Linear operators with wild dynamics

If $X$ is a separable infinite dimensional Banach space, we construct a bounded and linear operator $R$ on $X$ such that $$ A_R=\{x \in X, \|R^tx\| \rightarrow \infty\} $$ is not dense and has non empty interior with the additional property that $R$ can be written $I+K$, where $I$ is the identity and $K$ is a compact operator. This answers two recent questions of H\'ajek and Smith.


Introduction
Let X be a Banach space (we will sometimes distinguish the real and complex cases) and R a bounded linear operator on X. For x in X, let be the orbit of x under the action of R. If there exists x ∈ X such that O R (x) is dense in X, R is called a hypercyclic operator (and x a hypercyclic vector for R). This class of operator has been intensively studied in the last two decades, see [BM] for a nice survey on this topic. Furthermore, the orbits are connected with the invariant subset problem which asks if there exists an operator on X with non trivial invariant closed subset. Indeed, R does not have any trivial invariant closed subset if and only if for each x = 0, O R (x) is dense in X. In the space ℓ 1 , Read [R] constructed such an operator. But in the Hilbert space, the problem is still open. Here, we will be interested in the orbits which tend to infinity. The following is our main result.
Theorem 1.1. Let X be an infinite-dimensional separable, real or complex, Banach space. There exists a linear continuous operator R on X such that, if we put A) A R and B R have non-empty interior and {A R , B R } form a partition of X. B) R can be written I + K, with K a compact operator.
As a direct consequence, we obtain the following more general result.
Corollary 1.2. The conclusion of Theorem 1.1 remains true if one only assumes that X has a closed and separable subspace of infinite dimension which is complemented in X.
Proof. We write X = X 0 ⊕ Y where X 0 ⊂ X is a closed separable Banach space of infinite dimension and Y is a closed subspace of X. We then consider R = R 0 ⊕ I where R 0 is an operator on X 0 which satisfies Theorem 1.1. Since R t = R t 0 ⊕ I, we get A R = A R0 + Y , B R = B R0 + Y and it is clear that R has the required properties.
Theorem 1.1 is motivated by a conjecture of Prǎjiturǎ [P] which was recently solved negatively by Hájek and Smith [HS]: Conjecture 1.3. Let R be an operator on a Banach space, then A R is either empty or dense.
Hájek and Smith showed part A) of Theorem 1.1 when X has a symmetric basis. Therefore, they asked if the theorem could be true for a space which has only an unconditional basis (which is a weaker property than having a symmetric basis). We show that in fact no hypothesis of basis existence is necessary. Furthermore, a second question they asked was: is it possible to choose the operator R of the form I + K, with K a compact operator? The reason for this question is that they observed that for a compact operator K and more generally, for a strictly singular operator S (that is an operator which is not an isomorphism when restricted to any infinite-dimensional subspace), A S was either empty or dense. On the other hand, there exist some spaces with very few operators: Gowers and Maurey [GM] constructed a Banach space where each operator has the form λI + S, with S strictly singular and even more recently, Argyros and Haydon [AH] found a space where each operator has the form λI + K, with K compact. Thus, if some operators satisfying part A) of Theorem 1.1 exist on any separable Banach space, it must be true that they can take the form λI + K, and we show that it is indeed possible.
Note that if ( R t ) t 0 is unbounded, by the uniform boundedness principle, So for any operator R, A ′ R is either dense or empty. However, this does not show anything for A R since there exist some operators such that R t → ∞ but for all x, R t x ∞. For example it is the case for the following weighted backward shift [HS] defined on ℓ p (1 p < ∞): However, Müller and Vršovský [MV] showed that if R t grows fast enough, namely if Let us explain the organization of this paper. Section 2 is devoted to a preliminary result: we prove a separating lemma in finite dimension (which might be of independant interest). We then use it in section 3 to construct an operator satisfying Theorem 1.1 (we treat separately the complex and real cases). In section 4, we present a kind of version of the operator constructed by Hájek and Smith with a slightly simplified proof. In section 5, we make some comments on the two operators and we prove that in fact, there are many operators which fail Prǎjiturǎ's conjecture. Let us settle some notation: if X is a Banach space, we denote by L(X) the set of linear and bounded operators on X, by GL(X) ⊂ L(X) the set of invertible operators and by A X the set of operators which satisfy part A) of Theorem 1.1. Until the end of the paper X will always be assumed to be infinite dimensional (if X is finite dimensional, Prǎjiturǎ's conjecture is true, see [HS]).

An asymptotic separating lemma in finite dimension
Here d 2 is an integer and K = R or C.
We will do the proof in the real case. Let us denote by S = S R d the unit sphere of R d and put C = S ∩ H. We first prove the following lemma.
Lemma 2.2. There exists a constant K > 0 and a sequence (u n ) ⊂ C such that for each x ∈ C, there exists a sequence of integers (p n ) which goes to infinity, and satisfies for each n, p n n and x − u pn K n 1/d−1 . Proof of Lemma 2.2. There exists L > 0 such that for each k, we can find a 2 −k net H k of C with |H k | L(2 k ) d−1 where |H k | denotes the number of elements of H k . The sequence (u n ) is obtained by enumerating first the elements of H 1 , then all the elements of H 2 and so on. Let us fix x ∈ C and check the property of the lemma. For n ∈ N, there exists an integer k such that |H 1 | + . . . + |H k | n < |H 1 | + . . . + |H k | + |H k+1 |, in particular, we get where C d > 0 only depends on d. By the definition of H k , there exists an element u pn ∈ H k such that x−u pn 2 −k ( C d n ) 1/d−1 . Since k−1 |H 1 |+. . .+|H k−1 | < p n |H 1 | + . . . + |H k |, p n n, lim p n = ∞ and the lemma is shown.
Proof of Proposition 2.1. For each n, we can select a linear form f n such that f n (u n ) = 0, Kerf n ⊂ H and f n = n 1/2(d−1) . If x / ∈ H, then thus we have 1). For 2), we may assume that x = 0 and apply the lemma with y = x x ∈ C, we obtain a sequence (p n ) with p n n, lim p n = ∞ and y − u pn K n 1/d−1 . For each n we have |f pn (x)| = x |f pn (y − u pn )| f pn y − u pn x K n 1/2(d−1) x which proves 2) and achieves the proof.
For the complex case, the proof is almost the same. The only difference is that for ǫ > 0 the entropy of the unit sphere is of the order of 1 ǫ 2d−1 which gives a different estimate of x − u pn in the lemma, so we have also to change the exponent in f n in the proof of the proposition.

Construction of Operators
3.1. Complex case. In this sub-section, we prove the complex version of Theorem 1.1.
Recall that an operator N is nuclear if there exist some continuous linear functionals x * k and some vectors x k such that N can be written as These operators are obviously compact, being limits of finite rank operators.
To construct our operator, we will use the following theorem which says that a separable Banach space supports a sequence (e n ) n 1 which behaves more or less like a Schauder basis [OP].
Theorem 3.2. If X is a separable Banach space and ǫ > 0, one can find sequences (e n , e * n ) ⊂ X × X * such that: i) span(e n , n 1) is dense in X.
ii) e * n , e m = δ n,m where δ n,m = 0 if n = m and 1 if n = m. iii) sup e n e * n 1 + ǫ.
For our case, we won't need the precise estimate with ǫ. However, observe that by replacing e n by e n / e n and e * n by e n e * n we can suppose that e n = 1, K = sup e * n < ∞ and properties i) and ii) of Theorem 3.2 are still satisfied. We then consider a closed subset F ⊂ span(e 1 , e 2 ) which is a union of complex lines passing through 0 and the sequence (f k ) of linear forms associated on span(e 1 , e 2 ) in the asymptotic separating lemma (we just use the particular case of the lemma d = 2). Let us denote by P the projection defined on X onto span(e 1 , e 2 ) by P x = e * 1 , x e 1 + e * 2 , x e 2 . P is continuous with P 2K. Then, we put λ 1 = λ 2 = 1 and for k 3, λ k = e iπ m k where (m k ) is a sequence of positive integers such that m k |m k+1 and which goes very fast to infinity, namely We denote by c 00 the subspace generated by the sequence (e n ) n 1 . For x ∈ c 00 , x = n k=1 x k e k , we set This defines a bounded operator S on c 00 , indeed for x = n k=1 x k e k ∈ c 00 , Hence, S extends to a bounded operator on X (because c 00 is dense in X). We still denote this operator by S.
For x ∈ X, we now define an operator R by Note that the second sum is a bounded operator because ∞ k=3 1 then we will show that A R =Ã R and B R =B R so R will be the required operator (provided the relative interior of F in span(e 1 , e 2 ) is non-empty and F = span(e 1 , e 2 )). Since λ 2m k k = 1 and more generally λ 2mn k = 1 for n k (by the hypothesis of divisibility of the m k ), the operator S has the following property.
Proof. We first show that the sequence ( S 2m l ) is bounded. If we fix x = n k=1 x k e k such that x = 1, we have By density of c 00 in X, this implies that the sequence ( S 2m l ) is bounded.
To see the claim of the lemma, fix x ∈ X, ǫ > 0 and y ∈ c 00 such that y − x ǫ, we have Since for large enough l, S 2m l y = y, we have the conclusion.
We now explicit the iterates of R.
Lemma 3.5. For x ∈ X and t 1, we have Proof. The statement is true for t = 1, we proceed by induction and suppose it is true for t, we have λ l,t m l−1 f l (P x)e l , so P y = 0. Since λ 1 = λ 2 = 1, we have P S t x = P x, and since λ k λ k,t + 1 = λ k,t+1 , we deduce the formula for t + 1.
We now study the behaviour of λ k,t . We have the following easy fact.
Proof. i) is just the triangle inequality. ii) is obvious since λ k,t can be computed easily as a geometric sum: For iii), we get The desired lower bound follows from the inequalities : sin(y) 2 π y for y ∈ [0, π 2 ] and | sin(y)| |y| for y ∈ R.
We can now pass to the proof of the main theorem. Let x ∈ c 00 ∩Ã R , x = n k=1 x k e k . Fix k 3, we set δ k n = 1 if k n and δ k n = 0 if k > n. We have We claim thatÃ R ∩ c 00 is dense inÃ R . Since the last inequality involves only continuous functions, it stays true for x ∈Ã R and for k 3, t 1. Now, let us fix k 1 and m k−1 t m k , using the last inequality and the previous fact, we get This proves that R t x → ∞ whenever x ∈Ã R because by the separating lemma |f k (P x)| → ∞. For the proof of the claim, if x ∈Ã R , there exists (z k ) ⊂ c 00 such that x = lim z k . Write z k = z 1,k e 1 + z 2,k e 2 + y k with y k ∈ span(e l , l 3). If P x = x 1 e 1 + x 2 e 2 , the sequencez k = x 1 e 1 + x 2 e 2 + y k has the required properties. Now, let x ∈B R , then there exists a subsequence (k n ) such that |f kn (P x The first sum is equal to 0 according to the previous fact. Using the fact again, we see that the second term goes to 0 because |λ kn ,2m kn −1 | m kn −1 2. For the last sum, k k n + 1 so k n − 1 k − 2, and m kn−1 m k−2 , hence Since k 1 m k−2 m k−1 f k < ∞, the last term goes to 0. On the other hand, S 2m kn−1 x goes to x, this shows that lim R t x − x = 0 whenever x ∈B R .
Finally, let us check the last claim of our theorem, by a density argument we see that for x ∈ X and (e k ) and (e * k ) are bounded, N is nuclear and R = I + N .
Remark 3.7. The statement of our theorem can be sharpened: it is apparent from the definition of R that for any ǫ > 0, one can choose the sequence (m k ) to have R − I ǫ. This proves that I is in the norm closure of A X .
Remark 3.8. Although Corollary 1.2 does not apply to ℓ ∞ , we can see that our example works in ℓ ∞ . We thus conjecture that operators with the properties stated in Theorem 1.1 can be constructed in arbitrary Banach spaces.

Real case.
We now show the version of Theorem 1.1 for real spaces.
Theorem 3.9. Let X be a separable real Banach space. Then, there exists R ∈ A X such that R − I is nuclear.
We keep the same notations than in the complex case for F , (f k ), c 00 , (e k ),(e * k ), K and P . We define S on c 00 by the following formulas Se 1 = e 1 , Se 2 = e 2 and for k 2: Se 2k−1 = cos(θ k )e 2k−1 + sin(θ k )e 2k and Se 2k = − sin(θ k )e 2k−1 + cos(θ k )e 2k where θ k = π m k and (m k ) satisfy the same hypotheses of growth and divisibility as in the complex case. Thus in the (algebraic) basis (e k ), S is a block diagonal operator built from rotation matrices. We check that S is bounded on c 00 , so it extends to X and we check that S 2mn x → x. We then define R (the required operator) by The computation of the iterates of R gives l=0 sin(lθ k ) e 2k−1 . We thus have µ k,t = 0 ⇔ λ k,t = 0, µ k,t 2t and considering the functionals e * 2k−1 and e * 2k : So the inequalities for µ k,t are the same than λ k,t (up to a multiplicative constant) which allows to follow the previous proof (in the complex case) without other changes.

The construction of Hájek and Smith
We present a slightly modified version of the example of Hájek and Smith. The following theorem is of course a particular case of our above theorem. However, we will outline some differences between the two operators in the last section.
Theorem 4.1. Let X be a Banach space (real or complex) with a symmetric basis. Then, there exists R ∈ A X .
Recall that a basis (e k ) is said to be symmetric if for every permutation π of the integers, (e π(k) ) is a basis and is equivalent to (e k ). If for x = ∞ k=1 x k e k ∈ X, we set then . 0 is an equivalent norm on X which satisfies for every π, In the rest of this section, we suppose that our norm . has the two precedent properties (such a norm is sometimes called a symmetric norm).
In the complex case, our operator S was built with roots of unity, S will now be a cycle operator. As in the original paper of Hájek and Smith, we will prove some finite-dimensional estimates. However, Hájek and Smith make some estimates in ℓ p and c 0 and then combine it with a theorem of Tzafriri [T]. We will give a direct and short proof of these estimates.

Local estimates.
Let Z be a space of dimension H 1 with a symmetric basis (e 1 , . . . , e H ) such that H 4m (m and H are positive integers). We define an operator S on Z by Se i = e i+1 for 1 i H − 1 and Se H = e 1 . We will be interested in the behaviour of the sequence: w t = (I + . . . i=m+1 e i (this is motivated by the fact that (w t ) will appear in the study of the iterates of the operator in the general case). We have: ii) An easy computation shows that for 2m t H − 2m, w t has two bumps of height m and of length 2m − 1, one with positive terms, the other with negative terms: w t = e 1 +2e 2 + . . . + me m + (m − 1)e m+1 + . . . e 2m−1 − (e t+1 + 2e t+2 + . . . + me t+m + . . . e t+2m−1 ).
If we keep the coordinates which are superior to m 2 , we get w t m 2 e 1 + . . . + e 2m = m 2 w .
We have also shown that if u t is the decreasing rearrangement of |w t |, then for 2m t H − 2m and 2m s H − 2m, u t = u s . Another computation shows that for 1 s 2m and 2m t H − 2m, u s u t and hence that w s = u s u t = w t because the norm is symmetric. For H − 2m s H, we obtain the same estimates as in the case 1 s 2m. This shows that w t attains its maximum for 2m t H − 2m.

The operator of
Hájek and Smith. We assume that X is a Banach space with a symmetric and normalized basis (e n ) and write X = span(e 1 , e 2 ) ⊕ span(e n , n 3) = span(e 1 , e 2 ) ⊕ Y.
Next, we are going to define S on X (we give it the same name as in the finite dimensional case because it will be a sum of these operators on finite blocks). We choose some sequences (H k ) and (m k ) of positive increasing integers such that H k 4m k and H k |H k+1 . We define σ k the following permutation of order H k : σ k = (1, 2, . . . , H k ), that is σ k (i) = i + 1 for 1 i H k − 1 and σ k (H k ) = 1. We enumerate the basis vectors of Y as follows span(e n , n 3) = span(e i,1 , 1 i H 1 ) ⊕ span(e i,2 , 1 i H 2 ) ⊕ . . . So, each element x of X can be written and we put S has the following properties: Proof. i) is because the norm is symmetric. For ii), it is enough to check it for y ∈ Y . Fix ǫ > 0, there exists an integer N such that For k N , H k is a multiple of H N , so for 1 l N and 1 i H l , S H k (e i,l ) = e i,l , hence Since S is an isometry, we get for k N : S H k y − y 2ǫ.
We now define the operator R = R HS (we use the same notation and assumptions as in the previous section for F , (f k ) and P , and the definition of the setsÃ R andB R also remain unchanged). For x ∈ X, we put The iterates of R are given by (same proof as in previous section) Now, let us choose the various constants: first, we introduce a sequence of positive and increasing integers (a k ) such that ∞ k=1 f k a k < ∞ and 1 + a k | 1 + a k+1 .
Then, we put m k = a 1 . . . a k , H k = 2(m k + m k+1 ). We have The constant ǫ k is chosen such that v k,2m k = 1. The finite dimensional estimates give v k 2 m k . We deduce from this that R is well defined and bounded because The following lemma is now a reformulation of the finite dimensional estimates.
Proof. i) and ii) were proved before. For iii), use the triangle inequality and v k The proof of the theorem is now very close to the one we produced before. First, because the norm is symmetric and the (v k,t ) k 1 have disjoint supports, we get for k 1 and For each t, there exists k such that 2m k t 2m k+1 = H k − 2m k , thus Lemma 4.5 gives v k,t = 1 and we see that R t x → ∞ when x ∈Ã R . When x ∈B R , the proof is also in the same spirit as in section 3 (divide the sum into three parts) and if |f kn (P x)| → 0, check that R H kn −1 x → x using Lemma 4.5. Details are left to the reader.

Further results and remarks
5.1. Difference between the operators. We denote by σ(R) the spectrum of an operator R, by σ p (R) the set of its eigenvalues, by r(R) its spectral radius, and we only consider complex spaces. Let also D denote the unit closed complex disk and ∂D the unit complex circle. Let us recall that A X is the set of operators which satisfy part A) of Theorem 1.1 and that by Müller and Vršovský's Suppose that R ∈ A X , then we must have r(R) = 1. Indeed, if r(R) < 1, by the spectral radius formula, R k → 0 which of course contradicts A R = ∅. If r(R) > 1, then if we fix 1 < a < r(R), we get for k large enough R k a k (still with the spectral radius formula), thus ∞ k=1 1 R k < ∞ and A R is dense which is impossible. In particular, any operator R ∈ A X has a spectrum contained in the unit closed disk and its spectrum has to intersect the unit circle. More precisely, in the two studied examples (which are built with a strong analogy), we have: Proof. Our operator can be written as I + N , thus we have σ(R) = {1 + λ, λ ∈ σ(N )}. Since N is compact, σ(N ) = σ p (N ) ∪ {0}. Let λ ∈ σ p (N ) and If λ = 0, then x 1 = x 2 = 0, that is P x = 0 and we get that λ = λ k − 1 for some k 3. On the other hand, N e k = (λ k −1)e k for k 3, hence σ(N ) = σ p (N )∪{0} = {λ k − 1, k 1} and σ(R) = {λ k , k 1}. Now, the operator S HS of section 4 (associated to R HS ) has an infinite matrix which is block diagonal with identity on first block, and then some matrices of Frobenius which have X H k − 1 as characteristic polynomial on the k-th block (i.e span(e i,k , 1 i H k )) , so by taking an eigenvector associated to a H k root of the unity for the matrix of Frobenius on the k-th block and completing it with some zeros, we see that where U H k denotes the group of the H k -th roots of the unity. Since H k |H k+1 , G is itself an infinite group of ∂D as an union of increasing groups, thus G is dense in ∂D and since σ(R HS ) is a closed subset, we get ∂D ⊂ σ(R HS ).
The spectrum of R is thus a sequence that goes very fast to 1 whereas the spectrum of R HS is not countable.
Remark 5.2. With the notation above, we can show that if F (the set defined page 4) is more than one line, then 0 / ∈ σ p (N ), that is 1 / ∈ σ p (R). Indeed, if x = ∞ k=1 x k e k is such that N x = 0, then (λ k − 1)x k + f k (P x) m k−1 = 0 for k 3.
If P x = 0, then since |λ k − 1| π m k , we have As F is more than one line, we can find ǫ > 0 and a subsequence of (f k ) with |f k (P x)| ǫ. This is a contradiction because |x k | → 0 since (e k ) is normalized, so P x = 0 and x = 0. If F is just one line (which is a degenerated case for the operator R), then it is possible to have 0 ∈ σ p (N ). Indeed if F = span(e 1 ), then by Proposition 2.1, F ⊆ Kerf k for all k, so f k (e 1 ) = 0 and N e 1 = 0.
On the other hand, we have: Proposition 5.5. If X is a complex Banach space, then the set of operators which satisfy Prǎjiturǎ's conjecture, that is P = {R ∈ L(X), A R = ∅ or A R is dense in X} is dense (for the norm operator) in L(X). In particular, A X has empty interior.

Acknowledgements
I would like to thank my advisor Robert Deville for suggesting this topic and his help during the elaboration of this paper, Luis Sanchez for fruitful conversations and the anonymous referee for useful comments.