Some remarks on the dynamics of the almost Mathieu equation at critical coupling

We show that the quasi-periodic Schrödinger cocycle with a continuous potential is of parabolic type, with a unique invariant section, at all gap edges where the Lyapunov exponent vanishes. This applies, in particular, to the almost Mathieu equation with critical coupling. It also provides examples of real-analytic cocycles having a unique invariant section which is not smooth.


Introduction
In this note we consider the Schrödinger cocycle on T × R 2 given by F E : (x, y) → (x + ω, A E (x)y) * Dedicated to the memory of Russel A Johnson.
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where ω ∈ R\Q, (2, R) and v : T → R is a continuous function. In projective coordinates 1 r we can write F E as Since P 1 (R 2 ) ∼ = T we can view G E as a map of T 2 . We let I, if n = 0; A(x + nω) −1 · · · A(x − ω) −1 if n ≤ −1; and de ne the (maximal) Lyapunov exponent by Note that A n E (x) is the fundamental solution to the Schrödinger equation We say that the cocycle F E (for some xed parameter E) is uniformly hyperbolic if there exists a continuous splitting W + E (x) ⊕ W − E (x) = R 2 and constants C, γ > 0 such that the following holds for all x ∈ T and all n ≥ 1: |A n E (x)y| Ce −γn |y| for all y ∈ W − E (x); |A −n E (x)y| Ce −γn |y| for all y ∈ W + E (x).
In particular we have L(E) > 0 when F E is uniformly hyperbolic. We let σ = σ(v, ω) be the (closed) set of E for which F E fails to be uniformly hyperbolic. It is well-known [1] that this set coincides with the spectrum of the associated Schrödinger operator (H x u) n = −(u n+1 + u n−1 ) + v(x + nω)u n acting on ℓ 2 (Z) (since v is continuous and ω irrational, the spectrum of H x , as a set, is independent of x). This operator is bounded, and ∅ = σ ⊂ [min v − 2, max v + 2]. We shall denote (1.2) Thus, by de nition, F E is uniformly hyperbolic for all E < E 1 . Note that E 1 depends on v and ω. E 1 is often called the ground state energy. If E / ∈ σ, it follows from [2] that the subspaces W ± E are as smooth, as functions of x, as v; and they vary smoothly with E (recall that R\σ is open). Moreover, the splitting must be invariant under F E , i.e., In projective coordinates this implies that there are two continuous functions ϕ ± E : T → P 1 (R 2 ) such that G E (x, ϕ ± E (x)) = (x + ω, ϕ ± E (x + ω)) for all x ∈ T. It is also clear, due to uniform hyperbolicity, that the graphs of these two functions are the only G E -invariant curves. Furthermore, the Lebesgue measure on T, lifted to the graphs of ϕ ± E are the only ergodic and invariant Borel probability measures (see [3, proposition 6.2] for details).
If L(E) = 0 for some E (and thus E must be in σ) it follows from the classi cation in [4] that the cocycle F E is measurably conjugated to a cocycle B E which is either elliptic, weakly hyperbolic or parabolic (see [4] for the details). The latter case, which is the one relevant for the present article, means that there is a measurable function C : T → SL(2, R) and By far the most studied Schrödinger operator (and cocycle) is the so-called almost Mathieu operator, which is the one obtained by letting v(x) = λ cos(2πx), where λ is a constant. In this case we have a very good description of much of the spectral and dynamical properties (see, e.g., [5], and references therein). A very useful tool in this case is the so-called Aubry duality (see, for example, [6]); we will also make use of this duality in the present paper. We shall mainly be interested in the 'critical' case, i.e., the case when λ = 2. In this case the Lebesgue measure of the spectrum σ is zero; it can even be of zero Hausdorff dimension [7] (see also [8] for uniform upper bounds of the dimension). Plotting the spectrum σ as a function of the frequency ω gives rise to the famous Hofstadter's butter y. Not much seems to be known about the behaviour of the solutions of the almost Mathieu equation However, there can be no solutions in l 1 (Z) [9]; and typically no l 2 (Z) solutions [10].

Notations
In the formulations of our results below, we use the following notations: let π 1 and π 2 denote the projections π 1 (x, r) = x and π 2 (x, r) = r. Moreover, we denote by ω E (x, r) and α E (x, r) the ω-limit set and the α-limit set, respectively, of the point (x, r) under iterations of G E .
In some of the results we will need to assume that the frequency ω satis es a kind of (strong) Diophantine condition. Given an irrational number ω, let p n /q n denote the nth order continued fraction expansion of ω. We let P ⊂ T denote the set of ω ∈ T for which lim n→∞ q 1/n n exists and is nite. This set has full Lebesgue measure. See [11] for details.
Before stating our results, we mention that in all cases, except the ones which are speci cally for the almost Mathieu equation, we could have assumed that v : T d → R and ω = (ω 1 , . . . , ω d ) ∈ R d (d ≥ 1) is such that 1, ω 1 , . . . , ω d are rationally independent.

Dynamics at the lowest energy E 1
Since the proofs of the results are more elementary and transparent at the lowest (or highest) energy in σ, we begin by considering this case. The rst result of this paper is: Theorem 1. Assume that v : T → R is continuous and ω ∈ R\Q. Assume also that L(E 1 ) = 0. Then there exists an upper semi-continuous function ψ : T → (0, ∞) which is (at least) almost everywhere continuous, T log ψ(x)dx = 0, and whose graph Γ is

Remark 1.
(a) Note that π 2 (π −1 1 (x) ∩Γ) = {ψ(x)} at each point x ∈ T where ψ is continuous (that is, for almost every x ∈ T). We do not know if ψ is continuous everywhere. (b) Since all points are attracted to the closure of the graph of the almost everywhere continuous function ψ, it easily follows that the Lebesgue measure on T, lifted to the graph of ψ, is the only G E 1 -invariant and ergodic Borel probability measure (see [3] for details).
(Note that the projection onto the x-coordinate of any G E -invariant Borel measure must be the Lebesgue measure, due to the unique ergodicity of the shift x → x + ω on T.) Before stating a corollary of this result we note the following. Assume that ψ : T → (0, ∞) satis es G E (x, ψ(x)) = (x + ω, ψ(x + ω)). Let g(x) = log ψ(x) and let a n (x) = n k=0 g(x + kω) for n > 0, a 0 (x) = 0, and a n (x) = −a −n (x + nω) for n < 0. Then it is easy to verify that is a formal solution to the Schrödinger equation (1.1).

Corollary 1.
Assume that v : T → R is continuous and ω ∈ R\Q. Assume also that L(E 1 ) = 0. Let ψ : T → R be as in theorem 1, and let g, a n and u n be as above. Moreover, let X ⊂ T denote the sets of continuity points of ψ. Then:

Remark 2.
(a) A direct computation shows that satis es . Thus, A E 1 is of parabolic type. (b) Note that (c) follows directly from Atkinson's lemma (see, e.g., [12]), which states that lim inf n→±∞ |a n (x)| = 0 for a.e. x ∈ T since T g(x)dx = 0. (c) It is well-known that the equation −(w n+1 + w n−1 ) + v(x + (n − 1)ω)w n = E 1 w n (since E 1 ∈ σ) has a (non-trivial) bounded solution for some phase x 0 ∈ T (see, e.g., [[1], theorem 1.7]). We shall see (in section 3) that we must have w n = Cu n (x 0 ) for some constant C = 0. Thus, if this solution is bounded away from zero, it would follow from (c) that u n (x) is bounded for a.e. x ∈ T. (d) Note that (d) can be viewed as a version of the classical Gottschalk-Hedlund theorem (see, e.g., [13, theorem 2.9.4]). (e) In connection to this, we also recall a related result (which does not apply in our situation): if ( A n E (x 0 ) ) n≥0 is bounded for some E and some x 0 ∈ T, then the cocycle F E is continuously conjugated to a cocycle map taking values in SO(2, R) [14].
The remaining parts of corollary 1 will be proved in section 3 below. One can also consider the inverse problem, i.e., specify the invariant curveψ and ω and use them to de ne v (as we did in [15]). More precisely letψ: T → (0, ∞) be a continuous function such that T logψ(x)dx = 0, and de ne v(x) = exp(ψ(x + ω)) + exp(−ψ(x)). Then it is easy to verify that E 1 = 0 and L(E 1 ) = 0, and ψ =ψ. Furthermore, ifψ is chosen so that logψ is not a coboundary, i.e., the equation h(x + ω) − h(x) = logψ(x) has no continuous solution h, then the Gottschalk-Hedlund theorem implies that sup n≥0 |a n (x)| = ∞ for all x ∈ T (see, e.g., [16] and the references therein for more information on this topic). Thus, in this case it follows from corollary 1(d) that for all x ∈ T and all y ∈ R 2 \{0} we have inf n∈Z |A n E 1 (x)y| = 0 or sup n∈Z |A n E 1 (x)y| = ∞. The above argument shows, in particular, that any cylinder transformation (see, e.g., [16]) T(x, t) = (x + ω, t + g(x)) can be imbedded into a Schrödinger cocycle.
Next we consider the special case when v(x) = 2 cos(2πx). In this case it is well-known that L(E) = 0 for all E ∈ σ (see, e.g., [17, corollary 2]). In particular we have L(E 1 ) = 0. Thus the previous theorem applies for this v. In gure 1 we have numerically plotted an approximation of the function ψ; from these numerical investigations it looks as if ψ is continuous; but we do not know if this really is the case. However, we have (recall the de nition of the full-measure set P in subsection 1.1): and ω ∈ P. Then ψ / ∈ C 1+α (T) for any α > 1/2, where ψ is the function in theorem 1.

Remark 3.
(a) Since 2 cos(2πx) obviously is real-analytic, it follows immediately from [2], as we mentioned above, that for all E < E 1 the map G E has two real-analytic invariant curves which control all the dynamics. But, as we saw above, G E 1 is uniquely ergodic, and the measure is supported on the graph of ψ.
then it follows from [18] (see also [19]) that G E has real-analytic invariant curves for all . In this case G E 1 has two 'fractal' invariant graphs. See, [20]. (See also [21] for results for more general v.) (d) We recall the phenomenon with 'the last' invariant curve in certain Hamiltonian systems.
See, e.g., [22] and references therein. (e) If ω would satisfy a weaker Diophantine condition, the function ψ could be of higher, but still nite, regularity. The arithmetic condition on ω is needed when we solve the homological equation (4.2). However, we do not elaborate on this.
We will prove theorems 1 and 2 by combining previous results by Delyon [9], Herman [23] and Johnson [24]. In fact, the statements in theorem 1 follow immediately from the proposition below. This propositions will be proved in section 2.

Remark 4.
(a) These statements are close in spirit of [23,24]. Moreover, the rst part of the proposition is essentially a special case of [25, theorem 5.3] (which is based on [24, lemma 3.4]). However, we will provide an elementary proof of the statements in section 2 (the arguments become easier because we consider the lowest energy, E 1 , in the spectrum). (b) Note that, by the semi-continuity of ψ ± , the set M is closed.
We will prove corollary 1 and theorem 2 in sections 3 and 4, respectively.

Dynamics at other gap edges
We now consider the more general problem of describing the dynamics of F E (and its projective action G E ) at other gap edges of R\σ where the Lyapunov exponent vanishes.
By symmetry it is easy to check that the analogous picture to the one above holds for E 2 = max σ, i.e., for the highest energy in the spectrum. In particular, if v(x) = λ cos 2πx then The following theorem is a generalisation of theorems 1 and 2 to other gap edges.
Theorem 3. Assume that v : T → R is continuous and ω ∈ R\Q. Assume further that E * is a gap edge of a non-collapsed gap in R\σ, and that L(E * ) = 0. Then (a) there exists an upper semi-continuous function ψ : T → P 1 (R 2 ) which is (at least) almost everywhere continuous and whose graph Γ is and ω ∈ P, then the function ψ cannot be of class C 1+α for any α > 1/2.

Remark 5.
That ψ is semi-continuous means that, by viewing P 1 (R 2 ) as the circle T, there exists a lift ψ : This theorem is proved in section 5 below. In the proof we also apply results from Thieullen [4].

Open questions
We do not know if the function ψ in theorem 3 must be continuous. We also have the following related question: which is discontinuous almost everywhere and which attracts (in the projective action) all (or almost all) forward and backward iterations 1 ?
More generally, does there exists a smooth family of circle diffeomorphisms f x : T → T and irrational ω such that the map T : T 2 → T 2 given by T(x, y) = (x + ω, f x (y)) has an invariant graph y = ψ(x) which is discontinuous almost everywhere and which attract all (or almost all) forward and backward iterations? Remark 6. In [26] numerical investigations of the dynamics of G 0 (i.e., for E = 0), for v(x) = 2 cos(2πx), are presented. It should be noted that 0 ∈ σ, but E = 0 cannot be the endpoint of any spectral gap (see [26] for more details). The authors conjecture that F 0 is of parabolic type. If this is true the invariant section (for G 0 ) must be discontinuous (by a topological argument, due to the fact that the so-called bred rotation number is rational). We have made numerical computations on this model which seem to indicate(?) that 'for typical x' we have lim inf n→∞ A n E=0 (x) = Id (recall [4, lemma 1.3]). This would imply that points in the same bre, in projective coordinates, are not contracted to each other. Thus, if it indeed is true that the cocycle F 0 is of parabolic type, it is possible that an invariant section (in projective space) is not an attractor for G 0 (at least not in the sense asΓ is an attractor for G E * in theorem 3).

Monotonicity-proof of proposition 1.1
In this section we assume that v : T → R is a continuous function and ω ∈ R\Q. Recall the de nition of E 1 in (1.2). We shall use projective coordinates 1 r , r ∈ R ∪ {∞}. In [23, section 4.14] it is shown that for each E < E 1 , the two continuous functions ϕ ± E (the projectivization of W ± E ) satisfy ϕ ± E : T → (0, ∞). We recall that their graphs are G E -invariant, i.e., We shall denote the graphs by Γ ± E , i.e., It is clear that the two graphs cannot intersect. Moreover, they are connected to the Lyapunov Since F E is uniformly hyperbolic when E < E 1 it follows that for each Moreover, it is easy to check that the iterates are oriented as follows: The analogous result holds for backward iteration.
The following monotonicity result is essentially a special case of [24, lemma 3.4] (where the time-continuous Hill's equation is considered). For completeness we include an elementary proof in our setting.
. By the fact that ϕ + E ′′ (x) ≥ ϕ + E ′ (x) > 0 for all x, we see that the right-hand side is ≤ 0; but the left-hand side is > 0. This contradiction shows the statement.
The proof of (2) is similar. In the case when v(x) = v(−x) the statement follows immediately from (1) combined with remark 7.
By this monotonicity we have It also follows that exists for all x ∈ T, and ϕ − for all x ∈ T and all E < E 1 (in particular there is a constant c > 1 such that ψ ± (x) ∈ [1/c, c] for all x ∈ T). Moreover, since (2.1) holds for all E < E 1 , the graphs of ψ ± are G E 1 -invariant, i.e., for all x ∈ T.
(2.2) Furthermore, again by monotonicity, the function ψ + is upper semi-continuous, and ψ − is lower semi-continuous. We summarise these observations in From these facts it thus follows that the closed sets Note that the set M is G E 1 -invariant. We now show that the iterates of any point (x, r) under G E accumulate on M.
Proof. Recall the discussion on iterations of G E for E < E 1 in the beginning of this section. Fix x ∈ T. Since the set M is G E 1 -invariant we need only consider the cases −∞ < r < ψ − (x) and ψ + (x) < r ≤ ∞.
We rst assume that ψ Inductively we thus get ψ + (x + kω) < r k < ∞ for all k ≥ 1. Moreover, given any E < E 1 , let s k (E) = π 2 (G k E (x, r)). It is easy to inductively verify that r k < s k (E) for all k ≥ 1 and all E < E 1 . Indeed, we have s 1 Here we use that r k > ψ + (x + kω) > 0.
Next, note that we have ω E (x, r) = Γ + E ⊂ M E for all E < E 1 . Since ψ + (x + kω) < r k < s k (E) for all k ≥ 1 and all E < E 1 , it thus follows that ω E 1 We now assume that −∞ < r < ψ − (x). We claim that there exists k 0 > 0 such that ψ + (x + k 0 ω) < r k 0 ≤ ∞. Since ψ ± (x) > 0 it follows from (2.2) that if r ≤ 0 we have ψ + (x + ω) < De ne r k and s k (E) as above. Note . Fix such an E ′ < E 1 . If we would have r k > 0 for all k > 0 it would follow, as above, that ϕ − E ′ (x + kω) > s k (E ′ ) > r k > 0 for all k > 0; but we know that s j (E ′ ) > ϕ + E ′ for some j > 0. Therefore this is impossible. We conclude that ψ + (x + k 0 ω) < r k 0 ≤ ∞ for some k 0 .
That α E 1 (x, r) ⊂ M is proved similarly.

Corollary 2.4.
For all x ∈ T we have A n E 1 (x) → ∞ as n → ±∞.
Proof. If there were an x ∈ T, a constant C > 0 and a subsequence n k (either going to ∞ or −∞) such that A n k E 1 (x) < C for all k it would be impossible that all orbits under G E 1 accumulate on the set M (as the statement in the previous proposition yields).
The last statement follows from equation (2.2).

Proof of corollary 1
Assume that v : T → R is continuous and ω ∈ R\Q. Assume also that L(E 1 ) = 0. From corollary 2.4 we know that A n E 1 (x) → ∞ as n → ±∞ for all x ∈ T. Thus, recalling remark 7, it remains to prove statement (d) in corollary 1.
Let ψ = ψ + be as in proposition 1.1, and let a n (x), u n (x) be as in (1.3). Let X ⊂ T be the set of points where ψ is continuous.
By combining proposition 2.3 and lemma A.2 we see that for all x ∈ T we have lim n→∞ |A n (x)y| = ∞ for all y = 0 which do not correspond to the direction ψ − (x); and lim n→−∞ |A n (x)y| = ∞ for all y = 0 which do not correspond to the direction ψ + (x). From this we conclude that |A n (x)y| cannot be bounded for any y = 0 and x ∈ T such that ψ + (x) = ψ − (x).
Assume that that there is a constant c > 1 and x 0 ∈ T, y 0 ∈ R 2 \{0} such that 1/c < |A n (x 0 )y 0 | < c for all n ∈ Z. From the above observation we note that we must have ψ + (x 0 ) = ψ − (x 0 ), i.e., x 0 ∈ X (by proposition 2.5). Moreover, we must have y 0 = s 1 ψ(x 0 ) for some constant s = 0. Thus we have sup n∈Z |a n (x 0 )| < c ′ for some constant c ′ . Since the set X is invariant under the translation x → x + ω it now follows from lemma A.1 that sup n∈Z |a n (x)| ≤ 2c ′ for all x ∈ X. Since u n (x) = exp(a n (x)) this nishes the proof.
By d we shall denote the distance on the circle T; and an interval (a, b) ⊂ T means a counterclockwise oriented interval. We will slightly abuse the notation and write G E both for the map on T × P 1 (R 2 ) as well as the map on T × T.
Proof of Theorem 3. Assume that v : T → R is continuous and ω ∈ R\Q. We further assume that J := (E − , E + ) is a non-collapsed gap in R\σ (and thus the cocycle F E is uniformly hyperbolic for all E ∈ J) and that L(E ± ) = 0.
For E ∈ J we have the continuous G E -invariant sections ϕ ± E : T → P 1 (R 2 ) (the projectivizations of the subspaces W ± E ); we recall that they move continuously with E (within J). Moreover, we recall that for all E ∈ J we have: for each x ∈ T and each θ = ϕ − (x) d(π 2 (G n E (x, θ)), ϕ + E (x + nω)) → 0 as n → ∞; (5.1) and for each x ∈ T and each θ = ϕ + (x) Since P 1 (R 2 ) ∼ = T, each ϕ ± E (for E ∈ J) has a lift ϕ ± E : R → R, and we can choose the lifts so that (x, E) → ϕ ± E (x) are continuous on R × J. We focus on the dynamics at E + ; the analysis of E − is symmetric. By Johnson's monotonicity lemma [24, lemma 3.4] (see [25, theorem 5.3] for exactly our setting) it follows that ϕ + E (x) moves in the clockwise direction as E increases; and ϕ − E (x) moves in the counter clockwise direction. This means that ϕ + From this it follows that ψ ± (x) = lim EրE + ϕ ± E (x) exists for all x ∈ T. By monotonicity the lifts of ψ + are upper semi-continuous; and the lifts of ψ − are lower semi-continuous. It also follows that ψ ± : T → P 1 (R 2 ) are G E + -invariant sections. We note that for all x ∈ T and all E ∈ J.
Let M E be the closed strips Then we have M E ⊃ M E ′ for all E < E ′ in J, and We shall now show that ω E + (x, r) ⊂ M E + for all (x, θ) / ∈ M E + (clearly this holds for all Let θ k = π 2 (G E + (x 0 , θ 0 )) and s k (E) = π 2 (G E (x 0 , θ 0 )). Since (5.4) holds it follows that Moreover, by using the fact that ∂ E (π 2 (G E (x, θ))) < 0, combined with the fact that the graph of ψ + is G E + -invariant, it is easy to verify that [ψ + (x * + kω), for all k ≥ K(E). By recalling (5.2) and (5.3) we conclude that ω E + (x, θ) ⊂ M E + . Analogously, by considering backward iterations, one shows that α E + (x, θ) ⊂ M E + for all (x, θ) / ∈ M E + . Since α E + (x, θ), ω E + (x, r) ⊂ M E + for all (x, θ) ∈ T 2 , and since clearly M E + = T 2 , we must have A n E + (x) → ∞ as n → ±∞ for all x ∈ T. Since L(E + ) = 0, and since the graphs of ψ ± are G E + -invariant, it therefore follows from [4, proposition 1.6(ii)] that ψ + (x) = ψ − (x) for almost every x ∈ T. By semi-continuity we thus have that ψ + is continuous a.e.; and π −1 Next, from the fact that the graph of ψ + : T → P 1 (R 2 ) is invariant under G E + it follows that there is a function Z : 2T → R 2 , |Z(x)| = 1 for all x, and which is as smooth as ψ + , satisfying where c : T → R is positive (clearly the vector Z(x) corresponds to the direction ψ(x)). Since L(E + ) = 0 we have T log c(x)dx = 0. Moreover, Z(x) is 1-periodic if the degree of ψ is even; and Z(x) is 2-periodic and such that Z(x + 1) = −Z(x) for all x if the degree of ψ is odd.
. Thus the cocycle F E + is parabolic.
Since α E + (x, θ), ω E + (x, r) ⊂ M E + for all (x, θ) ∈ T 2 , and since lemma A.2 holds, it follows that for all x ∈ T we have lim n→∞ |A n (x)y| = ∞ for all y = 0 which do not correspond to the direction ψ − (x); and lim n→−∞ |A n (x)y| = ∞ for all y = 0 which do not correspond to the direction ψ + (x). Assume that there is a constant c > 1 and x 0 ∈ T, y 0 ∈ R 2 \{0} such that 1/c < |A n E + (x 0 )y 0 | < c for all n ∈ Z. Then we must have y 0 = sU(x 0 ) for some constant s = 0; and we must have ψ + (x 0 ) = ψ − (x 0 ), i.e., ψ + (and thus c) is continuous at x 0 . Thus we have sup n∈Z |a n (x 0 )| < ∞; and since the continuity points of c are invariant under translation it follows from lemma A.1 that sup n∈Z |a n (x)| < ∞ for a.e. x ∈ T. Hence sup n∈Z |U n (x)| < ∞ for a.e. x ∈ T.
It remains to show part (e) of theorem 3. We therefore assume that v(x) = 2 cos(2πx) and that ω ∈ P. The proof is essentially the same as that of theorem 2, and uses, as also mentioned above, the strategy in [27, remark 1.6]. Figure 2 gives an idea of what the graph of ψ + might look like in this case.
We shall argue by contradiction and thus assume that ψ + is C 1+α for some α > 1/2. The functions c (and hence log c) and Z above have the same smoothness. Let h : Since log c(x) is C 1+α (by assumption) and ω ∈ P, it follows [11] that h is α ′ -Hölder for any α ′ < α. Fix α ′ such that 1/2 < α ′ < α. If Z(x) has period 1, it follows that q 2 (x) also is of period 1. Letting n∈Z a n e 2πinx be the Fourier series of q 2 , the relation (5.5) gives us −(a n+1 + a n−1 ) + (2 cos(2πnω) + E + )a n = 0.
In both of these situations it follows from [9] that (a n ) / ∈ ℓ 1 (Z). But since q is α ′ -Hölder it follows (as in section 4) that the Fourier series of q is absolutely convergent, and thus (a n ) ∈ ℓ 1 (Z). This contradiction nishes the proof.
Lemma A.1. Assume that ω ∈ R\Q. Assume that f : T → R is such that the set X := {x ∈ T : f is continuous at x} is invariant under the translation x → x + ω (i.e., X = X + ω). If sup n≥0 n k=0 f (x 0 + kω) < M for some x 0 ∈ T and some constant M > 0, then sup n≥0 n k=0 f (x + kω) ≤ 2M for all x ∈ X.
Proof. Take x ∈ X. We argue by contradiction. Assume that N k=0 f (x + kω) > 2M for some N ≥ 0. Since the set X is invariant under the translation we know that f is continuous at the points x + jω (0 ≤ j ≤ N). Therefore we have N k=0 f (y + kω) > 2M for all y suf ciently close to x. Since ω is irrational it thus follows that there is T > 0 such that we get that the absolute value of the left-hand side is < 2M; and the absolute value of the right-hand side is > 2M. This contradiction nishes the proof.
The next lemma contains simple results from linear algebra. It gives information about the growth of vectors under assumptions on the associated projective action.
Proof. Assume, to derive a contradiction, that there exists a unit vector v / ∈ W(θ − ) and a constant C > 0 such that |A n k v| < C for all k ≥ 1. To get easier notation we assume that |A n v| < C for all n ≥ 1. Take a unit vector w / ∈ W(θ − ) such that α = ∠(v, w) > 0. Since each A n ∈ SL(2, R) we get sinα = |A n v A n w|sinα n , where α n = ∠(A n v, A n w). Since v, w / ∈ W(θ − ) it follows by assumption that sinα n → 0 as n → ∞. Since |A n v| is bounded we conclude that |A n w| → ∞ as n → ∞.
Let u n , |u n | = 1, be a vector which is contracted the most by A n . We note that |A n u n | → 0 as n → ∞. Let β n = ∠(v, u n ). Then sin β n = |A n u n A n v| sin(∠(A n v, A n u n )) → 0 as n → ∞. But this means that there is an arc [a, b], which contains the projectivization of v in its interior, but not containing θ − , such that | A n ([a, b])| → 0 as n → ∞.