As it is always the case with Sébastien’s expositions, he managed to communicate very clearly the ideas of a mathematically profound subject (and, by the way, this topic is not directly related to his excellent Bourbaki seminar talk from March 21st).
In the sequel, I’ll transcript my lecture notes for Sébastien’s talk. Of course, all errors and mistakes are my entire responsibility.
Let us warmup by giving a proof of the following theorem:
for all .Then, the sequence
converges as .
Remark 1 The origin can be replaced by any point because
As the reader might suspect, the fact that such an “innocent-looking” result was proved only in 1981 (in this paper here) indicates that its proof is not easy to find if we don’t use the “correct” setup.
Proof: The argument has two steps:
- the first step is to show that the distance of to the origin converges;
- the second step is to control the direction of .
The convergence of the distance
follows from the subadditivity of .
More precisely, since is a weak contraction, we have that
From this, it is not hard to see that converges to .
Indeed, given , we fix such that
Next, given , we write with . From the subadditivity of , we have that
for all sufficiently large (i.e., ). In other words, we have that
for any , that is, , as desired.
Next, let us control the direction of .
Observe that the case when is easy: by definition, the sequence converges to the origin , so that the proof of the theorem is complete in this situation.
Thus, it remains only to consider the case when , i.e., the sequence goes to with positive linear speed.
Fix a sequence such that as . Note that, by definition, for each , one has
as . This means that we can consider a sequence of “records” of the excursion of towards , i.e.,
for all .
Denote by a linear form on of norm so that . Because has norm and is a weak contraction, we see that
for all .
Since is a sequence of records, we conclude that
for all .
Now, we note that, up to taking a subsequence, one can assume that the sequence of linear forms of norm converges (in the weak- topology) to a linear form .
By construction, for all . Geometrically, this means that the sequence stays to the right of a sequence of parallel hyperplanes moving with linear speed to infinity.
Indeed, this is easily visualized in two dimensions: after rotating the kernel of in , we can suppose that . In this case, the inequality means that belongs to the half-plane .
From this geometric input, it is not hard to complete the proof of the theorem.
In fact, we have that, on one hand, for all , and, on the other hand, for each , we have that for all sufficiently large (because ).
From the strict convexity of the Euclidean ball , we obtain that belongs to the small lenticular regions whose geometries forces the direction of to be -close to the unit vector perpendicular to the kernel of . Since is arbitrary, we conclude that the directions of converge (to in our current situation).
Of course, this convergence together with the fact that as finishes the proof of the theorem.
Remark 2 Note that, except for the last part of the proof (where the geometry [strict convexity] of balls entered in the discussion), Karlsson’s argument can be generalized to abstract Banach spaces in the place of : for instance, the existence of the linear functionals follows from Hahn-Banach theorem, and the weak- convergence of a subsequence of is a consequence of Banach-Alaoglu theorem.
Remark 3 The statement of Theorem 1 is very sensitive on the choice of the norm. For example, this theorem is false for equipped with the supremum norm . In fact:
- Karlsson’s argument fails because the information
does not impose strong constraints in the direction of since the “ball”
is now a square, and the “lenticular region”
is a now rectangle of width and height .
- A weak -contraction such that does not converge can be constructed as follows. Let and . Consider a sequence of times such that . Using this sequence of times, we define so that follows a straight line segment in the direction for time , then a straight line segment in the direction for time , then a straight line segment in the direction for time , etc. (e.g., for , for , etc.); Since consists of straight line segments of slopes or , one can see that is a geodesic ray for the supremum norm (such that ); From , we define a weak -contraction by letting
where for . In this context, and, for an appropriate choice of times , one can check that does not converge because its direction keeps oscillating between and .
After this quick review of Karlsson’s proof of Kohlberg-Neyman theorem for weak contractions in Euclidean spaces, let us pass now to the study of weak contractions in metric spaces.
For this sake, we need the following tool (playing the role of “linear functionals”):
Definition 2 Let be a metric space and . We say that a function is a horofunction (or Busemann function) if there exists a sequence such that
Remark 4 By definition, . Furthermore, is a -Lipschitz function (because it is the limit of the -Lipschitz functions ).
Example 1 Consider the Euclidean space and let , . The level sets of the function are the Euclidean balls centered at : more precisely, takes the value on the circle of radius centered at . From this, one can see that converges to the linear functional . In this way, we see that linear functionals in Euclidean spaces are particular cases of horofunctions.
Example 2 Consider the plane equipped with the supremum norm . Let and . A direct inspection reveals that the level sets of are translations along the diagonal of the first quadrant of , i.e., where . In particular, converges to the horofunction .
Remark 5 It is possible to show that any horofunction on a normed vector space can be estimated from below by a linear functional. In particular, the level sets of horofunctions provide more information on the location of points than the level sets of linear functionals.
Theorem 3 (Karlsson (2002)) Let be a separable metric space, a weak contraction, and .Then,
and there exists a horofunction such that for all .
3. Iterated systems of contractions
Note that Karlsson’s theorem gives a geometrical description of the orbits of a single weak contraction, but one might wonder about the behavior of “random” compositions (cocycle) of several weak contractions.
In this direction, Gouëzel and Karlsson proved the following theorem:
Theorem 4 (Gouëzel-Karlsson) Let be a ergodic transformation, be a metric space and an (integrable) cocycle (where is the set of weak contractions of ).Denote by . Then, there exists such that, for -almost every , all , and some horofunction , we have that
Remark 6 There is no hope to get this kind of convergence result if we compose the weak contractions in the other way around in the definition of . In fact, this is not hard to see in the context of the composition of random (large) hyperbolic matrices in acting on the hyperbolic disk : since a large hyperbolic matrix tends to concentrate a big chunk of near a boundary point associated to the unstable direction of , we have that the compositions
of random large hyperbolic matrices will most likely take the origin of near the random boundary point , and, thus will not converge; on the other hand, the compositions
of random large hyperbolic matrices will most likely take the random point near the deterministic boundary point and this is why we expect convergence of .
Before saying a few words about the proof of Gouëzel-Karlsson theorem, let us put it into historical perspective by citing a couple of previous related results:
- Karlsson and Ledrappier showed in 2006 the validity of Gouëzel-Karlsson theorem in the special case of a cocycle of isometries (by exploiting in particular the fact that the composition of a horofunction and a isometry is still a horofunction);
- Karlsson and Margulis proved in 1999 the following slightly weaker version of Gouëzel-Karlsson theorem: under the same assumptions of Theorem 4, for each , there exists an horofunction such that
converges to a value in the interval as .
Let us now conclude this post with a sketch of proof of Theorem 4.
The first observation is that the sequence
is a subadditive cocycle in the sense that
Here, we used that takes its values in the set of weak contractions and was defined as (in this order).
In this setting, we can apply Kingman’s subadditive ergodic theorem to deduce the convergence of :
Theorem 5 (Kingman (1967)) Let be a (integrable) subadditive cocycle. Then,
almost surely and in . (Here, is independent of .)
Therefore, the proof of Gouëzel-Karlsson theorem will be complete once we construct a horofunction with the properties in the statement of Theorem 4.
Unfortunately, the information provided by Kingman’s theorem is not sufficient to build up the desired . For this reason, Gouëzel and Karlsson were “forced” to show the following improvement of Kingman’s subadditive ergodic theorem:
Then, for -almost every , there are as and as such that, for all and all ,
An important point of this theorem is that is independent from ! In particular, this fact can be exploited to build up a horofunction (obtained as the limit of a subsequence of suitable functions defined in the spirit of Karlsson’s proof of Kohlberg-Neyman theorem) such that
for all . Of course, since as , the previous estimate permits to conclude the proof of Gouëzel-Karlsson theorem.
Finally, the proof of Gouëzel-Karlsson improvement of Kingman’s theorem is similar in spirit to the usual proofs of Kingman’s result (based on the combinatorics of pieces of orbits of where the values of the cocycle fluctuates near a given value).
However, the technical details are somewhat intrincate and, for this reason, Sébastien decided that it was a better idea to explain why the naive approach based on the study of times where the cocycle attains its “records” (as in Karlsson’s proof of Kohlberg-Neyman theorem) does not work.
So, we close this post by following him in the explanation of two naive strategies that fail in proving Theorem 6.
We begin by taking a sequence as . By definition, for -almost every , we have that
Thus, it makes sense to consider a sequence of successive “records”, i.e.,
for all . Of course, the previous inequality can be rewritten as
which looks like the conclusion of Theorem 6 except that the argument of is instead of !
At this point, we have the impression to be “close” to show Theorem 6, but this is not the case. Indeed, suppose that we try to overcome the difficulty of the previous paragraph by making the “change of variables” (in hope of finding in the place of in the argument of ). It is not hard to check that is a subadditive cocycle with respect to . In particular, we can repeat our discussion with replaced by in order to find a sequence of “records”/good times for almost every .
By doing so, we have that almost every belongs to infinitely many of the subsets
Logically, we are not directly interested in the cocycle , but rather on . In other words, even though we got some information about the sets , we are really interested in the subsets
Here, it is tempting to conjecture that the fact that almost every belongs to infinitely many (from which case we would be able to deduce Theorem 6) is an immediate consequence of the fact that almost every belongs to infinitely many . Unfortunately, this conjecture is simply wrong if we do not have extra information about the structure of the subsets : for example, let be a fixed subset of positive but not full -measure; the ergodicity of implies that -almost every belongs to infinitely many , but it is obvious that the complement of is a subset of positive -measure subset consisting of which do not belong to a single !
In summary, a naive combination of the “records” strategy and a change of variables do not lead us anywhere close to showing Theorem 6, and, in fact, a more sophisticated combinatorial strategy is needed here.