalized to a longer chain, perhaps K − A1 − A2 − T where both A1 and A2 are non-scalar. We provide a clever observation of Troitsky [50] showing that this is not 4.10. SHARPNESS OF LOMONOSOV’S THEOREM 47 possible. Throughout this section, let T be Read’s operator, which is defined earlier in the chapter. We suppose, without loss of generality, that every term of the sequence d is even. Let us define A1 = T 2 . Clearly A1 is non-scalar and commutes with T . Also, let A2 to be the unique operator on !1 satisfying the following: A2 ei = Lemma 4.10.1. We have A2 fi =   e i  0   f i is even; otherwise. i is even; i  0 otherwise. Proof. Consider the cases indicated by Definition 4.3.3. In any of them, we have that fi can be written as a linear combination of vectors ei , ei−ran and ei−bn for some values of r and n. Notice that the indices i, i − ran and i − bn have the same parity since an and bn are even. The result follows. Corollary 4.10.2. The operator A2 commutes with A1 . Proof. Consider operators A1 A2 = T 2 A2 and A2 A1 = A2 T 2 . We have, by definition: T 2 A2 ei = and A2 T 2 ei =   T 2e = e i i+2  0   Ae 2 i+2 = ei+2  Ae 2 i+2 = 0 i is even; otherwise. i is even; otherwise. Therefore, we have A1 A2 = A2 A1 . The result follows. 4.10. SHARPNESS OF LOMONOSOV’S THEOREM 48 Now, simply define an operator K on !1 such that Kf0 =   f i  0 i = 0; otherwise. We have that K is a bounded finite-rank operator, and therefore compact by Proposition 3.1.14. It is easily seen that K commutes with A2 , completing the example. It is interesting to note that T 2 is seen to have a non-trivial invariant subspace by Lomonosov’s Theorem, provided that each term of d is even. In fact, as Troitsky [50] points out, if m ≥ 2 divides each term of d, then a similar argument can be used with T m in place of T 2 . Chapter 5 Conclusion There are a few questions which continue to linger, of which the following is most obvious: (1) Do bounded operators on infinite-dimensional separable Hilbert spaces have non-trivial invariant subspaces? Given the fact that nobody knows the answer to the first question, one might ask a second: (2) Is there at least a consensus among mathematicians regarding what the answer should be? Among the many open problems in mathematics, the invariant subspace problem is a somewhat exceptional in the fact that the answer to (2) is certainly “no.”1 When it comes to this problem, most mathematicians remain noncommittal, and it is quite rare for anyone to put forth a guess.2 While the powerful results of Lomonosov and Brown pull in one direction, the impressive counterexamples of Enflo and Read pull in the other. Enflo has noted in [20] that there are some major challenges in constructing counterexamples on reflexive Banach spaces (which include Hilbert spaces), and feels that this offers some weak evidence for a positive answer to the problem. To this day, 1 2 There are very few mathematicians who deny the truth of, for example, the twin prime conjecture. It is called the invariant subspace problem, after all, and not conjecture. 49 50 these challenges have not been overcome as there are still no examples on reflexive Banach spaces. On the other hand, some subsequent work in this area may suggest a negative answer. First, the problem on Banach spaces has been shown to be not only false, but very false; We have already mentioned Read’s example in [44] of a bounded linear operator so that non-zero vectors are hypercyclic. Also, in [11] Beauzamy cites an example of a bounded operator T on a Hilbert space that he has constructed with a hypercyclic vector x0 so that p(T )x0 is hypercyclic whenever p is a polynomial with complex coefficients (we have not been able to find this example in print). While it is still possible that certain vectors may not be T -cyclic, this is the closest anyone has come to a counterexample on a Hilbert space. Another valid and pertinent question is the following: (3) Why should we care about invariant subspaces in the first place? Part of the allure of the problem is simply that it is so simple to state, and yet disproportionally difficult to solve. However, there must be some more practical reason for people to direct so much time and energy towards it. There are very few (if any) instances where researchers assume the truth of the invariant subspace problem and derive important or surprising consequences. So, what is the point? History has shown that the study of the invariant subspaces of a certain class of operators is often one of the first steps towards a rich and useful structural theory for those operators, as is elegantly explained in [12]. One might consider, for example, the work of Ringrose [47] on compact operators, which was inspired by the results of Aronszajn and Smith. Also, as was briefly mentioned in Section 3.2, a very deep and rewarding study of quasitriangular operators began as a result of Arveson and Feldman’s invariant subspace theorem. This culminated in an impressive series of papers by Apostol and others which established a surprising link between quasitriangular and the more well- 51 known semi-Fredholm operators, see [1–5]. The work of Brown on operators related to normal operators also contained a wealth of information on the structure of subnormal and hyponormal operators and initiated an interesting line of research. As noted in [33], the understanding of invariant subspaces can also be applied to develop the theory of functional calculus. As a final question, we pose the following: (4) Will the invariant subspace problem ever be solved? It is, of course, impossible for us to judge whether the solution might be beyond the potential of human understanding. However, we tend to side with humanity on this particular issue. The future promises to bring many bright minds to fill the shoes of great operator theorists such as Enflo, Read, Lomonosov and Brown; Many of the secrets behind bounded operators will surely be unlocked. In the meantime, however, perhaps one of the most remarkable things about the invariant subspace problem is its ability to keep operator theorists humble. Appendix A Vector Spaces Vector spaces are the main objects of interest in functional analysis. In order to understand the content of this thesis, is important to have a good handle of the basic properties of vector spaces, norms and (to a lesser extent) inner products. Most of the definitions here can be found in any linear algebra textbook, such as [9]. Definition A.1. Given a field F, a set V is called a vector space over F if there are operations + : V × V → V and · : F × V → V such that for u, v, w ∈ V and a, b ∈ F the following hold: 1. u + v = v + u (commutativity of vector addition); 2. (u + v) + w = u + (v + w) (associativity of vector addition); 3. There exists 0 ∈ V such that for any x ∈ V we have 0 + x = x (vector additive identity); 4. For any x ∈ V there exists −x ∈ V such that x + (−x) = 0 (vector additive inverse); 5. a(bu) = (ab)u (associativity of scalar multiplication); 52 53 6. (a + b)u = au + bu (distributivity of scalar sums); 7. a(u + v) = au + av (distributivity of vector sums); 8. For 1 ∈ F, 1u = u (scalar multiplication by identity). If V is a vector space over F, the elements of V and F are referred to as vectors and scalars respectively. The operation + is called vector addition and · is called scalar multiplication. We say that V is a real or complex vector space if F = R or C respectively. Definition A.2. A subset W of a vector space V is called a subspace of V if W is closed under the operations of addition and scalar multiplication on V . Remark A.3. Every subspace W of a vector space V is a vector space under the operations of vector addition and scalar multiplication inherited from V . Definition A.4. Let S be a subset of a vector space V over F. A vector v ∈ V of the ! form v = ni=1 an vn where n ≥ 1, ai ∈ F, and vi ∈ S for 1 ≤ i ≤ n is called a linear combination of vectors in S. Definition A.5. If V is a vector space and S ⊆ V is non-empty, we let the linear span (or simply the span) of S be the set (S) of all linear combinations of vectors in S. In the case that S = ∅, we define (S) = {0}. Alternatively, (S) is the intersection of all subspaces of V containing S. Remark A.6. If S is a subset of a vector space V , then (S) is a subspace of V . Moreover, S is a subspace of V if, and only if, (S) = S. Definition A.7. The vectors of a subset S of a vector space V are said to be linearly dependent if there is some positive integer n, vectors v1 , v2 , . . . , vn ∈ S, and scalars ! a1 , a2 , . . . , an , not all zero, such that ni=1 an vn = 0. Otherwise, the vectors of S are said to be linearly independent. 54 Definition A.8. If V is a vector space, a subset β of V is called a basis for V if the vectors of β are linearly independent and (β) = V . The next Theorem ensures the existence of a vector space basis. The proof relies on the widely accepted (but somewhat controversial) ‘Axiom of Choice’ from the ZFC axioms of set theory. Theorem A.9. Given any linearly independent subset S of a vector space V there is a basis β for V such that S ⊆ β. In particular, every vector space has a basis. Theorem A.10. If V is a vector space, then any two bases of V have the same cardinality. Definition A.11. The dimension of a vector space V , denoted by dim(V ), is defined to be the cardinality of a basis β for V . If dim(V ) = n where n is an integer, we say that V is finite-dimensional or n-dimensional, otherwise V is said to be infinite-dimensional. Remark A.12. If the vectors of S ⊆ V are linearly independent, then |S| ≤ dim(V ). Remark A.13. If W is a subspace of V , then dim(W ) ≤ dim(V ). In particular, every subspace of a finite-dimensional vector space is finite-dimensional. Appendix B Norms and Inner Products In the most simple vector spaces, such as Rn , it is intuitive that each vector is determined by a length and a direction in relation to other vectors; however, the concepts of length and direction are not found in the basic definition of a vector space (Definition A.1). In fact, for certain vector spaces we are not able to define these concepts in a natural way. For this reason, we distinguish the vector spaces where length (norm) and direction relative to other vectors (inner product) can be defined to suit our intuition. Definition B.1. Given a vector space V over a field F = R or C, a norm on V is a function +·+ : V → R+ such that for u, v ∈ V and a ∈ F the following hold: 1. +u+ = 0 if, and only if, u is the zero vector (positive definiteness); 2. +au+ = |a| +u+ (positive scalability); 3. +u + v+ ≤ +u+ + +v+ (triangle inequality). A vector space V with a norm is called a normed space. Definition B.2. If V is a vector space over a field F = R or C, an inner product on V is a map (·, ·) : V × V → F such that for x, y, z ∈ V and a, b ∈ F the following hold: 55 B.1. SOME NORM TOPOLOGY 56 1. (x, x) ≥ 0 where equality holds if, and only if, x = 0 (positive definiteness); 2. (x, y) = (y, x) where the bar denotes complex conjugation (conjugate symmetry); 3. We have (ax + by, z) = a (x, z) + b (y, z) (linearity of the first argument). A vector space V with an inner product is called an inner product space. Note. The notation for an inner product should not be confused with that of a linear span (Definition A.5). An inner product has two arguments, both of which are vectors, while the linear span has one argument, a set. Definition B.3. For a subset S of an inner product space V , the orthogonal complement is defined to be the set S ⊥ = {x ∈ V : (x, y) = 0 for all y ∈ S} (read “S perp”). Remark B.4. If W is a subspace of V , then W ⊥ is a subspace of V . Remark B.5. If V is an inner product space, then the function +·+ : V → R+ defined . by +x+ = (x, x) is a norm on V . This is referred to as the norm induced by (·, ·), or simply the induced norm. Unless stated otherwise, the norm on an inner product space is the induced norm. B.1 Some Norm Topology Topological notions such as open, closed and compact sets are crucial to many areas of mathematics, including functional analysis. One of the most important topologies for a set is the metric topology induced by a distance function called a metric. Normed spaces inherit a natural distance function from their norm: the distance between vectors x and y is defined by +x − y+. Thus, a very natural topology for a normed space is the norm topology, also called the strong topology, induced by this distance function. B.1. SOME NORM TOPOLOGY 57 Definition B.6. Given a vector x0 ∈ X and a positive real number δ, we define the open ball of radius δ around x0 to be the set Bδ (x0 ) = {x : +x − x0 + < δ}. Definition B.7. Given a normed space V we say that a subset S of V is open if for every x0 ∈ S there exists δ > 0 such that Bδ (x0 ) ⊆ S. A subset F of V is said to be closed if V \F is open. Remark B.8. If {Uα }α and {Ui }ni=1 are collections of open sets, then ∪α Uα and ∩i Ui are open. Remark B.9. Every open ball is an open set of X. Definition B.10. Given a normed space V , we define the open unit ball and unit sphere of V to be the sets BV = B1 (0) and SV = {x ∈ V : +x+ = 1} respectively. Continuous functions are vital to many areas of mathematics. Through the study of continuous functions we are given a way of comparing, identifying, and understanding the topological properties of different mathematical objects. The following definition is stated for normed spaces in particular, but applies to general topological spaces. Definition B.11. Let V and W be normed vector spaces. A function f : V → W is continuous if f −1 (U ) is an open set of V whenever U is an open set of W . Definition B.12. Given a subset S of a normed space V a point x0 ∈ V is said to be a limit point of S if for every open set U containing x0 there exists some x ∈ S\{x0 } such that x ∈ U . We let S & denote the set of limit points of S. We call the set S = S ∪ S & the closure of S. Alternatively, S is the intersection of all closed sets containing S. Remark B.13. A subset S ⊆ V is closed if, and only if, S = S. Definition B.14. We say that a subset S of a normed space V is dense in V if S = V . B.1. SOME NORM TOPOLOGY 58 Definition B.15. Given a subset S of a normed vector space V , the closed linear span of S is the set [S] defined by [S] = (S). Alternatively, [S] is the intersection of all closed subspaces of V containing S. Remark B.16. For any subset S of a normed vector space V , the closed linear span of S is a subspace of V . Definition B.17. For a subset S of a normed space V a collection C is said to be a cover for S if S ⊆ ∪U ∈C U . In the case that each element of C is open in V , we say that C is an open cover for S. Definition B.18. We say a subset S of a normed space V is compact if every open cover of S has a finite subcover; That is, a finite subcollection which is also a cover for S. Definition B.19. A subset S of a normed space V is said to be relatively compact if S is compact. Definition B.20. A subset S of a normed space V is said to be bounded if there is a real number M such that +x+ ≤ M for every x ∈ S. A sequence (xn )∞ n=0 ⊆ V is bounded if the set S = {xn : n ≥ 0} is bounded. Definition B.21. Let (xn )∞ n=0 be a sequence in a normed space V . Given x ∈ V , we say that (xn )∞ n=0 converges (strongly) to x, written xn → x, if for every open set U containing x there exists some N ∈ N such that if n > N , then xn ∈ U . If (xn )∞ n=0 converges to some x ∈ V , then we simply say that (xn )∞ n=0 converges. Remark B.22. For a subset S ⊆ V , the closure of S is precisely the set of vectors x ∈ V such that there exists a sequence (xn )∞ n=0 ⊆ S converging to x. B.2. BANACH SPACES AND HILBERT SPACES B.2 59 Banach Spaces and Hilbert Spaces The real numbers have the important property of being complete. In real analysis, the property of completeness is usually introduced in terms of suprema and the ‘least upper bound property’ of R as an ordered field. Completeness can also be looked at in terms of the convergence of Cauchy sequences. In this way, the concept of completeness generalizes to arbitrary metric spaces and, in particular, normed spaces. Definition B.23. Given a normed space V we say that a sequence (xn )∞ n=0 ⊆ V is a Cauchy sequence if for every ε > 0 there exists an integer N such that if n, m > N , then +xn − xm + < ε. A normed space V is said to be complete if every Cauchy sequence in V converges. Definition B.24. Given a normed space V , the completion of V is the smallest complete normed space X containing V . Definition B.25. A normed space X which is complete is called a Banach space. Remark B.26. Every closed subspace of a Banach space is a Banach space. Example B.27. The following are examples of Banach spaces: 1. The n-dimensional real or complex space Rn or Cn with term wise vector addition .!n 2 and scalar multiplication and the Euclidean norm +x+ = i=1 |xi | for x ∈ X; 2. The space !p for 1 ≤ p < ∞ is the set of all sequences x = (xn )∞ n=0 ∈ R or C such ! p that n |xn | converges. Vector addition and scalar multiplication are defined term wise (in the usual way) and the norm on !p , sometimes called the p-norm, is .! defined by +x+ = p n |xn |p ; 3. The set of all sequences x = (xn )∞ n=0 in R or C of bounded modulus is denoted by !∞ , with the usual operations and with the uniform norm or supremum norm +x+ = supn {|xn |}; 60 B.2. BANACH SPACES AND HILBERT SPACES 4. Let C[0, 1] be the set of all real continuous functions on the interval [0, 1] ⊂ R. The set C[0, 1] becomes a Banach space under point wise addition and scalar multiplication and the supremum norm +f + = supx∈[0,1] {f (x)}. Definition B.28. A Banach space X is said to be separable if X has a countable dense subset. Example B.29. The spaces Rn , Cn , and !p for 1 ≤ p < ∞ (see Example B.27) are separable spaces. This follows from two basic facts: (1) Q is dense in R; and (2) The set of finitely non-zero sequences over Q is countable and dense in !p , 1 ≤ p < ∞. However, the space !∞ is non-separable. To see this, consider the set A of sequences having coordinates equal to 0 or 1. Clearly every such sequence is bounded, and so we have A ⊆ !∞ . The elements of A are in a 1-to-1 correspondence with the set of all subsets of N, and therefore A is uncountable.1 Also, any pair of distinct points x, y ∈ A satisfies +x − y+ ≥ 1. Therefore, the sets of the collection C = {B1/2 (x) : x ∈ A} are mutually disjoint. Suppose that S is any dense subset of !∞ . Then every set in C must contain an element of S, and therefore S is uncountable. Definition B.30. A complete inner product space is called a Hilbert space. Example B.31. The following are examples of Hilbert spaces: 1. Rn with the standard inner product (also called the dot product) (x, y) = for x, y ∈ Rn , which clearly induces the Euclidean norm; 2. Cn with the standard inner product (x, y) = induces the Euclidean norm; 1 !n i=1 xi yi !n n i=1 xi yi for x, y ∈ C , which again The collection of subsets of a set B is known as the power set of B, denoted by P(B). A famous result of Cantor states that |P(B)| > |B| for any set B. Thus, P(N) is uncountable. 61 B.2. BANACH SPACES AND HILBERT SPACES 3. The vector space !2 with the inner product (x, y) = ! n xn yn , where yn denotes the complex conjugate of yn . This inner product induces the 2-norm from Example B.27. Appendix C Linear Operators Loosely speaking, linear operators are functions between vector spaces which preserve the basic algebraic structure. One may think of linear operators as the natural categorical morphisms between vector spaces, similar to the homomorphisms in group theory, or continuous functions in topology. Definition C.1. Given vector spaces V and W over a common field F, a linear operator is a function T : V → W such that for all vectors u, v ∈ V and scalars a, b ∈ F we have T (au + bv) = aT (u) + bT (v). For x ∈ V we often write T x when referring to T (x). Definition C.2. Let V be a vector space. The operator IV : V → V defined by IV x = x for all x ∈ V is called the identity operator. We simply denote IV by I when the space is clear from context. Definition C.3. We let L(V, W ) denote the set of all linear operators mapping V to W , or simply L(V ) in the case that W = V . Remark C.4. The set L(V, W ) is a vector space under the usual point wise addition and scalar multiplication; ie. (aT1 + bT2 )x = aT1 x + bT2 x for all x. 62 63 Definition C.5. For a linear operator T : V → W we define the kernel (or nullspace) of T to be the set ker(T ) = {x ∈ V : T x = 0}. The range of T is defined by T (V ) = {T x : x ∈ V }. Remark C.6. If T : V → W is a linear operator, then ker(T ) is a subspace of V and T (V ) is a subspace of W . Definition C.7. Given a linear operator T : V → W , we define the rank of T to be the dimension of T (V ). We say that T is a finite-rank operator if T (V ) is finite-dimensional. The next definition introduces the important idea of a bounded operator, the main focus of this thesis. Definition C.8. Given normed spaces V and W the operator norm is defined by +T + = sup{+T x+ : +x+ ≤ 1} for T ∈ L(V, W ). An operator T : V → W is said to be bounded if +T + < ∞. The set of all bounded linear operators mapping V to W is denoted by B(V, W ), or B(V ) in the case that W = V . Note. Be careful to recognize that there are in fact 3 different ‘norms’ in the above definition: the norms on V and W as well as the operator norm on B(V, W ). Remark C.9. The operator norm satisfies the conditions of a norm on B(V, W ) (Definition B.1). If V and W are Banach spaces, then B(V, W ) is a Banach space under the operator norm and the operations inherited from L(V, W ). While all linear operators preserve the algebraic structure of normed spaces, only certain operators preserve the topological structure as well. As it turns out, these are precisely the bounded operators. Remark C.10. An operator T : V → W is bounded if, and only if, it is continuous. Remark C.11. If T is a bounded operator and (xn )∞ n=0 is a sequence such that xn → x, then T xn → T x. 64 Definition C.12. Given normed spaces V and W , an injective continuous linear operator T : V → W such that T −1 is continuous on the range of T is called an isomorphism. Note that this definition of an isomorphism does not match the usual categorical definition, since isomorphisms are typically surjective. This definition more closely resembles a topological imbedding, or categorcial section. Nonetheless, this definition agrees with “normed space custom.” [35, p. 31] Remark C.13. If T is a linear operator on a normed space V , then for all x ∈ V we have +T x+ ≤ +T + +x+. In fact, if T is bounded, then +T + is the smallest real number such that +T x+ ≤ +T + +x+ for all x ∈ V . Definition C.14. Given linear operators A and T on a vector space V , the operator AT on V is obtained by composing A with T . That is, AT x = A(T (x)) for x ∈ V . We define T 0 = I and for n ≥ 1 we let T n = T T n−1 . Remark C.15. For linear operators A and T on a normed space V , we have +AT + ≤ +A++T +. The following result is a well-known fact about bounded operators on dense subspaces. Proposition C.16. Let Y be a dense subspace of a Banach space X and let A : Y → Y be a linear operator. If A is bounded, then A extends to an operator T : X → X such that +T + = +A+ and T (y) = A(y) for all y ∈ Y . Appendix D Polynomials The study of polynomials and their roots is a fundamental topic in mathematics. In this thesis, we often encounter the idea of polynomial combinations of linear operators. Several basic facts about polynomials are required so that these encounters can go smoothly. Definition D.1. Let F be a field. A polynomial over F in variable t is a function ! n p : F → F of the form p(t) = ∞ n=0 an t where each an ∈ F and an #= 0 for only finitely many n. We let F[t] denote the set of all such polynomials.1 Definition D.2. The degree of a non-zero polynomial p(t) = !∞ n=0 an t n over F is defined to be the maximum index n ≥ 0 such that an #= 0, written deg(p). Polynomials of degree 0 are said to be constant. By convention, the degree of the zero polynomial is undefined. Definition D.3. We say λ ∈ F is a root of p if p(λ) = 0. Theorem D.4 (Factor Theorem). Given a field F, let p ∈ F[t]. A scalar λ is a root of p if, and only if, p can be written as p(t) = (t − λ)q(t) for some polynomial q over F. 1 Technically, the set F[t] is a ring under the usual operations of addition and multiplication of polynomials. 65 66 Definition D.5. If a polynomial p over F can be written as p(t) = r(t)q(t) for polynomials p and q, then we say that r divides p, written r | p. Remark D.6. If p(t) = q(t)r(t) where p, q and r are polynomials over F, then deg(p) = deg(r) + deg(q). The field of complex numbers has the nifty property of being algebraically closed. That is, every polynomial over C has a complex root. This result is known as the Fundamental Theorem of Algebra.2 Theorem D.7 (Fundamental Theorem of Algebra). Every non-constant polynomial over C has a root in C. These corollaries follow from the Fundamental Theorem of Algebra, Factor Theorem, and properties of real polynomials. Corollary D.8. If p is a polynomial over C, then there is a polynomial r of degree 1 so that r | p. Corollary D.9. If p is a polynomial over R, then there is a polynomial r of degree either 1 or 2 so that r | p. Given a polynomial p and an operator T : X → X, it is often useful to define an operator p(T ) based on p and T . Definition D.10. Given a linear operator T on a vector space V over F and polynomial ! n p ∈ F[t], we define an associated operator p(T ) = ∞ n=0 an T using point wise addition and scalar multiplication, where T n is defined as in Definition C.14. 2 The Fundamental Theorem of Algebra has an interesting history. Leibniz and Nikolaus Bernoulli both believed that they had found counterexamples in the early 18th century, until Euler proved them wrong. Famous mathematicians including d’Alembert, Euler, de Foncenex, Lagrange, Laplace, Wood, and Gauss thought that he had proofs, but there were a gaps. Finally, a rigorous proof was published by Argand in 1806, settling any controversy. 67 Remark D.11. Let T be a linear operator on a normed space V over F and let p be a polynomial over F. If T is bounded, then so is p(T ). Appendix E Properties of !1 The counterexample to the invariant subspace problem in Sections 4.2 - 4.9 require some special properties about the Banach space !1 . We include this here. Let (fi )∞ i=0 be the sequence of standard unit vectors in !1 . Every vector x in !1 ! 1 can be represented uniquely as a sum x = ∞ i=0 λi fi for scalars λi . For each m ≥ 0, we let Fm denote the finite-dimensional space ({fi : i = 0, 1, . . . , m}). We let F denote the set of all finitely non-zero sequences. That is, F = ∪m Fm . Clearly F is a dense subspace of !1 . The following gives a useful formula for calculating the norm of an operator on !1 . Proposition E.1. If A is a linear operator on !1 or F , then +A+ = sup{+Afi +}i . Proof. Consider an operator A on !1 or F . The bound +A+ ≥ sup{+Afi +}i is clear by definition of the operator norm. For the converse, let x = (x0 , x1 , . . . ) be any vector such that +x+ ≤ 1. Then we have the following bound * + ,* ∞ ∞ ∞ * * % % % * * +Ax+ = *A xi f i * ≤ |xi |+Afi + ≤ |xi | sup{+Afi +} * * i=0 1 i=0 i=0 This means that (fi )i is a Schauder basis for "1 . This is actually true for all sequence spaces "p , 1 ≤ p ≤ ∞. 68 69 = sup{+Afi +} ∞ % |xi | ≤ sup{+Afi +} (since +x+ ≤ 1). i=0 The result follows. The standard norm on !1 simply calculates the sum of |λi | where x = !∞ i=0 λi fi . We can also define a 1-norm with respect to a basis (ei )∞ i=0 for F different from (fi )i . Definition E.2. Let (ei )∞ i=0 be a basis for F . We may a define norm + · +e on F by +x+e = n % |λi | where x = i=0 n % λi ei for some n. i=0 For our purposes, this norm is referred to as the e-norm. Note that + ·+ e is defined on F rather than !1 . This is sometimes necessary in order to guarantee convergence. Given such a sequence (ei )∞ i=0 and m ≥ 0, let Em = ({e0 , e1 , . . . , em }). 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