Sophie Grivaux

Étienne Matheron

Quentin Menet

ORTHOGONALITY OF INVARIANT MEASURES FOR WEIGHTED SHIFTS by

Keywords: 2020 Mathematics Subject Classification. -47B01, 47A16, 28C20, 28A35 Weighted shifts, invariant measures, product measures, ℓp -spaces

We introduce and study the notion of orthogonality for two operators in the context of weighted backward shifts on ℓppZ`q, 1 ď p ă 8. Two continuous linear operators T1 and T2 acting on a Polish topological vector space X are said to be orthogonal if any two Borel probability measures m1 and m2 on X which are respectively T1 -invariant and T2invariant and satisfy m1pt0uq " m2pt0uq " 0 must be orthogonal. In this note, we provide several conditions on the weights u u u and v v v implying orthogonality or non-orthogonality of the associated weighted shifts Bu u u and Bv v v , and we investigate in some detail the case where the invariant measures are product measures.

To Gilles Godefroy, with affection and admiration 1. Introduction 1.1. Weighted shifts. -This note is a contribution to the study of the dynamics of weighted backward shifts acting on ℓ p -spaces, 1 ď p ă 8. More specifically, we will be concerned with invariant measures for such weighted shifts.

, which deals with the study of cyclic vectors for the unweighted backward shift B on ℓ 2 pZ `q.

, [20],

 [START_REF] Costakis | Topologically mixing hypercyclic operators[END_REF], [START_REF] Bernardes | Expansivity and shadowing in linear dynamics[END_REF], [START_REF] Bernardes | Shadowing and structural stability for operators[END_REF]. For instance, B w w w is hypercyclic, i.e. it admits a vector with a dense orbit, if and only if lim nÑ8 |w 1 ¨¨¨w n | " 8, and it is topologically mixing if and only if |w 1 ¨¨¨w n | Ñ 8 as n Ñ 8. See [START_REF] Grosse-Erdmann | Linear chaos[END_REF] for proofs of these results as well as extensions to a large class of sequence spaces. Observe that these characterizations are the same on any ℓ p . By contrast, chaoticity of a weighted shift on X " ℓ p depends explicitely on p (recall that an operator is chaotic if it is hypercyclic with a dense set of periodic points): B w w w is chaotic on ℓ p if and only ř 8

n"1 1 |w 1 ¨¨¨wn| p ă 8. For the class of weighted shift operators of ℓ p , chaos is an especially important notion since it turns out to be equivalent to strong dynamical properties such as frequent hypercyclicity (the existence of a vector whose orbit visits each non-empty open set along a set of integers having positive lower density) or U-frequent hypercyclicity (same as frequent hypercyclicity with lower density replaced by upper density). Indeed, by an important result of Bayart and Ruzsa [START_REF] Bayart | Difference sets and frequently hypercyclic weighted shifts[END_REF], frequent hypercyclicity and U-frequent hypercyclicity of a weighted shift B w w w on ℓ p are both equivalent to the condition ř 8

n"1 1 |w 1 ¨¨¨wn| p ă 8. We refer the reader to the books [START_REF] Bayart | Dynamics of linear operators[END_REF] and [START_REF] Grosse-Erdmann | Linear chaos[END_REF] for more on frequent hypercyclicity and related questions, and to the paper [START_REF] Charpentier | Chaos and frequent hypercyclicity for weighted shifts[END_REF] for a study of the relation between chaos and frequent hypercyclicity for weighted shifts on a large class of Fréchet sequence spaces.

The condition ř 8

n"1 1 |w 1 ¨¨¨wn| p ă 8 also turns out to be of special importance in the study of invariant measures for the weighted backward shift B w w w . Recall that a Borel probability measure m on X is said to be B w w w -invariant if m `B´1 w w w pAq ˘" mpAq for every Borel set A Ď X. Under the condition ř 8

n"1 1 |w 1 ¨¨¨wn| p ă 8, the operator B w w w acting on X " ℓ p admits plenty of invariant measures, of quite different kinds: discrete measures associated with periodic points, Gaussian measures with full support (with respect to some of which B w w w is ergodic or even strongly mixing), continuous measures with full support which are very far from being Gaussian, ... See [START_REF] Bayart | Frequently hypercyclic operators[END_REF] or [START_REF] Bayart | Dynamics of linear operators[END_REF]Chapter 5] for more information on (shift-invariant) Gaussian measures; and [START_REF] Grivaux | Some new examples of universal hypercyclic operators in the sense of Glasner and Weiss[END_REF], which testifies of the richness of the class of invariant measures for backward shifts by showing that under the condition ř 8

n"1 1 |w 1 ¨¨¨wn| p ă 8, the backward shift B w w w is universal for ergodic system in the sense of Glasner and Weiss [START_REF] Glasner | A universal hypercyclic representation[END_REF]: for every ergodic transformation T on a standard Lebesgue probability space pZ, B, µq, there exists a B w w w -invariant Borel probability measure m on X " ℓ p with full support such that the two dynamical systems pZ, B, µ; T q and pX, B X , m; B w w w q are isomorphic (where B X is the Borel σ-algebra of X).

Invariant measures for weighted backward shifts are thus far from being completely understood (and in some sense, they will never be). In this note, our aim is to contribute to their understanding by studying possible links between invariant measures for two different weighted shifts B u u u and B v v v . More specifically, we will be concerned with the notion of orthogonality of two weighted shifts.

1.2. Orthogonality. -Before giving the definition we will be playing with, we fix some notation. The scalar field is denoted by K. All measures on X " ℓ p pZ `q are understood to be Borel probability measures. Given a weight sequence w w w, we denote by P w w w pXq the set of all B w w w -invariant measures on X. If m and m 1 are two measures on X, we write m K m 1 to indicate that m and m 1 are orthogonal (i.e. mutually singular), and m ! m 1 to indicate that m is absolutely continuous with respect to m 1 . If m ! m 1 and m 1 ! m, we say that m and m 1 are equivalent and we write m " m 1 .

We now define orthogonality of two weighted backward shifts. Definition 1.1. -Two weighted backward shifts B u u u and B v v v on X " ℓ p are said to be orthogonal if the following holds true: whenever m u u u P P u u u pXq, m v v v P P v v v pXq and m u u u pt0uq " 0 " m v v v pt0uq, it follows that m u u u K m v v v .

Removing the δ 0 -parts of the measures, we see that this definition can be reformulated as follows: B u u u and B v v v are orthogonal if and only if whenever m u u u P P u u u pXq and m v v v P P v v v pXq are non-orthogonal, it must be that m u u u pt0uq ą 0 and m v v v pt0uq ą 0. Equivalently: there exists no (Borel probability) measure on X which is absolutely continuous with respect to both a B u u u -invariant measure and a B v v v -invariant measure, except the Dirac mass δ 0 . Thus, informally speaking, B u u u and B v v v are orthogonal if their invariant measures "have nothing to say to each other", unless they charge the singleton t0u. The main question we want to investigate is the following. Question 1.2. -Characterize the pairs of weight sequences pu u u, v v vq such that the associated weighted shifts B u u u and B v v v acting on X " ℓ p are orthogonal.

The motivation for looking at this question comes from a recent work of Charpentier, Ernst, Mestiri and Mouze [START_REF] Charpentier | Common frequent hypercyclicity[END_REF]. Let B be the unweighted backward shift acting on the complex Hilbert space ℓ 2 pZ `q, and let Λ Ď C. In [START_REF] Charpentier | Common frequent hypercyclicity[END_REF], it is shown that the operators λB, λ P Λ admit a common frequently hypercyclic vector if and only if the set t|λ|; λ P Λu is a countable relatively compact subset of p1, 8q. In particular, if a, b P C and |a|, |b| ą 1, then the operators aB and bB have a common frequently hypercyclic vector. This is by no means obvious if |a| ‰ |b|. So it is tempting to ask whether this result could be retrieved in a "soft" way by measure-theoretic arguments. Indeed, imagine that it were possible to find two measures m a and m b on ℓ 2 , with full support, invariant and ergodic for aB and bB respectively, such that one of them is absolutely continuous with respect to the other, say m a ! m b . By the pointwise ergodic theorem, it would follow that m a -almost every x P ℓ 2 is frequently hypercyclic for aB and m b -almost every x P ℓ 2 is frequently hypercyclic for bB; hence m a -almost every x would be frequently hypercyclic for both aB and bB since m a ! m b , so one could conclude in particular that aB and bB have a common frequently hypercyclic vector. However, such measures m a and m b simply do not exist: we will see below that the operators aB and bB are in fact orthogonal.

1.3. Organization of the paper. -In Section 2, we complement existing results by showing that a weighted backward shift B w w w on X " ℓ p pZ `q admits a non-trivial invariant measure if and only if ř 8

n"1 1 |w 1 ¨¨¨wn| p ă 8. In Section 3, we define orthogonality in a very general context, we give a simple condition ensuring orthogonality (Theorem 3.2), and we use this to show, among other things, that if T is any continuous linear operator acting on a Polish topological vector space X and a, b P K are such that |a| ‰ |b|, then the operators aT and bT are orthogonal (Corollary 3.6). We also deduce from our criterion that two weighted shifts that are "far from being similar" must be orthogonal (Theorem 3.10). In Section 4, we use periodic vectors to give examples of non-orthogonal weighted shifts; in particular, we show that there exist non-orthogonal weighted shifts which are not similar (Example 4.5). Section 5 is devoted to invariant product measures for weighted shifts acting on ℓ p , and their role in our study of orthogonality. We give a necessary and sufficient condition for two weighted shifts B u u u and B v v v to admit equivalent invariant Gaussian product measures (Corollary 5.8), and then we examine to what extent the existence of non-orthogonal invariant product measures implies the existence of equivalent Gaussian product measures (Theorem 5.13 and Proposition 5.22). Section 6 contains some additional facts. Some of these facts are actually used in earlier proofs, but we prefered to postpone them for the sake of fluid reading. We conclude the paper with a few natural questions.

Existence of non-trivial invariant measures

It is, of course, not very interesting to study orthogonality of two weighted shifts if one does not know at least that each of them admits non-trivial invariant measures, i.e. invariant measures different from the Dirac mass δ 0 . The following proposition says exactly when this happens for a weighted shift acting on ℓ p . Proposition 2.1. -Let B w w w be a weighted backward shift acting on X " ℓ p . Then B w w w admits a non-trivial invariant measure if and only if

8 ř n"1 1 |w 1 ¨¨¨wn| p ă 8.
One can give a short proof of Proposition 2.1 by using a (quite non-trivial) lemma inspired by [START_REF] Bayart | Difference sets and frequently hypercyclic weighted shifts[END_REF], which can be extracted from the proof of [11, []

M ě 0, › › › › › ÿ mPN nďmďn`M v m´n e m´n › › › › › ď 1 C 2 › › › › › ÿ mPN nďmďn`M v m´n x m v m e m´n › › › › › " 1 C 2 › › › › › ÿ 0ďkďM k`nPN v k x k`n v k`n e k › › › › › ď C 1 C 2 Ẅe then conclude thanks to [11, []Lemma 2.5].
Proof of Proposition 2.1. -As mentioned in the introduction, it is well-known that if ř 8

n"1 1 |w 1 ¨¨¨wn| p ă 8, then B w w w admits lots of non-trivial invariant measures. Perhaps the simplest such measure is the Dirac mass δ x at the fixed point x :" e 0 `ř8

n"1 1 w 1 ¨¨¨wn e n . Conversely, assume that B w w w admits a non-trivial invariant measure m. By the ergodic decomposition theorem, we may assume that m is an ergodic measure for B w w w . Since m ‰ δ 0 , we may choose a point u ‰ 0 in the support of m. Then B n w w w u P supppmq for all n ě 0 by the B w w w -invariance of m, so we may assume that xe 0 , uy ‰ 0. Let α ą 0 be such that |xe 0 , uy| ą α, and consider the open set U :" ␣ z P X; }z ´u} ă 1 and |xe 0 , zy| ą α ( .

Since U is a neighbourhood of u, we have mpU q ą 0. By the pointwise ergodic theorem, it follows that one can find x P X such that the set N Bw w w px, U q :" tn P Z `; B n w w w x P U u admits a positive density, i.e. lim N Ñ8

1 N `1 # ` 0, N X N Bw w w px, U q ˘exists and is (strictly) positive; in particular, N Bw w w px, U q has positive upper density. So we have found a vector x P ℓ p and a set of integers N with positive upper density such that }B n w w w x} ă 1 `}u} and |xe 0 , B n w w w xy| ą α for all n P N . By Lemma 2.2, this concludes the proof. Remark 2.3. -Proposition 2.1 is valid in a more general setting: the same proof shows that if X is a Banach sequence space for which pe n q ně0 is a boundedly complete unconditional basis, then a weighted shift B w w w acting on X admits a non-trivial invariant measure if and only if the series ř 1 w 1 ¨¨¨wn e n is convergent in X. We do not know if the analogous result holds for weighted shifts acting on c 0 pZ `q.

A general criterion for orthogonality

Although the definition of orthogonality was given in the introduction for pairs of weighted shifts only, it makes sense in a much more general context. If T is any continuous self-map of a Polish (i.e. separable and completely metrizable) space X, let us denote by P T pXq the set of all T -invariant (Borel probability) measures on X. Definition 3.1. -Let X be a Polish space, and let (P) be any property of measures on X. We say that two continuous self-maps T 1 , T 2 of X are orthogonal with respect to measures satisfying (P) if the following holds true: whenever m 1 P P T 1 pXq and m 2 P P T 2 pXq satisfy (P), it follows that m 1 K m 2 .

In accordance with Definition 1.1, if X is a Polish topological vector space and T 1 , T 2 are continuous linear operators on X, we will say that T 1 and T 2 are orthogonal if they are orthogonal with respect to measures not charging the singleton t0u.

We now present a general sufficient condition for orthogonality. Recall that if pA n q nPN is a sequence of subsets of X, then lim A n denotes the set of all x P X that belong to infinitely many A n 's.

Theorem 3.2. -Let X be a Polish space, and let Λ be a Borel subset of X. Let also T 1 and T 2 be two continuous self-maps of X. Assume that for any compact set K Ď X with K X Λ " H, there exists an infinite set I Ď N such that

(3.1) lim nPI T ´n 1 pKq X T ´n 2 pKq " H.
Then, T 1 and T 2 are orthogonal with respect to measures not charging Λ.

Proof. -Let m 1 P P T 1 pXq and m 2 P P T 2 pXq be such that m 1 pΛq " 0 " m 2 pΛq. We have to show that m 1 K m 2 , and this will be an easy consequence of the following fact.

Fact 3.3. -For any ε ą 0, one can find a Borel set E ε Ď X such that m 1 pE ε q ă ε and m 2 pE ε q ą 1 ´ε. , we see that m 1 pEq " 0 and m 2 pEq " 1, which shows that m 1 K m 2 .

Remark 3.4. -The assumption in Theorem 3.2 is here to ensure that for any compact set K Ď XzΛ and any measure m on X, it holds that inf ně1 m `T ´n 1 pKq X T ´n 2 pKq ˘" 0. However, it is very likely that (3.1) is a too strong assumption. More generally, it would be nice to have a "measure-free" characterization of the sequences of Borel sets pA n q ně1 Ď X such that inf ně1 mpA n q " 0 for every measure m on X. Note that there is a very simple characterization if "inf" is replaced by "lim": we have lim nÑ8 mpA n q " 0 for every measure m on X if and only if lim A n " H. On the other hand, for a sequence pA n q ně1 of non-empty Borel subsets of X, consider the following statements.

(i) There exists an infinite set I Ď N such that lim nPI A n " H.

(ii) inf ně1 mpA n q " 0 for every measure m on X.

(iii) inf ně1 mpA n q " 0 for every discrete measure m on X.

(iv) inf ně1 mpA n q " 0 for every finitely supported measure m on X.

(v) For any finite set F Ď X, there exist infinitely many n such that A n X F " H. Then (i) ùñ (ii) ùñ (iii) ðñ (iv) ðñ (v), but (v) does not imply (ii) and (ii) does not imply (i).

Proof. -(i) ùñ (ii) by Fatou's Lemma, and obviously (ii) ùñ (iii) ùñ (iv). Moreover, it is rather clear that in fact (iii) ðñ (iv), and that (v) ùñ (iv).

To prove that (iv) ùñ (v), assume that for some finite set F Ď X and some integer N , we have A n X F ‰ H for all n ą N . Choose a point a i P A i for i " 1, . . . , N . Then, the finitely supported measure m :"

1 2 ˆ1 N N ř i"1 δ a i `1 #F ř xPF δ
x ˙is such that inf ně1 mpA n q ą 0.

To show that (v) does not imply (ii), let X :" r0, 1s, and define

A n :" tt P r0, 1s; | sinp2πntq| ě 1{πu.

By Dirichlet's theorem, for any finite set F Ď r0, 1s one can find an increasing sequence of integers pn k q such that sinp2πn k tq Ñ 0 on F ; so (v) is satisfied. However, we have ş 1 0 | sinp2πntq| dt " 2{π for all n ě 1. Hence mpA n q ě 1{π for all n ě 1, where m is Lebesgue measure on r0, 1s.

Let us now show that (ii) does not imply (i). This example is due to N. de Rancourt. Consider the Cantor space ∆ " t0, 1u N identified with the power set of N. Let X :"

# J Ď N; ÿ nPJ 1 n ď 1 + ,
which is a closed subset of ∆ and hence a (compact) Polish space. For each n P N, let A n :" tJ P X; n P Ju.

It is clear that the sequence pA n q does not satisfy (i): indeed, for any infinite set I Ď N, one can find an infinite set J Ď I such that ř nPJ 1 n ď 1, and this J belongs to lim A n by definition.

However, pA n q satisfies (ii); in fact, for any measure m on X, we have ř 8

n"1 mpAnq n ă 8. Indeed, we have

8 ÿ n"1 mpA n q n " ż X ˜8 ÿ n"1 1 n 1 An pJq ¸dmpJq " ż X ˜ÿ nPJ 1 n ¸dmpJq ď 1.
Here is yet another example showing that (ii) does not imply (i), which is a kind of twin of the previous one. Let X :"

#

x P r0, πs;

8 ÿ n"1 | sinp2 n xq| n ď 1
+ ,
which is a closed subset of r0, πs. For each n ě 1, let

A n :" tx P X; | sinp2 n xq| ě 1{2u.
With some effort, one can prove the following: if pm i q iě1 is a sufficiently fast increasing sequence of integers and if we define

x :" π 2 8 ÿ i"1 1 2 m i , then x P A m k for every k ě 1.
It follows imediately that the sequence pA n q does not satisfy (i). However, one shows in the same way as above that pA n q satisfies (ii).

Remark 3.5. -With the notation of Theorem 3.2, consider the following statements.

(1) For any compact set K Ď XzΛ, it holds that lim T ´n 1 pKq X T ´n 2 pKq " H. (2) It is not possible to find x P X and an increasing sequence of integers pn k q such that the sequences pT n k 1 xq and pT n k 2 xq both converge to a limit not belonging to Λ. Then (2) ùñ (1), and ( 1)

ðñ (2) if Λ is a closed subset of X.
Proof. -It should be rather clear that (2) implies (1). Indeed, let K be a compact subset of XzΛ, and assume that lim T ´n 1 pKq X T ´n 2 pKq ‰ H. Then, there exists x P X and an increasing sequence of integers pn k q kě0 such that T n k 1 x P K and T n k 2 x P K for all k ě 0. Since K is compact, we may assume, upon extracting subsequences, that both sequences pT n k 1 xq and pT n k 2 xq converge, T n k 1 x Ñ u P K and T n k 2 x Ñ v P K; and since K Ď XzΛ, this shows that (2) is not satisfied.

Conversely, assume that Λ is closed and that (2) is not satisfied, i.e. there exist x P X and an increasing sequence of integers pn k q kě0 such that T n k 1 x Ñ u and T n k 2 x Ñ v, where u, v P XzΛ. Since XzΛ is open, one can find k 0 such that T n k 1 x, T n k 2 x P XzΛ for all k ě k 0 . Then the set K :" tT n k 1 x; k ě k 0 u Y tT n k 2 x; k ě k 0 u Y tu, vu is a compact subset of XzΛ and x P T ´nk 1 pKq X T ´nk 2 pKq for all k ě k 0 , so that (1) is not satisfied.

Here is a first application of Theorem 3.2.

Corollary 3.6. -Let X be a Polish topological vector space, and let T P LpXq. Let also a, b P K, and assume that aT and bT admit non-trivial invariant measures. Then, aT and bT are orthogonal if and only if |a| ‰ |b|. More precisely: aT and bT are orthogonal if |a| ‰ |b|; and they share a non-trivial invariant measure if |a| " |b|.

Proof. -To prove that aT and bT are orthogonal if |a| ‰ |b|, we use Theorem 3.2 and Remark 3.5 with Λ " t0u. Assume that there exists x P X and an increasing sequence of integers pn k q such that a n k T n k x Ñ u P X and b n k T n k x Ñ v P X. We must show that u " 0 or v " 0; and this follows from the continuity of the map pλ, zq Þ Ñ λz: indeed, assuming for example that |a| ą |b| and writing b

n k T n k x " pb{aq n k a n k T n k x, we see that b n k T n k x Ñ 0.
The converse will follow from Proposition 6.1 below.

Remark 3.7.

-If B w w w is a weighted shift acting on X " ℓ p pZ `q and a, b P K are such that |a| " |b| ": r with ř 8

n"1

1
pr n |w 1 ¨¨¨wn|q p ă 8, then the operators aB w w w and bB w w w admit equivalent ergodic Gaussian measures with full support. This will follow from Theorem 5.7 below. Corollary 3.6 leads to naive speculations regarding orthogonality of weighted shifts. Recall that two operators T 1 , T 2 acting on X are said to be similar if there exists an invertible operator J such that T 1 " JT 2 J ´1. By [22, [] When u u u and v v v have the form u u u " aw w w and v v v " bw w w for some weight sequence w w w and a, b P K, then u 1 ¨¨¨un v 1 ¨¨¨vn " `a b ˘n, and hence (3.2) holds true if and only if |a| " |b|. So it is tempting to "conjecture" that two weighted shifts B u u u and B v v v are orthogonal if and only if they are not similar. This is however not true, as will be seen in Section 4 below. Nevertheless, we now present two results showing that this is not that far from being true.

In what follows we denote by e j , j ě 0 the coordinate functionals on X " ℓ p pZ `q. And if x P X, we write x " ř 8 j"0 x j e j where x j " xe j , xy. Proposition 3.8. -Let B u u u and B v v v be two weighted shifts acting on X " ℓ p . Assume that B u u u and B v v v are not similar, i.e. that

(3.3) either lim ˇˇˇu 1 ¨¨¨u n v 1 ¨¨¨v n ˇˇˇ" 0 or lim ˇˇˇu 1 ¨¨¨u n v 1 ¨¨¨v n ˇˇˇ" 8.
Then, B u u u and B v v v are orthogonal with respect to measures not charging Λ e 0 :" kerpe 0 q.

Proof. -We apply Theorem 3.2 with Λ :" Λ e 0 . So let K be a compact subset of Xz kerpe 0 q. Choose some constants M ă 8 and γ ą 0 such that }z} ď M and |xe 0 , zy| ě γ for all z P K. If x " ř 8 j"0 x j e j P X then, for every n ě 1, we have

xe 0 , B n u u u xy " u 1 ¨¨¨u n x n and xe 0 , B n v v v xy " v 1 ¨¨¨v n x n ; so we see that B ´n u u u pKq Ď ␣ x P X; |x n | ¨|u 1 ¨¨¨u n | ě γ ( and B ´n v v v pKq Ď ␣ x P X; |x n | ¨|v 1 ¨¨¨v n | ď M ( .
Hence, the following implication holds :

B ´n u u u pKq X B ´n v v v pKq ‰ H ùñ ˇˇˇu 1 ¨¨¨u n v 1 ¨¨¨v n ˇˇˇě γ M Ït
follows that if lim ˇˇu 1 ¨¨¨un v 1 ¨¨¨vn ˇˇ" 0, then B ´n u u u pKq X B ´n u u u pKq " H for infinitely many n; and symmetrically, if lim ˇˇu 1 ¨¨¨un v 1 ¨¨¨vn ˇˇ" 8 then B ´n v v v pKq X B ´n u u u pKq " H for infinitely many n. By Theorem 3.2, this concludes the proof. Remark 3.9. -One may observe that for any weight sequence w w w and every n ě 1, we have B ´n w w w pΛ e 0 q " Λ e n . Hence, a measure m P P w w w pXq does not charge Λ e 0 if and only if it does not charge Λ e n . Therefore, m does not charge Λ e 0 if and only if it is supported on the set tx P X; xe n, xy ‰ 0 for all n ě 0u.

The next theorem shows that "full" orthogonality of B u u u and B v v v can be deduced from stronger assumptions on the weights u u u and v v v. Theorem 3.10. -Let B u u u and B v v v be two weighted shifts acting on X " ℓ p . Assume that either

(3.4) lim nÑ8 max 0ďdďN ˇˇˇu 1 ¨¨¨u n`d v 1 ¨¨¨v n`d ˇˇˇ" 0 for all N ě 0 or (3.5) lim nÑ8 min 0ďdďN ˇˇˇu 1 ¨¨¨u n`d v 1 ¨¨¨v n`d ˇˇˇ" 8 for all N ě 0.
Then, B u u u and B v v v are orthogonal.

Proof. -We apply Theorem 3.2 once again, with Λ :" t0u. So, let K be a compact subset of Xzt0u.

We first note that there exists an integer N such that max

B ´n u u u pKq Ď " x P X; max 0ďdďN |u 1`d ¨¨¨u n`d | ¨|x n`d | ě γ * and B ´n v v v pKq Ď " x P X; max 0ďdďN |v 1`d ¨¨¨v n`d | ¨|x n`d | ď M * ; so that B ´n u u u pKq X B ´n v v v pKq ‰ H ùñ max 0ďdďN ˇˇˇu 1`d ¨¨¨u n`d v 1`d ¨¨¨v n`d ˇˇˇě γ M Now, we have max 0ďdďN ˇˇu 1`d ¨¨¨u n`d v 1`d ¨¨¨v n`d ˇˇď C N max 0ďdďN ˇˇu 1 ¨¨¨u n`d v 1 ¨¨¨v n`d ˇˇfor some constant C N . Hence, if (3.4) is satisfied, then one can find an infinite set I Ď N such that max 0ďdďN ˇˇˇu 1`d ¨¨¨u n`d v 1`d ¨¨¨v n`d ˇˇˇÑ 0 as n Ñ 8 along I;
and it follows that B ´n u u u pKqXB ´n v v v pKq " H for all n P I sufficiently large. Symmetrically, if (3.5) is satisfied, then one can find an infinite set I Ď N such that B ´n v v v pKq X B ´n u u u pKq " H for all n P I sufficiently large. So Theorem 3.2 applies. Proof. -The assumption on u u u and v v v implies that for every N ě 0, there are constants c N ą 0 and

C N ă 8 such that max 0ďdďN ˇˇu 1 ¨¨¨u n`d v 1 ¨¨¨v n`d ˇˇď C N ˇˇu 1 ¨¨¨un v 1 ¨¨¨vn ˇˇand min 0ďdďN ˇˇu 1 ¨¨¨u n`d v 1 ¨¨¨v n`d ˇˇě c N ˇˇu 1 ¨¨¨un v 1 ¨¨¨vn ˇˇ. Hence, if B u u u and B v v v are not similar then (3.4) or (3.5) is satisfied.
Remark 3.12. -Theorem 3.10 can be reformulated in the following way: if -either, for every ε ą 0, the set A u u u,v v v,ε :"

! n P Z `; ˇˇu 1 ¨¨¨un v 1 ¨¨¨vn ˇˇă ε
) contains arbitrarily long intervals, -or, for every M ă 8, the set A 1 u u u,v v v,M :"

! n P Z `; ˇˇu 1 ¨¨¨un v 1 ¨¨¨vn ˇˇą M
) contains arbitrarily long intervals, then B u u u and B v v v are orthogonal.

Non-orthogonality via periodic points

In this section, our aim is to present some examples of non-orthogonal weighted shifts using measures supported by periodic orbits. We first note the following easy fact.

Fact 4.1. -Let X be a Polish topological vector space, and let T 1 , T 2 P LpXq. If T 1 and T 2 share a non-zero periodic point, then they are not orthogonal.

Proof. -Let x ‰ 0 be a common periodic point for T 1 and T 2 ; so

T d 1 x " x " T d 2 x for some d 1 , d 2 P N. For i " 1, 2, the measure m i :" 1 d i ř d i ´1 n"0 δ T n i x is T i -invariant
, and m i pt0uq " 0 since x ‰ 0. Moreover, m 1 and m 2 are not orthogonal since m 1 ptxuq ą 0 and m 2 ptxuq ą 0.

Theorem 4.2. -Let B u u u and B v v v be two weighted shifts on X " ℓ p pZ `q. Assume that ř 8 n"1 1 |u 1 ¨¨¨un| p ă 8 and ř 8 n"1 1 |v 1 ¨¨¨vn| p ă 8.
The following are equivalent.

(1) B u u u and B v v v share a non-zero periodic point;

(2) there exist d P N and 0 ď j ď d ´1 such that u 1`j ¨¨¨u dm`j " v 1`j ¨¨¨v dm`j for all m ě 1;

(3) there exist d P N and 0 ď j ď d ´1, and a non-zero scalar C such that u 1 ¨¨¨u dm`j "

C v 1 ¨¨¨v dm`j for all m ě 0 (where an empty product is declared to be equal to 1).

Proof. -( 1) ùñ (2) Assume that B u u u and B v v v share a non-zero periodic point x. Let d 1 be the period of x as a periodic point of T 1 , and let d 2 be the period of x as a periodic point of T 2 . Setting d :"

d 1 d 2 , we then have B d u u u x " x " B d v v v x (and hence B md u u u x " x " B md v v v x for all m ě 1.) Writing x " ř jě0
x j e j , we have

x j`md " 1 u j`1 ¨¨¨u j`md x j " 1 v j`1 ¨¨¨v j`md
x j for all j ě 0.

Since x ‰ 0, one can find 0 ď j ď d ´1 such that x j ‰ 0, and then u j`md ¨¨¨u j`1 " v j`md ¨¨¨v j`1 for all m ě 1.

(2) ùñ (3) This is clear: if ( 2) is satisfied for some d and 0 ď j ď d ´1, then (3) is satisfied with the same d and j and C :"

u 1 ¨¨¨u j v 1 ¨¨¨v j (so C " 1 if j " 0).
(3) ùñ (1) Assume that (3) holds true for some d, j, C. Since ř 8

n"1

1 |u 1 ¨¨¨un| p ă 8, the vector x :" 8 ÿ m"0 1 u 1 ¨¨¨u md`j e md`j " 1 C 8 ÿ m"0 1 v 1 ¨¨¨v md`j
e md`j P ℓ p is well-defined, and clearly

B d u u u x " x " B d v v v x. Thus, x is a non-zero common periodic point for B u u u and B v v v .
Corollary 4.3. -Let B u u u and B v v v be two weighted shifts on ℓ p pZ `q, with ř 8

n"1 1 |u 1 ¨¨¨un| p ă 8 and ř 8

n"1 1 |v 1 ¨¨¨vn| p ă 8. Assume that there exist d P N, 0 ď j ď d ´1 and a scalar C ‰ 0 such that u 1 ¨¨¨u dm`j " C v 1 ¨¨¨v dm`j for all m ě 0. Then B u u u and B v v v are not orthogonal.

Example 4.4. -Assume that u j " 2 for all j ě 1, v 1 " 3 and v j " 2 for all j ě 2. Then B u u u and B v v v acting on ℓ p share a non-zero fixed point, and hence they are not orthogonal.

In this example, the operators B u u u and B v v v are obviously similar. We now give an example of non-orthogonal weighted shifts which are not similar.

Example 4.5. -Let pr k q kě1 be an increasing sequence of integers, and let u u u and v v v be the two weight sequences defined as follows:

u j :" " 2 if j R t5r k `1, 5r k `4; k ě 1u 1 k if j " 5r k `1 or j " 5r k `4 for some k ě 1 and v j :" " 2 if j R t5r k `2, 5r k `3; k ě 1u 1 k if j " 5r k `2 or j " 5r k `3 for some k ě 1.
If pr k q is sufficiently fast increasing, then B u u u and B v v v acting on ℓ p share a non-zero periodic point and hence they are not orthogonal. However, we have

lim u 1 ¨¨¨u n v 1 ¨¨¨v n " 0 and lim u 1 ¨¨¨u n v 1 ¨¨¨v n " 8,
so that, in particular, B u u u and B v v v are not similar.

Proof. -It is clear that if pr k q is sufficiently fast increasing, then ř 8

n"0

1 |u 1 ¨¨¨un| p ă 8 and ř 8 n"0 1 |v 1 ¨¨¨vn| p ă 8.
Observe that if m is a positive integer and if we consider the largest k such that r k ď m, then either r k " m or 5r k `4 ă 5m. Since u 1 ¨¨¨u 5r k " v 1 ¨¨¨v 5r k by definition of the weights, it follows that

u 1 ¨¨¨u 5m " v 1 ¨¨¨v 5m
for all m ě 1.

Hence, B u u u and B v v v share a non-zero periodic point by Theorem 4.2. However, for any k ě 1 we have

u 1 ¨¨¨u 5r k `1 v 1 ¨¨¨v 5r k `1 " 1 2k and u 1 ¨¨¨u 5r k `3 v 1 ¨¨¨v 5r k `3 " 2k,
so that lim u 1 ¨¨¨un v 1 ¨¨¨vn " 0 and lim u 1 ¨¨¨un v 1 ¨¨¨vn " 8. -The question we consider in this section is the following: given two weighted shifts B u u u and B v v v on X " ℓ p pZ `q, when is it possible to find invariant measures m u u u and m v v v (for B u u u and B v v v respectively) not charging t0u which are not orthogonal and are also product measures?

The terminology requires some explanation since product measures should be defined on products of measurable spaces and X " ℓ p is not such a product space. However, ℓ p is contained in the product space Ω :" K Z `. It is well-known that the Borel σ-algebra of Ω (induced by the product topology) is identical with the product σ-algebra b ně0 B K . Moreover, ℓ p is a Borel subset of Ω when Ω is endowed with the product topology, and a subset of ℓ p is Borel in pℓ p , } ¨}p q if and only if is Borel in Ω. It follows that if µ is a Borel probability measure on Ω, then the restriction of µ to ℓ p is a Borel measure on ℓ p endowed with its usual topology (but not a probability measure unless µpℓ p q " 1); and conversely, any Borel probability measure on ℓ p can be considered as a Borel probability measure on Ω (supported on ℓ p ). We will say that a Borel probability measure m on ℓ p is a product measure on ℓ p if m is the restriction to ℓ p of a (probability) product measure µ on Ω, i.e. µ " Â ně0 µ n where each µ n is a Borel probability measure on K, such that µpℓ p q " 1. In this case, the measure µ is uniquely determined by m since µpBq " mpB X ℓ p q for every Borel set B Ď Ω, so we identify m and µ and simply write m " b ně0 µ n .

The following essentially obvious remark will be used repeatedly.

Fact 5.1. -Let X be a Borel subset of Ω " K Z `, and let µ be a Borel (probability) measure on Ω such that µpXq " 1. Set m :" µ |X . Let also T : Ω Ñ Ω be a Borel map such that T pXq Ď X.Then µ is T -invariant if and only if m is pT |X q -invariant.

Proof. -The measure µ is T -invariant if and only if µpT ´1pBqq " µpBq for every Borel set B Ď Ω. Now, we have µpBq " mpX X Bq and µpT ´1pBqq " mpX X T ´1pBqq " m `pT |X q ´1pX X Bq ˘because T pXq Ď X. So µ is T -invariant if and only if mpX X Bq " m `pT |X q ´1pX X Bq ˘for every Borel set B Ď Ω, which means exactly that m is pT |X qinvariant.

From this observation, it follows that if m is a product measure on X " ℓ p and if we denote by µ the measure m considered as a measure on Ω " K Z `, then m is invariant under some weighted shift B w w w if and only if µ is invariant under the natural extension of B w w w to Ω (defined by the same formula and also denoted by B w w w ). The next lemma says precisely when this happens. Lemma 5.2. -Let µ " b ně0 µ n be a product measure on Ω " K Z `, and let w w w be a weight sequence. Then, µ is B w w w -invariant if and only if, for each n ě 1, the measure µ n is the image of µ 0 under the map t Þ Ñ 1 w 1 ¨¨¨wn t, i.e. µ n pAq " µ 0 pw 1 ¨¨¨w n Aq for every Borel set A Ď K.

Proof. -For any Borel sets A 0 . . . , A N Ď K, denote by rA 0 , . . . , A N s Ď Ω the "cylinder set" defined as follows: rA 0 , . . . , A N s " ␣ t " pt j q jě0 P Ω; t j P A j for j " 0, . . . , N ( .

Since the cylinder sets generate the Borel σ-algebra of Ω, the measure µ is B w w w -invariant if and only if µ `B´1 w w w prA 0 , . . . , A N sq ˘" µ `rA 0 , . . . , A N s ˘ for all Borel sets A 0 , . . . , A N Ď K. Now, by definition of B w w w we have B ´1 w w w prA 0 , . . . , A N s ˘" ␣ t " pt j q jě0 P Ω; w j`1 t j`1 P A j for j " 0, . . . , N ( , so that B ´1 w w w prA 0 , . . . , A N s ˘" rK, p1{w 1 qA 0 , . . . , p1{w N `1qA N s . Since µ " b jě0 µ j , it follows that µ is B w w w -invariant if and only if N ź j"0 µ j pA j q " N ź j"0 µ j`1 `p1{w j`1 qA j for every N ě 0 and all Borel sets A 0 , . . . , A N Ď K. This is clearly equivalent to the fact that µ j`1 pp1{w j`1 qAq " µ j pAq for all j ě 0 and every Borel set A Ď K, which proves the lemma.

Corollary 5.3. -Let µ " b ně0 µ n be a product measure on Ω " K Z `, and let w w w be a weight sequence. Assume that µ 0 has a density p with respect to Lebesgue measure on K.

Then µ is B w w w -invariant if and only if, for each n ě 1, the measure µ n has a density p n given by p n ptq :"

|w 1 ¨¨¨w n | d ppw 1 ¨¨¨w n tq, where d " 1 if K " R and d " 2 if K " C.
Proof. -This follows immediately from Lemma 5.2

Corollary 5.4. -Let µ " b ně0 µ n be a product measure on Ω " K Z `, and let w w w be a weight sequence such that ř 8

n"0

1 |w 1 ¨¨¨wn| p ă 8.
If µ is B w w w -invariant and if the measure µ 0 is such that ş K |t| p dµ 0 ptq ă 8, then µpℓ p q " 1. Proof. -By Lemma 5.2, we have

ż Ω ˜8 ÿ n"0 |t n | p ¸dµptq " 8 ÿ n"0 ż K |t n | p dµ n pt n q " ˆżK |s| p dµ 0 psq ˙ˆ˜1 `8 ÿ n"1 1 |w 1 ¨¨¨w n | p ¸ă 8.
In particular, we see that ř 8 n"0 |t n | p ă 8 for µ -almost every t P Ω, i.e. µpℓ p q " 1.

To conclude this section, we point out the following general fact.

Fact 5.5. -Let B w w w be a weighted shift on ℓ p where w w w satisfies ř 8

n"1 1 |w 1 ¨¨¨wn| p ă 8, and let m " b ně0 µ n be a B w w w -invariant product measure on ℓ p . If the measure µ 0 has full support, then m is an ergodic measure with full support for B w w w .

Proof. -Let pξ n q ně0 be a sequence of independent K -valued random variables with law µ 0 . As a measure on K Z `, m is the distribution of the K Z `-valued random variable ξ :" ξ 0 e 0 `ř8

n"1 ξn w 1 ¨¨¨wn e n . Since mpℓ p q " 1, we have ξ P ℓ p almost surely, which means that the series defining ξ is almost surely convergent with respect to the ℓ p -norm. Moreover, since µ 0 has full support, the measure m has full support. By [1, []Proposition 2.5], it follows that m is an ergodic measure (in fact, a strongly mixing measure) for B w w w .

Non-orthogonality of product measures.

-A basic tool in the study of product measures is a famous classical result of Kakutani [START_REF] Kakutani | On equivalence of infinite product measures[END_REF] giving a necessary and sufficient condition for the equivalence of two infinite product probability measures ν " b ně0 ν n and ν 1 " b ně0 ν 1 n defined on a product of measurable spaces pΩ, Bq " b ně0 pΩ n , B n q. Let us first recall the definition of the so-called Hellinger integral Hpα, βq of two probability measures α, β on some measurable space pT, T q:

Hpα, βq " ż T c dα dτ c dβ dτ dτ,
where τ is any sigma-finite positive measure on pT, T q such that α and β are absolutely continuous with respect to τ . If we write α " dα dβ β `ν the Lebesgue-Nikodym decomposition of α with respect to β, then ´dα which shows that Hpα, βq is indeed independent of the choice of τ . Note that we always have 0 ď Hpα, βq ď 1 by the Cauchy-Schwarz inequality; and that Hpα, βq " 1 if and only if α " β, whereas Hpα, βq " 0 if and only if α K β.

Kakutani's Theorem as we will need it can now be stated as follows.

Theorem 5.6. -Let ν " b ně0 ν n and ν 1 " b ně0 ν 1 n be two product probability measures on pΩ, Bq " b ně0 pΩ n , B n q. The measures ν and ν 1 are non-orthogonal if and only if

8 ź n"0
Hpν n , ν 1 n q ą 0.

Moreover, under the assumption that ν n " ν 1 n for each n ě 0, the measures ν and ν 1 are either orthogonal or equivalent.

Note that this is not exactly what is proved in [START_REF] Kakutani | On equivalence of infinite product measures[END_REF]: the main result of [START_REF] Kakutani | On equivalence of infinite product measures[END_REF] states that under the assumption that ν n " ν 1 n for all n, the measures ν and ν 1 are equivalent if ś 8

n"0 Hpν n , ν 1 n q ą 0 and orthogonal otherwise. However, the above version of the theorem is certainly well-known (it is stated for example in [START_REF] Shepp | Distinguishing a sequence of random variables from a translate of itself[END_REF]), and it can be obtained by slight modifications of Kakutani's original proof; see Section 6 for more details.

Gaussian product measures.

-A particularly interesting class of product measures is that of Gaussian product measures. Our definition will be very restrictive: we will say that a product measure µ " b ně0 µ n on Ω " K Z `is a Gaussian product measure if for each n ě 0, the measure µ n is a Gaussian measure on K of the form N p0, σ 2 n q for some σ n ą 0, i.e. µ n is a centered Gaussian measure with covariance matrix σ 2 n I K . Accordingly, we say that a measure m on X " ℓ p is a Gaussian product measure if m is the restriction to ℓ p of a Gaussian product measure µ on Ω such that µpℓ p q " 1.

By Fernique's integrability theorem (see e.g. [START_REF] Bogachev | Gaussian measures[END_REF]), if m is a Gaussian measure on X " ℓ p , then m admits moments of all orders and in particular ş X }x} p dm ă 8. It follows that if µ " b ně0 µ n is a Gaussian product measure on Ω " K Z n"0 1 |w 1 ¨¨¨wn| p ă 8, and that if this holds, then any B w w w -invariant Gaussian product measure on K Z `is supported on ℓ p . (For the "only if" part, one could also have used Proposition 2.1 rather than Fernique's integrability theorem.) We refer the reader to [START_REF] Bayart | Frequently hypercyclic operators[END_REF], [START_REF] Bayart | Invariant Gaussian measures for operators on Banach spaces and linear dynamics[END_REF], [START_REF] Bayart | Mixing operators and small subsets of the circle[END_REF] and [START_REF] Bayart | Dynamics of linear operators[END_REF]Chapter 5] for an in-depth study of invariant Gaussian measures for operators on Hilbert or Banach spaces. We just point out here that if w w w is a weight sequence such that ř 8

n"0 1 |w 1 ¨¨¨wn| p ă 8 and if m is a B w w w -invariant Gaussian product measure on ℓ p , then m is an ergodic measure with full support for B w w w (see Fact 5.5).

The next theorem will provide a necessary and sufficient condition for two weighted backward shifts B u u u and B v v v on X " ℓ p to admit equivalent invariant Gaussian product measures.

Theorem 5.7. -Let u u u and v v v be two weight sequences, and let µ u u u " b ně0 µ u u u,n and µ v v v " b ně0 µ v v v,n be two Gaussian product measures on K Z `invariant under B u u u and B v v v respectively, with µ u u u,0 " N p0, σ 2 q and µ v v v,0 " N p0, σ 12 q. Then µ u u u and µ v v v are either equivalent or orthogonal, and they are equivalent if and only if (5.1)

8 ÿ n"1 ˆ1 ´σ1 σ |u 1 ¨¨¨u n | |v 1 ¨¨¨v n | ˙2 ă 8.
From this result, we immediately deduce Here is another consequence of Theorem 5.7.

Corollary 5.9. -Let Λ Λ Λ be a countable family of weight sequences. Assume that ´1 ´|u 1 ¨¨¨un| |v 1 ¨¨¨vn| ¯2 ă 8 for any u, v P Λ Λ Λ. Then, there exists a Gaussian measure with full support m on X " ℓ p with the following property: given any sequence of Borel sets pA i q iě0 Ď X with mpA i q ą 0, there exists x P X such that for every w w w P Λ and all i ě 0, the set N Bw w w px, A i q :" tn P N; B n w w w x P A i u has a positive density.

Proof. -For each w w w P Λ Λ Λ, let m w w w " b ně0 µ w w w,n be the B w w w -invariant Gaussian product measure on ℓ p defined by µ w w w,0 :" N p0, 1q. This is an ergodic measure with full support for B w w w . Since the measures m w w w , w w w P Λ Λ Λ are pairwise equivalent by Theorem 5.7 and since Λ is countable, the result follows from the pointwise ergodic theorem, taking m :" m w w w 0 for any w w w 0 P Λ.

Remark 5.10. -Taking as pA i q iě0 a countable basis of open sets for X in Corollary 5.9, we get a vector x P X which is frequently hypercyclic for all operators B w w w , w w w P Λ. However, it turns out that something much stronger holds true: if u u u and v v v are two weight sequences such that u 1 ¨¨¨un v 1 ¨¨¨vn has a non-zero limit as n Ñ 8, then B u u u and B v v v have in fact the same frequently hypercyclic vectors. This result, which is due to S. Charpentier and the third author, will be proved in Section 6.

Before proving Theorem 5.7, we use Corollary 5.8 to give an example of two nonorthogonal weighted shifts sharing no non-zero periodic points.

Example 5.11. -Let u u u and v v v be the weight sequences defined as follows:

u u u n :" 2 and v v v n :" 2 1 `εn 1 `εn´1 ,
where ε 0 " 0 and pε n q ně1 is a decreasing sequence of positive real numbers such that ř 8 n"1 ε 2 n ă 8. Then B u u u and B v v v acting on X " ℓ p share no non-zero periodic point, but they admit equivalent invariant Gaussian product measures (and hence they are not orthogonal).

Proof. -We have u 1 ¨¨¨u n " 2 n and v 1 ¨¨¨v n " 2 n p1 `εn q for all n ě 1. In particular, ř 8 n"1 1 pu 1 ¨¨¨unq p ă 8 and

ř 8 n"1 1 pv 1 ¨¨¨vnq p ă 8. Since u 1 ¨¨¨un v 1 ¨¨¨vn " 1 p1
`εnq is increasing, the conditions of Theorem 4.2 cannot be satisfied, so B u u u and B v v v do not share any non-zero periodic point. However

1 ´u1 ¨¨¨u n v 1 ¨¨¨v n " 1 ´1 1 `εn " ε n as n Ñ 8,
so B u u u and B v v v admit equivalent invariant Gaussian product measures by Corollary 5.8.

It is now time to prove Theorem 5.7.

Proof of Theorem 5.7.

-In what follows, we set d :" 1 if K " R and d :" 2 if K " C. By Kakutani's Theorem, we just have to show that ś 8 n"1 Hpµ u u u,n , µ v v v,n q ą 0 if and only if (5.1) is satisfied.

By Lemma 5.2, the measures µ u u u,n and µ v v v,n are uniquely determined by µ u u u,0 " N p0, σ 2 q and µ v v v,0 " N p0, σ 12 q. Explicitely,

µ u u u,n " N p0, σ 2 n q with σ 2 n " σ 2 |u 1 ¨¨¨u n | 2 and µ v v v,n " N p0, σ 12 n q with σ 12 n " σ 12 |v 1 ¨¨¨v n | 2
The computation of Hpµ u u u,n , µ v v v,n q now relies on the following fact. Fact 5.12. -If µ and µ 1 are two Gaussian measures on K of the form µ " N p0, σ 2 q and µ 1 " N p0, σ 12 q, then Hpµ, µ 1 q "

ˆ2σσ 1 σ 2 `σ12 ˙d{2 " ˆ2pσ 1 {σq 1 `pσ 1 {σq 2
˙d{2 Proof of Fact 5.12. -The measures µ and µ 1 are absolutely continuous with respect to Lebesgue measure on K with densities 1 p2πσ 2 q d{2 e ´|t| 2 {2σ 2 and 1 p2πσ 12 q d{2 e ´|t| 2 {2σ 12 . Hence,

Hpµ, µ 1 q " 1 p2πσσ 1 q d{2 ż R e ´1 4 ´1 σ 2 `1 σ 12 ¯|t| 2 dt " 1
p2πσσ 1 q d{2 ˆ˜4π 1 σ 2 `1 σ 12 ¸d{2 ¨
Going back to the measures m u u u " b ně0 µ u u u,n and m v v v " b ně0 µ v v v,n , let us set for each n ě 1:

λ n :" σ 1 n σ n " σ 1 σ |u 1 ¨¨¨u n | |v 1 ¨¨¨v n | By
Fact 5.12, we have

Hpµ u u u,n , µ v v v,n q " ˆ2 λ n 1 `λ2 n ˙d{2 Since 0 ď Hpµ u u u,n , µ v v v,n q ď 1 for all n ě 1, it follows that 8 ź n"1
Hpµ u u u,n , µ v v v,n q ą 0 if and only if

8 ÿ n"1 ˜1 ´ˆ2 λ n 1 `λ2 n ˙d{2 ¸ă 8.
This can be satisfied only if 2 λn 1`λ 2 n Ñ 1 as n Ñ 8; and an examination of the function Ψpλq :" 2 λ 1`λ 2 shows that this is equivalent to the fact that λ n Ñ 1. In this case, writing λ n " 1 `un and using Taylor's formula, we see that

1 ´ˆ2 λ n 1 `λ2 n ˙d{2 " d 4 u 2 n " d 4 p1 ´λn q 2 .
So we get that

8 ź n"1
Hpµ u u u,n , µ v v v,n q ą 0 if and only if

8 ÿ n"1 p1 ´λn q 2 ă 8,
which is (5.1). Theorem 5.7 is proved.

Products of absolutely continuous measures.

-It is natural to believe that some condition strictly weaker than (b) of Corollary 5.8 might still yield the existence of (non-Gaussian) equivalent non-trivial invariant product measures for the backward shifts B u u u and B v v v . The next theorem shows that this is in fact not the case, at least for product measures whose marginals are absolutely continuous with respect to Lebesgue measure. We point out that it is possible to deduce parts (1ii) and (2) of this theorem from a Theorem of Shepp [START_REF] Shepp | Distinguishing a sequence of random variables from a translate of itself[END_REF] concerning (non-)orthogonality of translates of a product measure on R Z `, namely [21, Theorem 1]. However, we are going to give here a self-contained proof which looks rather different from what is done in [START_REF] Shepp | Distinguishing a sequence of random variables from a translate of itself[END_REF] (except, of course, for the use of Kakutani's Theorem), and we will indicate in Section 6 how Shepp's result can be used in our context; see Theorem 6.11.

For simplicity, we will assume that K " R and that all the weights are positive.

Theorem 5.13. -Let u u u and v v v be two positive weight sequences. Assume that the backward shifts B u u u and B v v v acting on the real space X " ℓ p admit invariant product measures m u u u " b ně0 µ u u u,n and m v v v " b ně0 µ v v v,n such that µ u u u,0 and µ v v v,0 are absolutely continuous with respect to Lebesgue measure on R, µ u u u,0 " pptqdt and µ v v v,0 " qptqdt.

(1) If m u u u and m v v v are not orthogonal, then: (i) the quotient u 1 ¨¨¨un v 1 ¨¨¨vn has a limit a P p0, 8q as n Ñ 8, and qptq " appatq almost everywhere;

(ii) we have

8 ř n"1 ´1 ´1 a u 1 ¨¨¨un v 1 ¨¨¨vn ¯2 ă 8.
(2) Assume that the function f :" ? p is continuous on Rzt0u and C 1 -smooth except at a finite number of points with tf 1 ptq P L 2 pRq, and that qptq " appatq for some a P p0, 8q.

Then, m u u u and m v v v are non-orthogonal if and only if

8 ř n"1 ´1 ´1 a u 1 ¨¨¨un v 1 ¨¨¨vn ¯2 ă 8.
(3) Assume that f " ? p is C 1 -smooth except at a finite number of points with tf 1 ptq P L 2 pRq, that p has at least one discontinuity point in Rzt0u, and that qptq " appatq for some a P p0, 8q. Then, m u u u and m v v v are non-orthogonal if and only if

8 ř n"1 ˇˇ1 ´1 a u 1 ¨¨¨un v 1 ¨¨¨vn ˇˇă 8.
From part (1) of this theorem and Theorem 5.7, we immediately deduce Corollary 5.14. -Let B u u u and B v v v be two shifts with positive weights acting on the real space ℓ p . The following are equivalent.

(a) B u u u and B v v v are not orthogonal with respect to product measures whose marginals are absolutely continuous with respect to Lebesgue measure on R.

(b) There exists κ ą 0 such that ř 8 n"1 ´1 ´κ u 1 ¨¨¨un v 1 ¨¨¨vn ¯2 ă 8. (c) B u u u and B v v v admit equivalent invariant Gaussian product measures.

We point out that Corollary 5.14 is also valid in the complex case and for not necessarily positive weights; see Corollary 6.12.

For the proof of Theorem 5.13, we will need the following lemma. (a) The function Ph is continuous and even, with Phpαq ď }h} 2 2 for avery α P R and Php0q " }h} 2 2 . (b) If h belongs to the Sobolev space W 1,2 pRq, then Ph is C 2 -smooth on R, with pPhq 1 p0q " 0 and pPhq 2 p0q ă 0. In particular, }h} 2 2 ´Phpαq " cα 2 as α Ñ 0, for some constant c ą 0. (c) If h R W 1,2 pRq, then p}h} 2 2 ´Phpαqq{α 2 Ñ 8 as α Ñ 0. (d) Assume that h is C 1 -smooth except at finitely many points, with h 1 P L 2 pRq, and that h has at least one discontinuity point. Then Ph is left-differentiable and rightdifferentiable at 0, with pPhq 1 p0 ´q ą 0 and pPhq 1 p0 `q ă 0.

This lemma has the following immediate consequence.

Corollary 5.16. -For any non-zero real-valued h P L 2 pRq, there exists a constant c ą 0 such that }h} 2 2 ´Phpαq ě cα 2 for α sufficiently close to 0. Note that using Plancherel's formula, it can be seen that this can be stated equivalently as follows: for any g P L 2 pRqzt0u with real-valued Fourier transform, there exists a constant c such that ş R |gptq| 2 `1 ´cospαtq ˘dt ě cα 2 for α sufficiently close to 0. Proof of Lemma 5.15. -For notational brevity, let us set F :" Ph.

(a) We have F pαq " xh, τ α hy L 2 pRq , where τ α hpxq :" hpx `αq, so F is continuous by continuity of the map α Þ Ñ τ α h from R into L 2 pRq. It is clear that F is even and that F p0q " }h} 2 2 . Finally, F pαq ď }h} 2 2 by the Cauchy-Schwarz inequality.

(b) Since

h P W 1,2 pRq, the map α Þ Ñ τ α h is C 1 -smooth from R into L 2 pRq, with derivative α Þ Ñ τ α h 1 . Since F pαq " xh, τ α hy L 2 pRq , it follows that F is C 1 -smooth on R, with F 1 pαq " xh, τ α h 1 y L 2 pRq " ż R hpuqh 1 pu `αq du " ż R hpu ´αqh 1 puq du.
Hence, the same argument shows that F is in fact C 2 -smooth on R, with F 2 pαq " ´żR h 1 pu ´αqh 1 puq du.

We have F 1 p0q " 0 because F is even; and

F 2 p0q " ´żR h 12 ă 0.
The strict inequality holds because if we had F 2 p0q " 0, then h 1 would be equal to 0 almost everywhere, so h would be 0 since it belongs to W 1,2 pRq. (c) Note that

F pαq " xh, τ α hy L 2 pRq " 1 2 `}h} 2 2 `}τ α h} 2 ´}h ´τα h} 2 2 ˘" }h} 2 2 ´1 2 }h ´τα h} 2 2 , so that }h} 2 2 ´F pαq " 1 2 }h ´τα h} 2 2 . Now since h R W 1,2 pRq, we have lim αÑ0 › › ταh´h α › › " 8
. Indeed, otherwise one could find a sequence pα n q tending to 0 such that τα n h´h αn has a weak limit u P L 2 pRq, and this would give that h P W 1,2 pRq with h 1 " u. This proves (c).

(d) Since F is an even function, it is enough to show that F is right-differentiable at 0 with F 1 p0 `q ă 0.

Let u 0 ă ¨¨¨ă u N be the discontinuity points of h. Also, let I 0 :" p´8, u 0 q, I N `1 :" pu n , 8q and I k :" pu k´1 , u k q for 1 ď k ď N . Note that by assumption on h, the restriction of h to each interval I k belongs to the Sobolev space W 1,2 pI k q. In particular, h has a left limit hpu ḱ q and a right limit hpu k q at each point u k .

For every α ą 0, we have

F pαq " N ÿ k"0 ż I k hpuqhpu `αq du ": N `1 ÿ k"0 F k pαq.
We consider separately the functions F 0 , F N `1 and F k for 1 ď k ď N . Let us start with F k , 1 ď k ď N . For 0 ă α ă u k ´uk´1 , we write

F k pαq " ż u k ´α u k´1 hpuqhpu `αq du `ż u k u k ´α hpuqhpu `αq du ": F k,1 pαq `Fk,2 pαq.
Consider the open triangle Ω :" tpα, βq; u k´1 ă β ă u k and 0 ă α ă u k ´βu, and the function G : Ω Ñ R defined by Gpα, βq :" ş β u k´1 hpuqhpu `αq du. Since the restriction of h to I k " pu k´1 , u k q belongs to the Sobolev space W 1,2 pI k q, the map G is C 1 -smooth on Ω, with B α Gpα, βq " ş β u k´1 hpuqh 1 pu `αq du and B β Gpα, βq " hpβqhpβ `αq. Moreover, G is continuous on Ω and its partial derivatives extend continuously to Ω, with B α Gpα, u k ´αq " ş u k ´α u k´1 hpuqh 1 pu `αq du and B β Gpα, u k ´αq " hpu k ´αqhpu ḱ q for 0 ă α ă u k ´uk´1 . It follows that F k,1 pαq " Gpα, u k ´αq is C 1 -smooth on p0, u k ´uk´1 q, with F 1 k,1 pαq " ş u k ´α u k´1 hpuqh 1 pu `αq du ´hpu k ´αqhpu ḱ q. To see this, observe that for any α 0 P p0, u k ´uk´1 q, the function G n pαq :" Gpα, u k ´α ´1 n q is well-defined in a neighbourhood of α 0 if n is large enough, that G n pαq Ñ F k,1 pαq uniformly in a neighbourhood of α 0 as n Ñ 8, and that G 1 n pαq Ñ ş u k ´α u k´1 hpuqh 1 pu `αq du ´hpu k ´αqhpu ḱ q uniformly in a neighbourhood of α 0 . Since F 1 k,1 pαq Ñ ş u k u k´1 hpuqh 1 puq du ´hpu ḱ q 2 as 0 `, we deduce that F k,1 is right-differentiable at 0 with

F 1 k,1 p0 `q " ż u k u k´1 hpuqh 1 puq du ´hpu ḱ q 2 " " 1 2 hpuq 2 ȷ u k u k´1 ´hpu ḱ q 2 " ´1 2
`hpu ḱ q 2 `hpu k´1 q 2 ˘.

Similar arguments show that F k,2 pαq "

ş u k u k ´α hpuqhpu `αq du is C 1 -smooth on p0, cq for some c ą 0, with F 1 k,2 pαq " ş u k u k
´α hpuqh 1 pu`αq du`hpu k ´αqhpu k q; and it follows that F k,2 is right-differentiable at 0 with F 1 k,2 p0 `q " hpu ḱ qhpu k q. Hence, F k is right-differentiable at 0, with

F 1 k p0 `q " hpu ḱ qhpu k q ´1 2 `hpu ḱ q 2 `hpu k´1 q 2 ˘.
One shows in the same way that F 0 pαq " ş u 0 ´8 hpuqhpu `αq du and F N `1pαq " ş 8 u N hpuqhpu `αq du are right-differentiable at 0, with

F 1 0 p0 `q " hpu 0 qhpu 0 q ´1 2 hpu 0 q 2 and F 1 N `1p0 `q " ´1 2 hpu Ǹ q 2 .
Altogether, we see that F " ř N `1 k"0 F k is right-differentiable at 0, with

F 1 p0 `q " hpu 0 qhpu 0 q ´1 2 hpu 0 q 2 `N ÿ k"1 ˆhpu ḱ qhpu k q ´1 2 `hpu ḱ q 2 `hpu k´1 q 2 ˘˙´1 2 hpu Ǹ q 2 " N ÿ k"0 hpu ḱ qhpu k q ´1 2 N ÿ k"0 `hpu ḱ q 2 `hpu k q 2 " ´1 2 N ÿ k"0
`hpu k q ´hpu ḱ q ˘2 ă 0.

We can now give the proof of Theorem 5.13.

Proof of Theorem 5.13.

-In what follows, we set f :" ? p and g :" ? q. Also, for each n ě 1, let

λ n :" u 1 ¨¨¨u n v 1 ¨¨¨v n
The following simple computation will be essential for the proof.

Fact 5.17. -For every n ě 1, we have

Hpµ u u u,n , µ v v v,n q " Ψpλ n q,
where Ψ : p0, 8q Ñ R is the function defined by

Ψpλq :" ? λ ż R f pλtqgptq dt.
Proof of Fact 5.17. -Denoting by τ the Lebesgue measure on R, we have dµ u u u,n dτ ptq " u 1 ¨¨¨u n ppu 1 ¨¨¨u n tq and dµ v v v,n dτ ptq " v 1 ¨¨¨v n qpv 1 ¨¨¨v n tq.

Hence

Hpµ u u u,n , µ v v v,n q " ż R ? u 1 ¨¨¨u n f pu 1 ¨¨¨u n tq ˆ?v 1 ¨¨¨v n gpv 1 ¨¨¨v n tq dt " c u 1 ¨¨¨u n v 1 ¨¨¨v n ż R f ˆu1 ¨¨¨u n v 1 ¨¨¨v n s ˙gpsq ds " Ψpλ n q.
By Fact 5.17 and Kakutani's Theorem, we know that m u u u and m v v v are not orthogonal if and only if (5.2)

8 ÿ n"1 p1 ´Ψpλ n qq ă 8;
and in that case, we have in particular that (5.3) Ψpλ n q Ñ 1 as n Ñ 8.

Proof of (1i) in Theorem 5.13. -We first note that we must have lim λ n ą 0 and lim λ n ă 8. Indeed, otherwise B u u u and B v v v are orthogonal with respect to measures not charging kerpe 0 q by Proposition 3.8. The measures µ u u u and µ v v v have this property since µ u u u,0 and µ v v v,0 do not charge t0u, so this cannot happen. Now, the key point is the following fact.

Fact 5.18. -The function Ψ is continuous on p0, 8q with 0 ď Ψpλq ď 1, there is at most one λ P p0, 8q such that Ψpλq " 1, and if Ψpλq " 1 then qptq " λppλtq almost everywhere.. If Ψpλq " 1 then, by the equality case in Cauchy-Schwarz's inequality and since ş R p " ş R q " 1, we must have qptq " λppλtq almost everywhere. It follows easily that there can be at most one λ such that Ψpλq " 1. Indeed, assume that Ψpλq " 1 " Ψpλ 1 q for some 0 ă λ ă λ 1 . Then λppλtq " λ 1 ppλ 1 tq almost everywhere. Setting c :" λ 1 {λ ą 1, it follows that ppxq " c ppcxq almost everywhere. Hence ppxq " c n ppc n xq almost everywhere for every n P Z. So, with I n :" pc n , c n`1 s, we obtain that Since the intervals I n form a partition of p0, 8q and ş 8 0 ppxq dx ă 8, this implies that ppxq " 0 almost everywhere on p0, 8q. Similarly ppxq " 0 almost everywhere on p´8, 0q, hence ppxq " 0 almost everywhere on R, which is a contradiction since ş R p " 1. Finally, to prove that Ψ is continuous, we note that Ψpλq " ? λ xf λ , gy L 2 pRq , where f λ ptq :" f pλtq. Since f P L 2 pRq, the map λ Þ Ñ f λ is continuous from p0, 8q into L 2 pRq; so Ψ is indeed continuous.

It is now easy to conclude the proof of (1i). Since Ψpλ n q Ñ 1 and Ψ is continuous on p0, 8q, any cluster point λ P p0, 8q of the sequence pλ n q must be such that Ψpλq " 1; hence, pλ n q has at most one cluster point in p0, 8q by Fact 5.18. Since 0 ă lim λ n ď lim λ n ă 8, it follows that pλ n q has a limit a P p0, 8q; and by Fact 5.18 again, we have (Ψpaq " 1 and) qptq " appatq almost everywhere.

Proof of (1ii) in Theorem 5.13. -Let Φ : p0, 8q Ñ R and Θ : p0, 8q Ñ R be defined by Φpλq :" ż R f ptqf pλtq dt and Θpλq :" ? λ Φpλq.

Since gptq " ? af patq almost everywhere by (1i), we have

Ψpλq " ? λ ż R f pλtq ˆ?af patq dt " a λ{a Φpλ{aq " Θpλ{aq.
So, we see that 0 ď Θpλq ď 1 for every λ P p0, 8q, and by (5.2):

(5.4)

8 ÿ n"1 `1 ´Θpλ n {aq ˘ă 8.
From that, we have to deduce that (5.5)

8 ÿ n"1 ˆ1 ´1 a λ n ˙2 ă 8.
To do this, we perform a change of variable in order to apply Lemma 5.15. If we set λ " e α , then

Θpλq " e α{2 ˆż `8

0 f ptqf pe α tq dt `ż 0 ´8 f ptqf pe α tq dt " e α{2 ż R f pe u qf pe α`u q e u du `eα{2 ż R f p´e u qf p´e α`u q e u du " ż R h `puqh `pu `αq du `żR h ´puqh ´pu `αq du,
where h `and h ´are the functions defined on R by h `pxq :" f pe x qe x{2 and h ´pxq :" f p´e x qe x{2 .

Note that the functions h `and h ´belong to L 2 pRq, and that

}h `}2 2 `}h ´}2 2 " ż R f 2 " 1.
So, with the notation of Lemma 5.15, we have (5.6) Θpλq " Ph ``logpλq ˘`Ph ´`logpλq ˘.

Since at least one of h ˘is non-zero, it follows that there exists a constant c ą 0 such that 1 ´Θpλq ě c logpλq 2 for λ sufficiently close to 1. Hence 1 ´Θpλq ě cpλ ´1q 2 for λ sufficiently close to 1 (and some other constant c ą 0), so that (5.5) is indeed a direct consequence of (5.4) since we already know that λ n Ñ a. This concludes the proof of (1ii).

Proof of (2) in Theorem 5.13. -By Kakutani's Theorem and with the notation of the proof of (1ii), we have to show that

8 ÿ n"1 `1 ´Θpλ n {aq ˘ă 8 if and only if 8 ÿ n"1 ˆ1 ´1 a u 1 ¨¨¨u n v 1 ¨¨¨v n ˙2 ă 8.
By (1ii), the "only if" implication is already known. To prove the converse, we keep the notation of the proof of (1ii). So we have as above

Θpλq " Ph ``logpλq ˘`Ph ´`logpλq ˘,
where the functions h `and h ´are defined by

h ˘pxq " f p˘e x qe x{2 .
Note that by assumption on f , the functions h `and h ´are continuous on R and C 1 -smooth except at a finite number of points. Moreover, h 1 ˘P L 2 pRq. Indeed, we have h 1 ˘pxq " ˘f 1 p˘e x qe 3x{2 `1 2 f p˘e x qe x{2 almost everywhere, and both terms belong to

L 2 pRq since ż R f 1 p˘e x q 2 e 3x dx " ż ˘p0,8q t 2 f 1 ptq 2 dt ă 8 and ż R f p˘e x q 2 e x dx " ż ˘p0,8q
f ptq 2 dt ă 8.

So h ˘belongs to the Sobolev space W 1,2 pRq. Hence, by Lemma 5.15 and since }h `}2 2 }h 2

´}2 " 1, we know that the function F :" Ph ``P h ´is C 2 -smooth on R, with F p0q " 1, F 1 p0q " 0 and F 2 p0q ă 0. It follows that there is a constant c ą 0 such that 1 ´Θpλq " cpλ ´1q 2 as λ Ñ 1; and this concludes the proof of the "if" implication in [START_REF] Bayart | Frequently hypercyclic operators[END_REF].

Proof of (3) in Theorem 5.13. -We keep the notation of the proofs of (1ii) and (2). This time, we have to show that (5.7)

8 ÿ n"1 `1 ´Θpλ n {aq ˘ă 8 if and only if 8 ÿ n"1 ˇˇˇ1 ´1 a λ n ˇˇˇă 8.
As above, let F :" Ph ``P h ´, so that Θpλq " F `logpλq ˘. By assumption on p, the functions h `pxq " f pe x qe x{2 and h ´pxq " f p´e x qe x{2 are C 1 -smooth except at finitely many points, with h 1 ˘P L 2 pRq, and at least one of them has a discontinuity point. By Lemma 5.15, it follows that there exists two constants c ą 0 and c 1 ă 8 such that c |α| ď 1 ´F pαq ď c 1 |α| for α sufficiently close to 0. Hence c |1 ´λ| ď 1 ´Θpλq ď c 1 |1 ´λ| for λ sufficiently close to 1, and (5.7) follows.

The proof of Theorem 5.13 is now complete.

? λ ş R f ptqf pλtq dt at λ " 1.
It is perhaps worth noticing that if Θ happens to be differentiable at λ " 1, then we necessarily have Θ 1 p1q " 0. This is because Θ is "symmetric with respect to 1", i.e. Θp1{λq " Θpλq.

Remark 5.20. -Part (3) of Theorem 5.13 may be applied for example if µ u u u,0 is the uniform distribution over some bounded interval I Ď R and µ v v v,0 is the uniform distribution over 1 a I.

Remark 5.21. -Part (2) of Theorem 5.7 remains valid under the following assumption on f " ? p: the restrictions of f to the intervals p0, 8q and p´8, 0q have locally integrable derivatives (in the distribution sense), and tf 1 ptq P L 2 pRq.

Proof. -Looking back at the proof of (2), we see that the only thing to check is that the functions h `and h ´belong to the Sobolev space W 1,2 pRq. We do this for h `and, for notational simplicity, we set h :" h `and we denote by f the restriction of f to p0, 8q. So we have to show that if f P L 2 p0, 8q has a distributional derivative f 1 P L loc 1 p0, 8q such that tf ptq P L 2 p0, 8q, then hpxq " f pe x qe x{2 P W 1,2 pRq.

Since f P L 2 p0, 8q, it is clear that h P L 2 pRq. To find the distributional derivative of h, let us fix a test function φ P DpRq. Let ψ P Dp0, 8q be the function defined by the relation φpxq " ψpe x qe x{2 , i.e. ψptq " 1 ? t φplogptqq. Writing φ 1 pxq " e 3x{2 ψ 1 pe x q `1 2 e x{2 ψpe x q and using the change of variable x " logptq, an elementary computation reveals that

ż R hpxq φ 1 pxq dx " ´ż 8 0 ˆe3x{2 f 1 pe x q `1 2 e x{2 f pe x q ˙φpxq dx.
This means that h has a distributional derivative h 1 P L loc 1 pRq given (as expected) by h 1 pxq " e 3x{2 f 1 pe x q `1 2 e x{2 f pe x q. Since f P L 2 p0, 8q and tf 1 ptq P L 2 p0, 8q, we see that h 1 P L 2 pRq, hence h P W 1,2 pRq. 5.5. Products of discrete measures. -Theorem 5.13 [START_REF] Bayart | Invariant Gaussian measures for operators on Banach spaces and linear dynamics[END_REF] shows that for weighted shifts acting on ℓ p , to admit non-orthogonal invariant measures which are products of absolutely continuous measures having some "singularities" is in fact a strictly stronger requirement on the weights than to admit equivalent invariant Gaussian product measures. The next proposition goes along the same lines, in an even more dramatic way. Proposition 5.22. -Let u u u and v v v be two weight sequences.

(a) If B u u u and B v v v admit non-orthogonal invariant product measures m u u u " b ně0 µ u u u,n and m v v v " b ně0 µ v v v,n such that µ u u u,0 and µ v v v,0 have non-zero discrete parts and do not charge t0u, then ˇˇu 1 ¨¨¨un v 1 ¨¨¨vn ˇˇis eventually constant (and hence B u u u and B v v v admit equivalent invariant Gaussian product measures). (b) Assume that B u u u and B v v v admit equivalent non-trivial invariant product measures or, more generally, non-trivial invariant product measures m u u u " b ně0 µ u u u,n and m v v v " b ně0 µ v v v,n such that µ u u u,n and µ v v v,n have the same support for each n ě 0. Let S :" supppµ u u u,0 q " supppµ v v v,0 q. If either S is compact or 0 R S, then ˇˇu 1 ¨¨¨un v 1 ¨¨¨vn ˇˇ" 1 for all n ě 1; and hence u u u " v v v if u u u and v v v are positive weight sequences.

Proof. -For each n ě 1, let us set as usual λ n :" u 1 ¨¨¨un v 1 ¨¨¨vn (a) Since the measures µ u u u,n and µ v v v,n do not charge t0u, one can find a countable multiplicative subgroup S of Kzt0u such that the discrete parts of all measures µ u u u,n and µ v v v,n are supported on S and S contains the set tu n ; n ě 1u Y tv n ; n ě 1u. Denote by τ S the counting measure on S. Let also τ c be a continuous measure such that the continuous parts of all measures µ u u u,n and µ v v v,n are absolutely continuous with respect to τ c . Then, one can take τ :" τ c `τS to compute Hpµ u u u,n , µ v v v,n q, for every n ě 0.

For each n ě 0, write

µ u u u,n " α n τ c `pn τ S and µ v v v,n " β n τ c `qn τ S
for some non-negative measurable functions α n , p n , β n , q n . Without loss of generality, we may assume that α n 1 S " β n 1 S " 0 and p n 1 KzS " q n 1 KzS " 0, so that we also have µ u u u,n " pα n `pn qτ and µ v v v,n " pβ n `qn qτ.

Hence,

Hpµ u u u,n , µ v v v,n q " ż K ? α n `pn a β n `qn dτ " x ? α n `pn , a β n `qn y L 2 pτ q .
Now, by Kakutani's Theorem we know that Hpµ u u u,n , µ v v v,n q Ñ 1 as n Ñ 8. Since } ?

α n `pn } L 2 pτ q " } a β n `qn } L 2 pτ q " 1, it follows that › › ? α n `pn ´aβ n `qn › › L 2 pτ q Ñ 0. Moreover, by assumption on α n , β n , p n , q n , we have

› › ? α n `pn ´aβ n `qn › › 2 L 2 pτ q " › › ? α n ´aβ n › › 2 L 2 pτcq `› › ? p n ´?q n › › 2 L 2 pτ S q
. Hence, we have in particular that } ? p n ´?q n } 2 L 2 pτ S q Ñ 0 as n Ñ 8; and since τ S is the counting measure on S, this means that (5.8)

ÿ sPS ´ap n psq ´aq n psq ¯2 Ñ 0 as n Ñ 8.
So far, we have not used the fact that the measures m u u u and m v v v are invariant under B u u u and B v v v . Set p :" p 0 and q :" q 0 . Then, the discrete parts of µ u u u,0 and µ v v v,0 are respectively ř sPS ppsq δ s and ř sPS qpsq δ s . So, by the invariance properties and since S is a multiplicative group containing u 1 ¨¨¨u n and v 1 ¨¨¨v n , the discrete parts of µ u u u,n and µ v v v,n (for n ě 1) are respectively ř sPS ppu 1 ¨¨¨u n sq δ s and ř sPS qpv 1 ¨¨¨v n sq δ s ; in other words, we have p n psq " ppu 1 ¨¨¨u n sq and q n psq " qpv 1 ¨¨¨v n sq for all s P S. Hence, if we set f :" ? p and g :" ? q, (5.8) can be re-written as follows:

ÿ sPS `f pu 1 ¨¨¨u n sq ´gpv 1 ¨¨¨v n sq ˘2 Ñ 0 as n Ñ 8
or, equivalently, ÿ sPS `f pλ n sq ´gpsq ˘2 Ñ 0.

It follows in particular, f pλ n sq Ñ gpsq for all s P S. So, taking any s such that gpsq ą 0 (such an s exists since the discrete part of µ v v v,0 is non-zero), we see that one can find ε ą 0, e.g. ε :" gpsq{2, such that f pλ n sq ě ε for all n sufficiently large. It follows that the sequence pλ n q can take only finitely many distinct values: indeed, otherwise f ps 1 q ě ε for infinitely many s 1 P S, a contradiction since f P ℓ 2 pSq. Now, assume that |λ n | is not eventually constant. Then, since λ n takes only finitely many values, λ n P S and f pλ n sq Ñ gpsq for all s P S, we see that one can find λ, λ 1 P S with |λ| ‰ |λ 1 | such that f pλsq " f pλ 1 sq for all s P S. Setting α :" λ{λ 1 , we then have f pα k sq " f psq for any s P S and all k P N; and taking any s such that f psq ą 0 (again, such an s exists), we obtain a contradiction since f P ℓ 2 pSq and the α k , k P N are pairwise distinct. This concludes the proof of (a). By the invariance properties, we have supppµ u u u,n q " 1 u 1 ¨¨¨un S and supppµ v v v,n q " 1 v 1 ¨¨¨vn S, so that λ n S " S. Hence λ r n S " S for all r P Z. If S is compact, it follows that S " t0u by letting r Ñ `8; hence µ u u u,0 " δ 0 " µ v v v,0 , which is the required contradiction. If 0 R S, then we get S " H by letting r Ñ ´8, which is another contradiction. n"1 1 pv 1 ¨¨¨vnq p ă 8. The weighted shifts B u u u and B v v v acting on ℓ p are not orthogonal with respect to product measures whose marginals have non-zero discrete parts and do not charge t0u if and only if u 1 ¨¨¨un v 1 ¨¨¨vn is eventually constant; and in that case they admit equivalent invariant product measures with purely discrete marginals not charging t0u.

Proof. -The "only if" part follows from Proposition 5.22 (a). Conversely, assume that λ n :" u 1 ¨¨¨un v 1 ¨¨¨vn is eventually constant, say λ n " λ for n ą n 0 . Let S be the multiplicative subgroup of p0, 8q generated by the set tu n ; n ě 1u Y tv n ; ě 1u. Let p : S Ñ R be a strictly positive probability density function such that ř sPS ppsq s p ă 8, and let q : S Ñ R `be the probability density function defined by qpsq :" ppλsq. Finally, let µ u u u,0 :" ř sPS ppsqδ s and µ v v v,0 :" ř sPS qpsqδ s , and denote by m u u u " ' ně0 µ u u u,n and m v v v " ' ně0 µ v v v,n the associated B u u u -invariant and B v v v -invariant measures on K Z `. By Corollary 5.4 and the assumption on u u u, v v v and p, the measures m u u u and m v v v are supported on ℓ p , and they have purely discrete marginals not charging t0u. Finally, by the proof of Proposition 5.22 (a), we have Hpµ u u u,n , µ v v v,n q " ř sPS a ppλ n sqqpsq for all n ě 0, so that Hpµ u u u,n , µ v v v,n q " ř sPS qpsq " 1 for all n ą n 0 . Since µ u u u,n " µ v v v,n for all n ě 0, it follows that m u u u " m v v v . This concludes the proof. Remark 5.24. -In Proposition 5.22 (b), one cannot replace "equivalent" by "nonorthogonal". For example, let u u u and v v v be defined as follows: u n " 2 for all n ě 1, v 1 " 1, v 2 " 4 and v n " 2 for all n ě 3. Then u u u ‰ v v v (!), and yet B u u u and B v v v admit non-orthogonal product measure m u u u " b ně0 µ u u u,n and m v v v " b ně0 µ v v v,n for which µ u u u,0 " µ v v v,0 has a compact support not containing 0; for example, µ u u u,0 " µ v v v,0 could be the uniform distribution on the interval r1, 3s.

6. Additional facts 6.1. Invariant measures with symmetry properties. -It is clear that if an operator T acting on a Polish topological vector space X admits a non-trivial invariant measure m, then the measure r m defined by r mpAq :" 1 2 pmpAq `mp´Aqq is a non-trivial invariant measure for both T and ´T , and that r m is additionally symmetric, i.e. r mp´Aq " r mpAq for every Borel set A Ď X. The next proposition goes along the same lines. Proposition 6.1. -Let X be a Polish space, and let T be a continuous self-map of X admitting an invariant measure m. Let also G be a compact abelian group acting continuously on X, and assume that T g " βpgqT for all g P G, where β : G Ñ G is a continuous map invariant under the Haar measure of G. Let r m be the measure on X defined by

r mpAq :" ż G mpg ´1Aq dg,
where dg denotes integration with respect to the Haar measure of G. Then r m is Ginvariant and both gT -invariant and T g -invariant for every g P G.

Proof. -It is clear that r m is G -invariant. Moreover, if h P G then, for any bounded Borel function f : X Ñ R `, we have

ż X f ˝pβphqT q d r m " ż X ˆżG f pβphqT pgxqq dg ˙dmpxq " ż X ˆżG f pβphqβpgqT pxqq dg ˙dmpxq " ż X ˆżG f pβphqgT pxqq dg ˙dmpxq " ż X ˆżG f pgT pxqq dg ˙dmpxq " ż G ˆżX f pguq dmpuq ˙dg " ż X f d r m.
Since β is necessarily onto (because the Haar measure of G has full support), it follows that r m if gT -invariant for every g P G; and hence also T g -invariant since T g " βpgqT .

Corollary 6.2. -Let X be a Polish space, and let T 1 , T 2 be two continuous self-maps of X. Let also G be a compact abelian group acting continuously on X, and assume that T i g " gT i for all g P G and i " 1, 2. Finally, let x 0 P X be a fixed point for the action of G. If T 1 and T 2 admit equivalent (resp. non-orthogonal) invariant measures m 1 , m 2 not charging tx 0 u, then they also admit equivalent (resp. non-orthogonal) invariant measures not charging tx 0 u which are additionally G -invariant.

Proof. -Let Ă m 1 and Ă m 2 be defined as in Proposition 6.

1. It is clear that Ă m 1 , Ă m 2 do not charge tx 0 u. So we just have to check that if m 1 " m 2 then Ă m 1 " Ă m 2 , and that if Ă m 1 K Ă m 2 then m 1 K m 2 .
Assume that m 1 " m 2 . Let A Ď X be a Borel set such that Ă m 1 pAq " 0. Then m 1 pgAq " 0 for almost every g P G; hence m 2 pgAq " 0 for almost every g P G since m 2 ! m 1 , and hence Ă m 2 pAq " 0. This shows that Ă m

2 ! Ă m 1 ; and similarly Ă m 1 ! Ă m 2 . Assume that Ă m 2 K Ă m 1 .
Let A Ď X be a Borel set such that Ă m 1 pAq " 0 and Ă m 2 pXzAq " 0. Then m 1 pgAq " 0 for almost every g P G and m 2 pXzgAq " m 2 pgpXzAqq " 0 for almost every g P G. So one can find at least one g such that m 1 pgAq " 0 " m 2 pXzgAq, which shows that m 1 K m 2 . Corollary 6.3. -Let X be a Polish topological vector space, and let T P LpXq. Assume that κT admits a non-trivial invariant measure for some κ P K. Then, there exists a non-trivial measure which is aT -invariant for all a P K such that |a| " |κ|.

Proof. -Apply Proposition 6.1 with the group G :" tω P K; |ω| " 1u acting by multiplication on X. Corollary 6.4. -Let B u u u be a backward shift acting on ℓ p . If B u u u admits a non-trivial invariant measure, then there exists a measure m on ℓ p with full support which is invariant for all backward shifts B v v v such that |v n | " |u n | for all n ě 1.

Proof. -Since ř 8

n"1 1 |u 1 ¨¨¨un| p ă 8, we know that B u u u admit an invariant measure m with full support (which can even be taken to be Gaussian and ergodic for B u u u ). Let G :"

! pω n q ně0 P K Z `; |ω n | " 1 for all n ě 0 ) .
This is a compact abelian group acting continuously on X :" K Z `by coordinatewise multiplication. Moreover, if we set T :" B u u u acting on X, we have B u u u g " σpgqB u u u for every g P G, where σ : G Ñ G is (the restriction to G of) the canonical backward shift. Hence, considering m as a Borel measure on X, we may apply Proposition 6.1. This gives a measure r m on K Z `which is B v v v -invariant for every weight sequence v v v such that |v n | " |u n | for all n ě 1. Moreover, since B u u u is supported on ℓ p and ℓ p is G -invariant, r m is supported on ℓ p ; and since m has full support, it is readily checked that r m has full support.

Remark 6.5. -It is well-known (see [START_REF] Shields | Weighted shift operators and analytic function theory[END_REF]) that two weight sequences u u u and v v v are such that |u n | " |v n | for all n ě 1 if and only if the backward shifts B u u u and B v v v (acting on any ℓ p ) are unitarily similar, i.e. there exists an isometry J of ℓ p such that B v v v " JB u u u J ´1.

6.2. Product measures charging small subspaces. -The following proposition says in essence that whether or not a product measure K Z `invariant under a weighted shift B w w w is supported on some "small" subspace of K Z `depends on the rate of growth of the products w 1 ¨¨¨w n .

Proposition 6.6. -Let µ 0 be a Borel probability measure on K, let w w w be a sequence of non-zero scalars, and let µ w w w " b ně0 µ n be the B w w w -invariant measure on K Z `defined by µ 0 , i.e. µ n pAq " µ 0 pw 1 ¨¨¨w n Aq for each n ě 1 and every Borel set A Ď K. then µ w w w pℓ 1 q " 1. (c) If |w 1 ¨¨¨w n | Ñ 8 and µ 0 ‰ δ 0 , then µ w w w pc 0 q " 0. If |w 1 ¨¨¨w n | Ñ 8 and µ 0 does not have compact support, then µ w w w pℓ 8 q " 0.

Proof. -(a) This is Corollary 5.4.

(b) Since ℓ 1 is a tail subset of K Z `, we have µ w w w pℓ 1 q " 0 or 1, by Kolmogorov's 0 -1 law. Let A :"

␣ x P K Z `; |x n | ď ε n for all n ě 1 ( .
Obviously A Ď ℓ 1 , and we have µ w w w pAq "

8 ź n"1 µ n `|t| ď ε n " 8 ź n"1 µ 0 `|t| ď |w 1 ¨¨¨w n | ε n ˘.
Since ř 8 n"1 `1 ´µ0 p|t| ď |w 1 ¨¨¨w n | ε n q ˘ă 8 by assumption, it follows that µ w w w pAq ą 0, and hence µ w w w pℓ 1 q " 1. n"0 α 2 n " 8. Then, one can find a sequence of real numbers pβ n q such that ř 8 n"0 β 2 n ă 8, β n α n ě 0 for all n and ř 8 n"0 β n α n " 8. Now, consider the measure τ :" 1 2 p r m `r m 1 q. By our assumption, the sequence pe nq ně0 is orthogonal and bounded in L 2 pΩ, τ q. Since ř 8 n"0 β 2 n ă 8, it follows that the series ř β n e n is convergent in L 2 pΩ, τ q. Hence, there is an increasing sequence of integers pN k q kě0 such that if we define f k :" ř N k n"0 e n, then the sequence of linear functionals pf k q converges τ -a.e. on Ω. So, the linear subspace

E :" ␣ x P Ω; the sequence pf k pxqq is convergent ( Ď Ω
is such that τ pEq " 1 and hence r mpEq " 1. On the other hand, we have f k pα α αq " ř N k n"0 β n α n , so the sequence pf k pα α αqq is not convergent, i.e. α α α R E; and since E is a linear subspace of Ω, it follows that E X pE `α α αq " H. So we have τ pE `α α αq " 0, and hence r m 1 α α α pEq " r m 1 pE `α α αq " 0. Since r m and r m 1 α α α are not orthogonal, this is a contradiction. Now, let us consider the general case. Let c :" ş R x dr µpxq and c 1 :" ş R x dr µ 1 pxq, so that ş Ω e n d r m " c and ş Ω e n d r m 1 " c 1 for all n ě 0. Then ş Ω e n d r m c " 0 and ş Ω e n d r m 1 c 1 " 0 for all n ě 0, where c :" pc, c, . . . q and c 1 :" pc 1 , c 1 , . . . q. Since the measures r m c and `r m 1 c 1 ˘α α α`c´c 1 " `r m 1 α α α ˘c are not orthogonal, it follows that ř 8 n"0 pα n ´pc 1 ´cqq 2 ă 8.

From Theorem 6.10, it is essentially a formal matter to deduce the following theorem, which is not far from saying that if two weighted shifts are not orthogonal with respect to product measures whose marginals do not charge t0u, then they admit equivalent invariant Gaussian product measures. Theorem 6.11. -Let u u u and v v v be two sequences of non-zero scalars, and let m u u u " b ně0 µ n,u u u and m v v v " b ně0 µ n,v v v be two product measures on K Z `, respectively B u u u -invariant and B v v v -invariant, with µ 0,u u u pt0uq " 0 " µ 0,v v v pt0uq. Assume that the measures m u u u and m v v v are not orthogonal. Finally, let λ n :" u 1 ¨¨¨un v 1 ¨¨¨vn for all n ě 1. (1) If there exists a P Kzt0u such that µ 0,v v v pAq " µ 0,u u u paAq for every Borel set A Ď K, then ř Proof. -(1) Since µ 0,u u u pt0uq " 0 " µ 0,v v v pt0uq, we view µ 0,u u u and µ 0,v v v as measures on G :" Kzt0u, and hence we view m u u u and m v v v as measures on G Z `. Note that G Z `is an abelian group under entry-wise multiplication. Consider the measures m " b ně0 µ n and m 1 " b ně0 µ 1 n where µ n :" µ 0,u u u and µ 1 n :" µ 0,v v v for all n ě 0. If we set U :" p1, u 1 , u 1 u 2 , . . . q, V :" p1, v 1 , v 1 v 2 , . . . q and a :" pa, a, a, . . . q, then we have for every Borel set B Ď G Z which concludes the proof.

(2) The proof is similar, using Theorem 6.10 (2).

Corollary 6.12. -Let B u u u and B v v v be two weighted shifts acting on ℓ p . The following are equivalent.

(a) B u u u and B v v v are not orthogonal with respect to product measures whose marginals are absolutely continuous with respect to Lebesgue measure on K. (b) B u u u and B v v v are not orthogonal with respect to product measures whose marginals do not charge t0u and are such that log |t| P L 2 .

(c) There exists κ ą 0 such that ř 8 n"1 ´1 ´κ ˇˇu 1 ¨¨¨un v 1 ¨¨¨vn ˇˇ¯2 ă 8. (d) B u u u and B v v v admit equivalent invariant Gaussian product measures.

Proof. -By Theorem 6.11 and keeping the same notation, we just have to check that if µ 0,u u u and µ 0,v v v are absolutely continuous with respect to Lebesgue measure on K, then there exists a P Kzt0u such that µ 0,v v v pAq " µ 0,u u u paAq for every Borel set A Ď K. When K " R this is done in the proof of (1i) in Theorem 5.13. We repeat the argument here, in a way that makes it work when K " C as well. Write µ 0,u u u " pptqdt and µ 0,v v v " qptqdt. Then Ψpλ n q Ñ 1 as n Ñ 8, where Ψ : Kzt0u Ñ r0, 1s is the function defined by Ψpλq :" |λ| d{2 ş K f pλtqgptq dt and d " 1 or 2 depending on whether K " R or C. Moreover, since B u u u and B v v v must be similar by Proposition 3.8, the sequence pλ n q has a cluster point a P Kzt0u. Then Ψpaq " 1 by continuity of Ψ, and it follows that qptq " |a| d ppatq almost everywhere. 6.5. Kakutani's Theorem. -In this section, our aim is to prove the following variant of Kakutani's Theorem from [START_REF] Kakutani | On equivalence of infinite product measures[END_REF]. Theorem 6.13. -Let ν " b ně0 ν n and ν 1 " b ně0 ν 1 n be two product probability measures on Ω " ś ně0 Ω n . The measures ν and ν 1 are orthogonal if and only if ś 8 n"0 Hpν n , ν 1 n q " 0. Moreover, if ś 8 n"0 Hpν n , ν 1 n q ą 0 and ν n ! ν 1 n for all n, then ν ! ν 1 .

Note that this variant was used in the proofs of Theorem 5.13 and Proposition 5.22. We give a rather detailed proof since we were not able to locate one in the literature. However, this is essentially a "copy and paste" of Kakutani's original proof of his theorem.

Proof of Theorem 6.13. -For each n ě 0, let us choose a Radon-Nikodym derivative of ν n with respect to ν (i) Assume that ś 8 n"0 Hpν n , ν 1 n q ą 0, and let us show that the measures ν and ν 1 are not orthogonal, with ν ! ν 1 if ν n ! ν 1 n for all n. For each k ě 0, let F k :" b k n"0 f n , i.e. F k : Ω Ñ R `is the function defined by

F k pωq :" k ź n"0 f n pω n q.
We claim that F k converges in L 2 pΩ, ν 1 q to some F : Ω Ñ R `. Indeed, if 0 ď p ă q, then }F q ´Fp } 2 L 2 pν 1 q " }F p } 2

L 2 pν 1 q › › ›1 ´bq n"p`1 f n › › › 2 L 2 pν 1 q ď › › ›1 ´bq n"p`1 f n › › › 2 L 2 pν 1 q ď 2 ˜1 ´q ź n"p`1
Hpν n , ν 1 n q ¸; so we see that pF k q kě0 is a Cauchy sequence in L 2 pΩ, νq.

By L 2 -convergence and since ş Ωn f 2 n dν 1 n ě ´şΩn f n dν 1 n ¯2 for all n ě 0, we have Hpν n , ν 1 n q 2 ą 0.

Moreover, for any cylinder set A " rA 0 , . . . , A N s Ď Ω, we have where the last inequality follows from the fact that ş Ωn f 2 n dν 1 n ď 1 for every n ą N . Therefore, the positive measure F 2 ν 1 is non-zero and absolutely continuous with respect to both ν 1 and ν, and hence ν and ν 1 are not orthogonal. Finally, if ν n ! ν 1 n for all n, i.e. α n " 0, then we see that νpAq " ş A F 2 dν 1 for all cylinder sets A, so ν ! ν 1 . (ii) Now, assume that ś 8 n"0 Hpν n , ν 1 n q " 0, and let us show that ν K ν 1 . It is enough to show that for any ε ą 0, one can find a measurable set E Ď Ω such that ν 1 pEq ă ε and νpΩzEq ă ε.

Let ε ą 0, and let us choose N ě 0 such that ś N n"0 Hpν n , ν 1 n q ă ε, i.e. ż Ω F N dν 1 ă ε.

Let us also choose, for each n P 0, N , a measurable set A n Ď Ω n such that ν 1 n pA n q " 0 and α n pΩ n zA n q " 0, and let E :" tω P Ω; F N pωq ě 1u Y tω P Ω; ω n P A n for some n P 0, N u.

We have on the one hand ν 1 pEq " ν 1 pF N ě 1q ă ε; and, on the other hand, νpΩzEq " b 8

n"0 pp n ν 1 n `αn q `tF N ă 1u X rΩ 0 zA 0 , . . . , Ω N zA N s " b N n"0 pp n ν 1 n q `tpω 0 , . . . , ω N q; a p 0 pω 0 q ¨¨¨p N pω N q ă 1u " This concludes the proof.

Some questions

Of course, the main open question that remains at the end of this work is the one from which we started. In Section 3, we saw that orthogonality might be related to non-similarity, in a way which is not yet clear. In particular, in view of Theorem 3.10, the following question is natural. In the same spirit, Example 4.5 shows that non-similarity does not imply orthogonality. Conversely, it would be quite interesting to know whether similarity implies nonorthogonality. Does it follow that B u u u and B w w w are non-orthogonal?

  0ďdďN |z d | ą 0 for every z P K. Indeed, the compact set K is covered by the open sets O d :" tz P X; |z d | ą 0u. Choose γ ą 0 such that max 0ďdďN |z d | ě γ for every z P K, and M ă 8 such that }z} ď M for every z P K. As in the proof of Proposition 3.8, we get that for every n ě 1,

Corollary 3 .

 3 11. -Assume that the weight sequences u u u ad v v v are bounded below, i.e. inf ně1 |u n | ą 0 and inf ně1 |u n | ą 0. If B u u u and B v v v are not similar, then they are orthogonal.

Remark 4 . 6 .

 46 -In the above example, the weights are not bounded below. This is not accidental: indeed, if two weighted shifts B u u u and B v v v share a non-zero periodic point and if inf ně1 |u n | ą 0 and inf ně1 |v n | ą 0, then B u u u and B v v v are necessarily similar by Corollary 3.11. (Alternatively, this follows easily from Theorem 4.2.) 5. Non-orthogonality via product measures 5.1. Invariant product measures for weighted shifts.

  dτ " 0 τ -a.e. since ν K β,

Corollary 5 . 8 . 8 n"0 1 |u 1 ¨¨¨un| p ă 8 and ř 8 n"0 1 |v 1 8 ř n" 1 ´1 ´κ |u 1 ¨¨¨un| |v 1

 5881811811 -Let B u u u and B v v v be two weighted shifts on X " ℓ p , where the weight sequences u u u and v v v are such that ř ¨¨¨vn| p ă 8. The following are equivalent.(a) B u u u and B v v v admit equivalent invariant Gaussian product measures.(b) There exists a constant κ ą 0 such that ¨¨¨vn| ¯2 ă 8.

8 ř n" 1 1|w 1 8 ř

 8118 ¨¨¨wn| p ă 8 for all w w w P Λ Λ Λ and

Lemma 5 . 15 .

 515 -Let h P L 2 pRq be non-zero and real-valued, and let Ph : R Ñ R be the function defined by Phpαq :" ż R hpxqhpx `αq dx.

Proof of Fact 5 . 18 .

 518 -It is clear that Ψpλq ě 0. Moreover, by the Cauchy-Schwarz inequality

ż

  In ppxq dx " ż I 0 c n ppc n xq dx " ż I 0 ppxq dx for all n P Z.

Remark 5 .

 5 19. -The proofs of parts (2) and (3) of Theorem 5.13 rely on the local properties of the function Θpλq "

  (b) Let us fix n ě 1 and, towards a contradiction, assume that |λ n | ‰ 1, say |λ n | ă 1.

Corollary 5 . 8 n" 1 1pu 1 ¨¨¨unq p ă 8 and ř 8

 5818 23. -Let u u u and v v v be positive weight sequences such that ř

( a ) 8 n" 1 1|w 1 ¨¨¨wn| p ă 8 , 8 ÿ n" 1 µ 0

 a818810 Let p P r1, 8q. If ş K |t| p dµ 0 ptq ă 8 and ř then µ w w w pℓ p q " 1. (b) If there exists a summable sequence of positive real numbers pε n q such that `|t| ą |w 1 ¨¨¨w n | ε n ˘ă 8,

8 n"0 α 2 n ă 8 , i.e. 8 ÿ n" 1 plog |a ´1λ n |q 2 ă 8 ;

 88818 u pBq " mpUBq and m v v v pBq " m 1 pVBq " mpaVBq. Now, consider the map L : G Z `Ñ R Z `defined by Lpt 0 , t 1 , . . . q :" `log |t 0 |, log |t 1 |, . . . ˘. Denote by r m u u u , r m v v v , r m and r m 1 the images of m u u u , m v v v , m and m 1 under this map L. Then r m " b ně0 r µ n and r m 1 " b ně0 r µ 1 n , where r µ n and r µ 1 n are the images of µ 0,u u u and µ 0,v v v under the map t Þ Ñ log |t|, for every n ě 0. Moreover, since L is a group homomorphism, we have for every Borel set B Ď R Z `: r m u u u pBq " r m `B `LpUq ˘and r m u u u pBq " r m `B `LpaVq ˘. Now, the measures r m u u u " m u u u ˝L´1 and r m v v v " m v v v ˝L´1 are not orthogonal since m u u u and m v v v are not orthogonal. So their translates by ´LpaVq are not orthogonal either. In other words, if we set α α α :" LpUV ´1a ´1q " p0, log |a ´1λ 1 |, log |a ´1λ 2 |, . . . q, then the measure r m and r m α α α are not orthogonal. By Theorem 6.10 (1), it follows that ř

F 2

 2 dν 1 ,

ΩF

  ż ? p 0 b¨¨¨bp N ă1 p 0 b ¨¨¨b p N dν 1 ď ż Ω ? p 0 b ¨¨¨b p N dν 1 " ż N dν 1 ă ε.

Question 7 . 1 .

 71 -Find a characterization of the pairs of weight sequences pu u u, v v vq such that the weighted shifts B u u u and B v v v acting on ℓ p are orthogonal.

Question 7 . 2 .N ě 0 or lim nÑ8 min 0ďdďN ˇˇˇu 1 ¨¨¨u n`d v 1

 721 -Is it true that if B u u u and B v v v admit non-trivial invariant measures and are orthogonal, then eitherlim nÑ8 max 0ďdďN ˇˇˇu 1 ¨¨¨u n`d v 1 ¨¨¨v n`dˇˇˇ" 0 for all ¨¨¨v n`d ˇˇˇ" 8 for all N ě 0?

Question 7 . 3 .

 73 -Suppose that B u u u and B v v v admit non-trivial invariant measures and are similar, i.e.0 ă lim ˇˇˇu 1 ¨¨¨u n v 1 ¨¨¨v n ˇˇˇď lim ˇˇˇu 1 ¨¨¨u n v 1 ¨¨¨v n ˇˇˇă 8.

  Theorem 2.1]. Since this lemma might be useful in other situations, we state it explicitely. Let us denote by e 0 the first coordinate functional on X " ℓ p .

Lemma 2.2. -Let B w w w be a weighted shift acting on X " ℓ p pZ `q. Assume that one can find a vector x P ℓ p and a set of integers N such that (i) N has positive upper density,

(ii

) sup nPN }B n w w w x} ă 8, (iii) inf nPN |xe 0 , B n w w w xy| ą 0. Then, one can conclude that 8 ř n"1 1 |w 1 ¨¨¨wn| p ă 8. Sketch of the proof of Lemma 2.2. -Write x " ř 8 n"0 x n e n . Let C 1 :" sup nPN }B n w w w x} and C 2 :" inf nPN |xe 0 , B n w w w xy| " inf nPN |w 1 ¨¨¨w n x n |. If we let v n :" pw 1 ¨¨¨w n q ´1, we have for every n P N and every integer

  Proof of Fact 3.3. -By regularity of the measures m 1 and m 2 and since m 1 and m 2 do not charge Λ, there exists a compact set K Ď XzΛ such that m 1 pKq ą 1 ´ε{2 and m 2 pKq ą 1 ´ε{2. Let I be an infinite subset of N such that (3.1) holds true.

			Then
	m 1	`T ´n 1 pKq X T ´n 2 pKq ˘Ñ 0 as n Ñ 8 along I by Fatou's Lemma; so one can find an
	integer n ě 1 such that m 1	`T ´n 1 pKqXT ´n 2 pKq ˘ă ε{2. Since m 1

`T ´n 1 pKq ˘" m 1 pKq ą 1έ {2, it follows that m 1 `T ´n 2 pKq ˘ă ε. On the other hand, m 2 `T ´n 2 pKq ˘" m 2 pKq ą 1 ´ε; so we may take E :" T ´n 2 pKq. If we now define E :" Ť kPN Ş nPN E 2 ´n k

  `, then µ is supported on ℓ p if and only if ş |t| p dµ 0 ptq for all n ě 1 by Lemma 5.2. Altogether, we see that a weighted shift B w w w acting on ℓ p admits an invariant Gaussian product measure if and only if ř 8

Ω

`ř8

n"0 |t n | p ˘dµptq ă 8. Moreover, if µ is invariant under some weighted shift B w w w , then

ş Ω |t n | p dµptq " 1 |w 1 ¨¨¨wn| p ş K

  1 n , i.e. a measurable function p n : Ω n Ñ R `such that ν n " p n ν 1 n `αn , where α n is a positive measure orthogonal to ν 1 n . Let also f n :"

							?	p n . With this notation,
	we have	Hpν n , ν 1 n q "	ż	?	p n dν 1 n "	ż	f n dν 1 n .
			Ωn			Ωn

The third author is a Research Associate of the

(c) Assume that µ w w w pc 0 q ą 0. Then, there exists a compact set K Ď c 0 such that µ w w w pKq ą 0. By the well-known description of the compact subsets of c 0 , one can find a sequence of positive real numbers pε n q tending to 0 such that

As in the proof of (b), it follows that

In particular, µ 0 p|t| ą |w 1 ¨¨¨w n | ε n q Ñ 0 as n Ñ 8. If µ 0 ‰ δ 0 , this implies that no subsequence of p|w 1 ¨¨¨w n |ε n q can tend to 0; and hence |w 1 ¨¨¨w n | Ñ 8.

The second part of (c) is proved in a similar way.

Corollary 6.7. -For any probability measure µ 0 on K, one can find a sequence of positive real numbers w w w such that µ w w w pℓ 1 q " 1.

Proof. -This is clear by (b): choose a sequence of positive numbers pX n q such that ř 8 n"1 µ 0 p|t| ą X n q ă 8, and then take w w w such that 2 ´nw 1 ¨¨¨w n ě X n for all n ě 1. Corollary 6.8. -If the measure µ 0 is such that ř 8 n"1 µ 0 p|t| ą C n q ă 8 for some constant C, then µ w w w pℓ 1 q " 1 for any weight sequence w w w such that lim |w 1 ¨¨¨w n | 1{n ą C. In particular, if µ 0 p|t| ą Xq " O `logpXq ´α˘a s X Ñ 8 for some constant α ą 1, then µ w w w pℓ 1 q " 1 for any weight sequence w w w such that lim |w n | 1{n ą 1.

Proof. -Take ε n :" pC{ρq n in (b), where ρ :" lim |w n | 1{n . 6.3. When frequently hypercyclic vectors are the same. -As mentioned in Remark 5.10, the following result is due to S. Charpentier and the third author. We thank S. Charpentier for allowing us to include it here. Proposition 6.9. -Let u u u and v v v be two weight sequences such that B u u u and B v v v are frequently hypercyclic on ℓ p . If u 1 ¨¨¨un v 1 ¨¨¨vn has a non-zero limit as n Ñ 8, then B u u u and B v v v have the same frequently hypercyclic vectors.

Proof. -In what follows, we set a :" lim nÑ8 u 1 ¨¨¨un v 1 ¨¨¨vn P Kzt0u. Let x be a frequently vector for B u u u ; we want to show that x is also a frequenty hypercyclic vector for B v v v . So, we fix y " ř d k"0 y k e k P c 00 and ε ą 0, and our task is to show that the set N Bv v v `x, Bpy, εq ˘has positive lower density.

Since x is a frequently hypercyclic vector for B u u u , it is enough to find a vector z P ℓ p and α ą 0 such that any sufficiently large n P N Bu u u `x, Bpz, αq ˘belongs to N Bv v v `x, Bpy, εq ˘. We consider z :"

Note that K ă 8 since for every n, k ě 0, we have

Let us show that if n P N Bu u u `x, Bpz, ε 2K q ˘is large enough, then n P N Bv v v `x, Bpy, εq ˘. We have

where zpnq :"

Therefore, if n P N Bu u u `x, Bpz, ε 2K q ˘, we get

and the desired result follows because

6.4. Using Shepp's Theorem. -In this section, we show how results like Shepp's theorem from [START_REF] Shepp | Distinguishing a sequence of random variables from a translate of itself[END_REF] mentioned before Theorem 5.13 can be used in the context of weighted shifts. Specifically, we will make use of the following theorem. For any measure m on X " R Z `and α α α P X, let us denote by m α α α the translate of m by α α α, which is the measure on X defined by m α α α pAq :" mpA `α α αq.

Theorem 6.10. -Let r µ and r µ 1 be two measures on R, and let r m and r m 1 be the product measures on R Z `with marginals r µ n :" r µ and r µ 1 n :" r µ 1 . Let also α α α " pα n q ně0 P R Z `, and assume that the measures r m and r m 1 α α α are not orthogonal.

µ and r µ 1 have a moment of order 2, then ř 8 n"0 pα n ´αq 2 ă 8 for some α P R. Proof. -( 1) is (the first third of) [START_REF] Shepp | Distinguishing a sequence of random variables from a translate of itself[END_REF]Theorem 1].

To prove (2), we use a method devised by Dudley [START_REF] Dudley | Singularity of measures on linear spaces[END_REF] in order to generalize Shepp's theorem. Denote by e n, n ě 0 the coordinate linear functionals on Ω :" R Z `. The assumption on r µ and r µ 1 implies that e n belongs to L 2 pΩ, r mq X L 2 pΩ, r m 1 q with L 2 -norms Note that a positive answer to Question 7.2 would imply a positive answer to Question 7.3. A very special case of the latter has a positive answer by Corollary 6.4: if B u u u and B v v v admit non-trivial invariant measures and are unitarily similar, then they are nonorthogonal.

Next, in view of Theorem 5.13, Proposition 5.22 and Theorem 6.11, it is natural to ask whether the existence of equivalent non-trivial invariant product measures always implies the existence of equivalent invariant Gaussian product measures.

Question 7.4. -Is it true in general that if B u u u and B v v v admit equivalent non-trivial invariant product measures m u u u " b ně0 µ u u u,n and m v v v " bµ v v v,n , then they admit equivalent invariant Gaussian product measures? From a more general point of view, the following question seems also natural. Question 7.5. -If T 1 and T 2 are non-orthogonal operators, does it follow that they admit equivalent non-trivial invariant measures?

In the same spirit, one may ask Question 7.6. -When do two weighted shifts B u u u and B v v v share a common non-trivial invariant measure? Finally, the following seems to be unknown. Question 7.7. -Let B w w w be a weighted shift acting on c 0 pZ `q. Is it true that B w w w admits a non-trivial invariant measure if and only if |w 1 ¨¨¨w n | Ñ 8 as n Ñ 8?