A concise proof of the Multiplicative Ergodic Theorem on Banach spaces

We give a streamlined proof of the multiplicative ergodic theorem for quasi-compact operators on Banach spaces with a separable dual.


Introduction
The multiplicative ergodic theorem (MET) is a very powerful result in ergodic theory establishing the existence of generalized eigenspaces for stationary compositions of linear operators. It is of great interest in many areas of mathematics, including analysis, geometry and applications. The MET was first established by Oseledets [9] in the context of matrix cocycles. The decomposition into generalized eigenspaces is called the Oseledets splitting.
After the original version, the MET was proved by a different method by Raghunathan [10]. The result was subsequently generalized to compact operators on Hilbert spaces by Ruelle [11]. Mañé [8] proved a version for compact operators on Banach spaces under some continuity assumptions on the base dynamics and the dependence of the operator on the base point. Thieullen [12] extended this to quasi-compact operators. Recently, Lian and Lu [7] proved a version in the context of linear operators on separable Banach spaces, in which the continuity assumption was relaxed to a measurablity condition.
We give a streamlined alternative, hopefully simpler, proof of Lian and Lu's result, under a mild additional assumption (the separability of the dual space, rather than the space itself). At the same time, we remove the assumption of injectivity of the operators. An important feature of the present approach is its constructive nature. Indeed, it provides a robust way of approximating the Oseledets splitting, following what could be considered a power method type strategy. This makes the work also relevant from an applications perspective.
The approach of this work is similar in spirit to that of Raghunathan, in that we primarily work with the 'slow Oseledets spaces'. Mañé's proof works hard to build the fast space, as do the subsequent works based on Mañé's template. These proofs rely on injectivity of the operators; some of them make use of natural extensions to extend the result to non-invertible operators -this was the strategy in [12], and it was also used by Doan in [1] to extend [7] to the non-invertible context. In contrast, we establish the non-invertible version first and recover the (semi-)invertible one, including the 'fast spaces', straightforwardly using duality. Another key simplifying feature of our method is that we prove measurability at the end of the proof, rather than working to ensure that all intermediate constructions are measurable.
While Raghunathan's proof uses singular value decomposition and hence relies on the notion of orthogonality, we study instead collections of vectors with maximal volume growth. Another important difference with Raghunathan's approach is that instead of dealing with the exterior algebra, we work with the Grassmannian. We claim this is more natural since subspaces correspond to rank one elements of the exterior algebra (those that can be expressed as v 1 ∧ . . . ∧ v k ). In the Euclidean setting, rank one elements naturally appear as eigenvectors of Λ k (A * A), but this does not seem to generalize to the Banach space case. Section 2 contains volume calculations for bounded linear maps T on a Banach space X. We establish an asymptotic equivalence between kdimensional volume growth under T and T * , as well as other measures of volume growth. Section 3 deals with random dynamical systems, or linear cocycles. The non-invertible Oseledets multiplicative ergodic theorem is proved under the assumptions that X * is separable and a quasi-compactness condition holds. Separability allows one to show all the necessary measurability conditions in an elementary fashion.
The recent semi-invertible version of the MET established in [4], building on [2,3], and from which the invertible version of the MET can be recovered, is also presented here. This result is obtained via duality: complementary spaces to the Oseledets filtration arise as annihilators of the corresponding Oseledets filtration for the dual cocycle.

Volume calculations in Banach spaces
Let X be a Banach space with norm · . As usual, given a non-empty subset A of X and a point x ∈ X, we define d(x, A) = inf y∈A d(x, y). We denote by B X and S X the unit ball and unit sphere in X, respectively. The linear span of a collection C of vectors in X will be denoted by lin(C) with the convention that lin(∅) = {0}. The dual of X will be denoted by X * .
We define the k-dimensional volume of a collection of vectors (v 1 , . . . , v k ) by It is easy to see that vol In the case where the normed space is Euclidean this notion corresponds with the standard notion of k-dimensional volume. Notice that vol k (v 1 , . . . , v k ) is not generally invariant under permutation of the vectors.
Given a bounded linear map T from X to X, we define Lemma 1 (Submultiplicativity). Let T : X → X and S : X → X be linear maps.
Proof. Let v 1 , . . . , v k ∈ X. Then one checks from the definition that for any collection of coefficients (α ij ) j<i , the following holds Since the linear spans in the definition of volume are finite-dimensional spaces, the minima are attained so that . . . T (w k ) and set u j = T (w j )/ T (w j ) . Using (1), we have Taking a supremum over v 1 , . . . , v k in the unit ball of X, one obtains the bound D k (S • T ) ≤ D k (S)D k (T ) as required.
We now proceed to compare volume estimates for a linear operator T and its dual. We introduce a third quantity to which we compare both D k (T ) and D k (T * ). Given linear functionals θ 1 , . . . , θ k ∈ X * and points x 1 , . . . , x k ∈ X, we let U ((θ i ), (x j )) be the matrix with entries U ij = θ i (T (x j )) and define E k (T ) = sup det U (θ i ), (x j ) : θ i = 1 and x j = 1 for all i, j Lemma 3 (Relation between volumes for T and T * ). For all k > 0, there exist positive constants c k and C k with the following property: For every bounded linear map T from a Banach space X to itself, Proof. The statement will follow from the following inequalities: The second inequality of (2) is proved as follows: Let x 1 , . . . , x k and θ 1 , . . . , θ k all be of norm 1 in X and X * respectively. Let . Note that U = U ((θ i ), (y j )) may be obtained from U = U ((θ i ), (x j )) by column operations that leave the determinant unchanged. Notice also that |U ij | = |θ i (T y j )| ≤ α j . From the definition of a determinant, we see that det U = det U ≤ k!α 1 . . . α k . This inequality holds for all choices of θ i in the unit sphere of X * . Now, maximizing over choices of x j in the unit sphere of X, we obtain the desired result.
The second inequality of (3) may be obtained analogously. We let β i = d(T * θ i , lin(T * θ 1 , . . . , T * θ i−1 )) and choose linear combinations φ i of the θ i for which the minimum is obtained. The matrix U = U ((φ i ), (x j )) is obtained by row operations from U and the |U i,j | = To show the first inequality of (2), fix x 1 , . . . , x k of norm 1. As before, let α j = d(T x j , lin(T x 1 , . . . , T x j−1 )). By the Hahn-Banach theorem, there exist linear functionals ( Maximizing over the choice of (x j ), we obtain E k (T ) ≥ D k (T ) as required. Finally, for the first inequality of (3), we argue as follows. Let > 0 be arbitrary and let θ 1 , . . . , θ k belong to the unit sphere of X * . We may assume that T * θ 1 , . . . , T * θ k are linearly independent -otherwise the inequality is trivial. Let φ i = T * θ i − k<i a ik T * θ k be such that φ i = d(T * θ i , lin({T * θ k : k < i}). We shall pick x 1 , . . . , x k inductively in such a way that det((φ i (x j )) i,j≤l ) is at least l i=1 ( φ i − ) for each 1 ≤ l ≤ k. Suppose x 1 , . . . , x l−1 have been chosen. Then since det((ψ i (x j )) i,j<l ) is non-zero, the rows span R l−1 . Hence there exist (b i ) i<l such that ψ l := φ l + i<l b i φ i satisfies ψ l (x j ) = 0 for all j < l. By assumption, ψ l ≥ φ l . Pick x l ∈ S X such that ψ l (x l ) > ψ l − . Then the matrix with a row for ψ l and a column for x l adjoined has determinant at least l i=1 ( φ i − ). The corresponding matrix with φ l replacing ψ l has the same determinant, completing the induction. Maximizing over the choice of (θ j ) j≤k , letting shrink to 0, and observing that det((φ i (x j )) i,j≤k ) = det((T * θ i (x j )) i,j≤k ) completes the proof.
A fourth quantity that will play a crucial role in what follows is Lemma 4 (Relation between determinants and F k ). Let T be a bounded linear map from a Banach space X to itself. Then . We may assume E k (T ) > 0 as otherwise the inequality is trivial. Let θ 1 , . . . , θ k be elements of the unit sphere of X * and x 1 , . . . , x k be elements of the unit sphere of X. Let U be the matrix with entries θ i (T x j ). Assume that det U = 0. Since x 1 , . . . , x k span a k-dimensional space, there exists a v = a 1 x 1 +. . .+a k x k of norm 1 such that T v ≤ F k (T ). By the triangle inequality, one of the a's, say a j 0 , must be at least 1 k . Letx j = x j for j = j 0 andx j 0 = v and setŨ to be the matrix with entries θ i (Tx j ). By properties of determinants, we see Finally, the j 0 th column ofŪ has a single non-zero entry that is at most For the other inequality, we may suppose that T has kernel of codimension at least k, otherwise F k (T ) = 0 and there is nothing to prove. Let θ 1 , . . . , θ k−1 and x 1 , . . . , x k−1 be arbitrary. Let ∆ be the determinant of the matrix with entries θ i (T x j ). Let V be a k-dimensional subspace such that V ∩ker T = {0}. Let W = lin(T x 1 , . . . , T x k−1 ). Let z be a point in the unit sphere of T (V ) such that d(z, W ) = 1 (the existence follows from a theorem of Gohberg and Krein, see Lemma 211 of [5]). Let v ∈ V ∩S X be such that T (v) is a multiple of z. Let θ k be a linear functional of norm 1 such that θ k | W = 0 and θ k (z) = 1 and let

Corollary 5. [of Lemmas 3 and 4]
For each k > 0, the quantities D k (T ), D k (T * ), E k (T ) and i≤k F i (T ) agree up to multiplicative factors that are independent of the bounded linear map T and the Banach space X.
By definition, for each natural number k, one can find sequences . We now show that we can find infinite sequences (θ i ) and (x j ) so that, for each k, Lemma 6 (Existence of consistent sequences). For any linear map T , there exist (θ i ) i≥1 in S X * and (x j ) j≥1 in S X such that for all k, Proof. The proof is by induction: suppose (θ i ) i<k and (x j ) j<k have been chosen and satisfy the desired inequalities at stage k − 1. Then pick Using the result of Gohberg and Krein, let Averaging the inequalities, we get If |a k | > 1 2 , the first inequality yields T x ≥ 1 4 F k (T ). If |a k | ≤ 1 2 , then j<k a j x j ≥ 1 2 and the second inequality combined with the inductive hypothesis gives T x ≥ 1 Lemma 7 (Lower bound on volume growth in a subspace of finite codimension). For any natural numbers k > m, there Proof. Let > 0. Let P be a projection from X to V of norm at most √ m + (such a projection exists by Corollary III.B.11 in the book of Wojtaszczyk [13]). Then, 1 − P ≤ √ m + + 1. Let x 1 , . . . , x k be a sequence of vectors in X of norm 1. The proof of Lemma 3 shows that there exist , by multilinearity of the determinant. At most m of the j can be 0, as otherwise more than m vectors lie in a common m-dimensional space, so that at least k − m of them lie in V . Hence, there exist vectors y 1 , . . . , y m in S X and y m+1 , . . . , Using the proof of Lemma 3 again, we deduce that This completes the proof.

Random dynamical systems
A closed subspace Y of X is called complemented if there exists a closed subspace Z such that X is the direct sum of Y and Z, written X = Y ⊕ Z. That is, for every x ∈ X, there exist y ∈ Y and z ∈ Z such that x = y + z, and this decomposition is unique. The Grassmannian G(X) is the set of closed complemented subspaces of X. We denote by G k (X) the collection of closed k-codimensional subspaces of X (these are automatically complemented). We equip In this section, we consider random dynamical systems. These consist of a tuple R = (Ω, F, P, σ, X, L), where (Ω, F, P) is a probability space; σ is a measure preserving transformation of Ω; X is a separable Banach space; the generator L : Ω → B(X, X) is strongly measurable (that is for fixed x ∈ X, ω → L ω x is (F, B X )-measurable); and log L ω is integrable.
This gives rise to a cocycle of bounded linear operators L (n) ω on X, defined by L (n) ω (x) = L σ n−1 ω • · · · • L ω x. We will consider F and P to be fixed, and thus refer to a random dynamical system as R = (Ω, σ, X, L). We say R is ergodic whenever σ is ergodic.
Proof. Fix a dense sequence x 1 , x 2 , . . . in the unit ball of X. By strong measurability, for each fixed x, ω → L ω x is measurable. Then for each j 1 , . . . , j i , we have that f j i |j 1 ,...,j i−1 (ω) := inf q 1 ,..., Remark 9 (Random dynamical systems with random domains). Let R = (Ω, σ, X, L) be a random dynamical system acting on a separable Banach space. Suppose there exists a measurable family of closed, finite codimensional subspaces V = {V (ω)} ω∈Ω which is equivariant under L (that is, L ω (V (ω)) ⊂ V (σω)). Then, one can construct a new random dynamical system R| V by restricting the generator L to V . The results of this section apply to this more general type of random dynamical systems as well.
Indeed, the fact that the domains of L ω are the same is only used in Lemma 8, when a countable dense sequence in S X is chosen. In the case of the restriction R| V , the choice of a sequence of measurable functions ω → u j (ω) so that (u j (ω)) j>0 is dense in V (ω) ∩ S X can be made as follows.
First, for fixed v ∈ X, ω → d(v, V (ω)) is a measurable function, as it is the composition of continuous and measurable functions. Fix a dense sequence v 1 , v 2 , . . . ∈ S X . Now for each j, set u 0 j (ω) = v j , and let u k+1 For each j, this is a measurable convergent sequence and hence the limit point u ∞ j (ω) is measurable, and belongs to V (ω). The sequence (u ∞ j (ω)) is dense in V (ω) ∩ S X because there are v j arbitrarily close to all points of V (ω) ∩ S X .
One may also consider the quotient spaces Q(ω) := X/V (ω), equipped with the quotient norm, x Q(ω) = d(x, V (ω)), and the inducedL ω : Q(ω) → Q(σω). The measurability results of Lemma 8 also apply in this case. Indeed, if v 1 , v 2 , . . . is a countable dense sequence in S X , and {u i (ω)} i∈N is a sequence of measurable functions dense in V (ω) ∩ S X as constructed above, then measurability of ω → d(L ω (v i ), V (ω)) is straightforward to establish, and hence ω → L ω is measurable.
When R is ergodic, Lemma 8 (or Remark 9) combined with Kingman's sub-additive ergodic theorem ensures the existence of the maximal Lyapunov exponent of R, defined by λ(R) := lim n→∞ 1 n log L (n) ω , for P-a.e. ω ∈ Ω. Similarly, using the fact that the index of compactness is also sub-multiplicative and bounded above by the norm, we have existence of the index of compactness of R, defined by with the property that κ(R) ≤ λ(R). Furthermore, we have the following.
Lemma 10. Given an ergodic random dynamical system R, there exist constants ∆ k = ∆ k (R) such that for almost every ω ∈ Ω, Proof. The first claim follows from Kingman's sub-additive ergodic theorem, via Lemma 8 (or Remark 9) and Lemma 1. The remaining two claims are consequences of Corollary 5.
The µ k 's of the previous lemma are called the Lyapunov exponents of R. When µ k > κ(R), µ k is called an exceptional Lyapunov exponent.
Theorem 11 (Lyapunov exponents and index of compactness). Let R be a random dynamical system with ergodic base acting on a separable Banach space X. Then • For any ρ > κ(R), there are only finitely many exponents that exceed ρ; • If σ is invertible, the dual random dynamical system R * and R have the same Lyapunov exponents.
Proof. That the µ i are decreasing follows from Lemma 10 and the observation that F k (T ) ≤ F k−1 (T ). That the system and its dual have the same exponents follows from Lemma 3 together with the simple result (in [2,Lemma 8.2]) that if (f n ) is sub-additive and satisfies f n (ω)/n → A almost everywhere, then one has f n (σ −n ω)/n → A also. It remains to show that for ρ > κ, the system has at most finitely many exponents that exceed ρ. Let κ < α < β < ρ. Since log L ω is integrable, there exists a 0 < δ < (β − α)/2|α| such that if P(E) < δ, then E log L ω dP(ω) < (β − α)/2. By the sub-additive ergodic theorem, there exists L > 0 such that P(ic(L ω B X may be covered by finitely many balls of size e αL . Let r be chosen large enough so that P(G) > 1 − δ, where G (the good set) is defined by ω B X may be covered with e rL balls of size e αL }. We split the orbit of ω into blocks: if σ i ω ∈ G, then the block length is L; otherwise, if σ i ω is bad, we take a block of length 1. Consider the following iterative process: start with a ball of radius 1. Then look at the current iterate of ω, σ i ω. If it is good, replace the current crop of balls by e rL times as many balls of size e αL times their current size. If the iterate of ω is bad, replace the balls by balls inflated by a factor of L σ i ω . In this way, one obtains a collection of balls that covers the push-forward of the unit ball.
We claim that for almost all ω, for sufficiently large N , L (N ) ω (B X ) is covered by at most e rN balls of size e βN . For large N , the combined inflation through the bad steps is less than e (β−α)N/2 . If α ≥ 0, then for large N , through the good steps, the balls are inflated by a factor at most e αN . If α < 0, then combining the good blocks, the balls are scaled by a factor of e α(1−δ)N < e (α+β)N/2 or smaller. In both cases, we see that overall, balls are scaled by at most e βN . The splitting only takes place in the good blocks, and yields at most e rN balls. Now suppose that µ k > ρ. For almost all ω, we have that for all large N , we have D k (L It is not hard to see that all of these points belong to the image of the unit ball of X under L (N ) ω . Further, from the definition of D i , one can check that these points are mutually separated by at least 2e βN , so that one requires at least e kN (ρ−β) /(2k) k balls to cover L (N ) ω (B X ). Hence we obtain e kN (ρ−β) (2k) k ≤ e rN . Since this holds for all large N , we deduce k ≤ r/(ρ−β) as required.
Lemma 12 (Measurability II). Suppose that X is a Banach space with separable dual. Suppose further that R is an ergodic random dynamical system acting on X.
Assume there exist λ > λ ∈ R and d ∈ N such that for P-almost every ω, there is a d-codimensional subspace V (ω) of X such that for all v ∈ V (ω), lim sup n→∞ Proof. Given V ∈ G d (X) and vectors w 1 , . . . , w d such that V ⊕lin(w 1 , . . . , w d ) = X, define a neighbourhood of V by Since G d (X) is separable, one can fix a countable collection of V 's dense in G d (X) and fix for each such V a complementary sequence of w's. By restricting to rational γ's, we obtain a countable collection of neighbourhoods which generates the Borel σ-algebra on G d (X). (To see this, notice that each open set is the union of the neighbourhoods in the collection that it contains). Hence to show the desired measurability, it suffices to show that for each N = N V ;w 1 ,...,w d ;γ , {ω : V (ω) ∈ N } is measurable.
Fix a dense set v 1 , v 2 , . . . in the unit sphere of V . We claim that (provided ω lies in the set of full measure on which the growth rates are ≤ λ and ≥ λ and where the dimension is correct), V (ω) lies in N if and only if the following condition holds: For each rational > 0, there is n 0 > 0 such that for each n ≥ n 0 , for each j ∈ N, there exist rational a j 1 , . . . , a j d in In the second case, there exists η > 0 such that for every a 1 , . . . , Hence, in both cases the condition above fails.
Since this condition is obtained by taking countable unions and intersections of measurable sets, the measurability of {ω : V (ω) ∈ N } is demonstrated.
Theorem 14 (The Oseledets filtration). Let R be an ergodic random dynamical system acting on a Banach space X with separable dual. Suppose that κ(R) < λ(R). Then there exist 1 ≤ r ≤ ∞ 1 and: • a sequence of exceptional Lyapunov exponents λ(R) = λ 1 > λ 2 > . . . > λ r > κ(R); • a sequence m 1 , m 2 , . . . , m r of positive integers; and • an equivariant measurable filtration . . be as in Theorem 11. Let λ 1 > λ 2 > . . . be the decreasing enumeration of the distinct µ-values that exceed κ(R) (if this an infinite sequence, then Theorem 11 establishes that λ i → κ(R)). The fact that λ(R) = λ 1 is straightforward from the definitions. Let m be the number of times that λ occurs in the sequence (µ i ) and let We now turn to the construction of V (ω). For a fixed ω, let the sequences (θ −1 (ω). All of these depend on the choice of θ's and x's. No claim of uniqueness or measurability is made. The space V (n) (ω) is an approximate slow space. The proof will go by the following steps: (a) For almost all ω, for arbitrary > 0 and for sufficiently large n, L is a Cauchy sequence for almost all ω -we define the limit to be V (ω); (c) The V (ω) are equivariant: (e) For all a > 0 and > 0, there exists n 0 so that for all n ≥ n 0 and all x ∈ S X such that d(x, V +1 (ω)) ≥ a, one has L (g) ω → V (ω) is measurable; (h) The restriction, R , of R to V (ω) has the same exponents as R with the initial M −1 exponents removed.
Proof of (a). Note that by construction For an arbitrary x ∈ V (n) ∩ S X , let φ ∈ S X * be such that φ(L j )) 1≤i,j≤M −1 , we see that the x column has all 0 entries except for the 1 + M −1 -st (by definition of V (n) (ω)), and so we arrive at the bound (uniform over x ∈ S X ∩ V (n) ), . The conclusion follows from Lemma 10. Proof of (b). Let us assume that n 0 is chosen large enough that for all n ≥ n 0 , the following conditions are satisfied: L ω , L σ n ω are less than e n ; L x ≤ e (λ + )n L σ n ω ≤ e (λ +2 )n .
Hence, each point in the unit sphere of V (n) (ω) is exponentially close to V (n+1) (ω). Since the two spaces have the same codimension, one obtains a similar inequality in the opposite direction (possibly with the constant increased by a factor of at most 2) -see Lemma 233 of [5]. This establishes that V (n) (ω) is a Cauchy sequence. Proof of (c). We argue essentially as in (b). For large n, we take v ∈ V (n+1) (ω) ∩ S X . We write L ω (v) as u + w with u ∈ V (n) (σ(ω)) and w ∈ Y σ(ω) w e λ −1 n w (here means 'is smaller up to sub-exponential factors'). Combining the inequalities as before, one obtains a bound w e −(λ −1 −λ )n . Taking a limit, we obtain L ω V (ω) ⊂ V (σ(ω)) as required. Proof of (d). Let x ∈ V +1 (ω), with x = 1. For large n, if x is written as u n + v n with u n ∈ V (n) +1 (ω) and v n ∈ Y (n) (ω), then v n ≥ 1 2 d(x, V +1 (ω)) and u n ≤ 1 + v n . By (6), L (n) ω u n ≤ e (λ +1 + )n (1 + v n ) for large n, while Lemma 6 gives L (n) ω v n ≥ e (λ − )n v n for large n. The conclusion follows. The proof of (e) is the same, using the uniformity in Lemma 6.
For the inductive part, notice that the case = 1 is trivial. Suppose that the claims have been established for all k < .
The next corollary provides a splitting when the base σ is invertible.
Corollary 15 (The Oseledets splitting). Let R be a random dynamical system as in Theorem 14. Suppose that the base σ is invertible. Then there exist 1 ≤ r ≤ ∞ and exceptional Lyapunov exponents and multiplicities as in Theorem 14. Furthermore, there is an equivariant measurable direct sum decomposition X = Z 1 (ω) ⊕ · · · ⊕ Z r (ω) ⊕ V ∞ (ω), such that for P-a.e. ω, dim Z i (ω) = m i and v ∈ Z i (ω) \ {0} implies lim n→∞ The deduction of this in general starting from a non-invertible filtration theorem, including uniqueness of the complement, appears in [4]. We give a simple proof here, based on duality, in the special case where X is reflexive. We make use of the following facts valid for reflexive Banach spaces. If X is reflexive and Θ is a closed subspace of X * of codimension k, then its annihilator, Θ • is k-dimensional. Further if θ is a bounded functional such that θ| Θ • = 0, then θ ∈ Θ.