Estimating Hausdorff measure for Denjoy maps

By employing the recurrence method worked out in Pawelec (2019 arXiv: 1911.12992), we provide effective lower estimates of the proper–dimensional Hausdorff measure of minimal sets of circle homeomorphisms that are not conjugate to any rotation.


Introduction
In this paper we deal with orientation preserving homeomorphisms and, more specifically, diffeomorphisms of the unit circle S 1 . A classical Poincaré's result states that for every such homeomorphism f there is a unique number α ∈ [0, 1) such that the map f : S 1 → S 1 and the rotation R α (t) := t + α (mod 1) are semi-conjugate, i.e. h • f = R α • h, for some continuous surjection h : S 1 → S 1 . The number α is called the rotation number of f . It can be defined more explicitly without bringing up semi-conjugacies but we do not need such definition in our paper. We do need semi-conjugacy.
Much later, Denjoy proved that if the derivative of f is of bounded variation, then there is a (full) conjugacy between f and R α , i.e. the map h is a homeomorphism. However, if the derivative f is only δ-Hölder continuous for some 0 < δ < 1, then the semi-conjugacy cannot be made into a full conjugacy. We then call the homeomorphism (or diffeomorphism) f a Denjoy map.
The reason for the non-existence of the full conjugacy is the presence of wandering intervals. Then the complement of the (forward and backward) orbit of that interval is the minimal set for the considered system f : S 1 → S 1 . It is a compact, perfect, nowhere dense subset of S 1 , thus a Cantor set. In this paper we study the Hausdorff dimension and the Hausdorff measure of this unique minimal set of f . We will denote it by Ω(f ) or, even simpler, just by Ω.
Up to our knowledge, the, up to date, best estimates of the Hausdorff and box-counting dimensions of the minimal set of a Danjoy map were obtained by B. Kra and J. Schmeling in [2]. They proved a general lower bound of the box-counting dimension, depending only on the smoothness of the derivative of f , namely that dim B (Ω(f )) ≥ δ. They also showed that the lower bound of the Hausdorff dimension depends additionally on the Diophantine class ν ≥ 1 of the rotation number α; see Section 2 for the definition. More precisely, they proved that dim H (Ω) ≥ δ/ν. Moreover, for the case of so-called classical Denjoy maps (see Example 2), they showed that both the Hausdorff and box-counting dimensions are in fact equal to their lower bounds, resp. δ/ν and δ, and so they coincide (only) for ν = 1.
In this paper we push these results further. We apply a rather non-standard method of estimating the Hausdorff measure from below developed by the first named author in [3]. There it is shown that given a continuous self-map T : X → X of a compact metric space X preserving a Borel probability measure µ, a good knowledge of recurrence properties of the map T leads to lower bounds of the density of µ with respect to a Hausdorff measure H; in the case when µ is absolutely continuous with respect to H, this density is the Radon-Nikodym derivative dµ/dH. This bound entails in turn bounds on H(X).
Applying this method we arrive at an effective bound from below on the Hausdorff measure, in the relevant Hausdorff dimension, of the minimal set of a diffeomorphism f : S 1 → S 1 . It depends on the Diophantine properties of the rotation number and on the lengths of the of the wandering intervals of f , which as we know, are the connected components of S 1 \ Ω(f ). See Theorem 10 for a precise statement. We would like to emphasize that the estimates we obtain are effectively computable as we show in several examples in the last section of the paper.
Furthermore, our results allow us to recover the lower bounds on the Hausdorff dimension of Ω obtained in [2]. In addition, for ν = 1 we provide a simple upper estimate of the Hausdorff dimension of Ω and it coincides with the lower bound. In some cases we can even get a meaningful upper estimate of the Hausdorff measure. For example, our results show that for the classical Denjoy map the relevant Hausdorff measure is finite and positive and we provide with concrete bounds.

Setting, Definitions and some Tools Used
In this section we introduce notation and bring up the tools which we will use in the subsequent sections. Throughout the paper we always assume that α ∈ (0, 1), the rotation number of a considered circle diffeomorphism, is irrational, and that this diffeomorphism is exactly C 1+δ smooth with some δ ∈ (0, 1). We denote by x is the norm (arc-wise distance from the rightmost point) of a point x on the circle. Lastly, we always assume that the total length of a circle is equal to 1.

Construction of Denjoy Maps.
We start with recalling briefly the standard construction of Denjoy maps. The notation we implement here follows [2]. A precise construction may be found e.g. in [1].
Definition 1. Given a number δ ∈ (0, 1), a Denjoy sequence of class δ is any sequence { n } n∈Z of non-negative numbers satisfying the following two conditions.  where c δ is the normalizing constant making the sum n∈Z n equal to 1.
Observe that the classical sequence trivially satisfies the requirements of Definition 1. Such a sequence may be considered as a best one in the class of Denjoy sequences.
Also, using Riemann's ζ function we may state the value of the normalising constant: Now, fix a δ ∈ (0, 1) and take any Denjoy sequence ( n ) n∈Z of class δ. To obtain a Denjoy map, we start with a circle of length 1 with a map that is the rotation by α. Then, we blow up subsequently every point on the orbit of 0, i.e. point of the form ±nα, n ∈ N 0 , to an open interval J n of length n , rescaling at every step the remainder of the circle to have its length equal equal to 1.
Then we map J n diffeomorphically onto J n+1 with derivative equal to 1 at the endpoints. This map has a continuous extension, which we denote by f , to the entire circle S 1 , and, because of (2.2), this extension can be chosen to be of the class C 1+δ .
The minimal set (whose Hausdorff dimension and Hausdorff measure we want to estimate) of the map f is then equal to Ω = Ω δ α = S 1 \ n∈Z J n . (2.5) Observe that Ω also depends on the choice of the sequence ( n ) n∈Z .

Figure 1. Denjoy map construction
Remark 3. We want to note that in the construction of a Denjoy map f , the sum of lengths of n , n ∈ Z, might actually be smaller than 1, cf. (2.1). However, in such a situation the minimal set of f would have Hausdorff dimension trivially equal to 1 and positive Lebesgue measure. This is why we have removed such sequences from our considerations.

Remark 4.
Observe that every Denjoy map can be obtained by such a construction. The non-existence of the full conjugation gives rise to the minimal set, whose omitted intervals must follow the rotation by α. Take the longest such interval, call it J 0 , and then the semiconjugation with the rotation about α, defines the sequence ( n ) n∈Z . The property (2.2) comes from the smoothness of the diffeomorphism. As we mentioned in the previous remark, the lengths n need not sum up to 1, but -again -we remove such case from our consideration.
We will denote by h : Ω → S 1 the semi-conjugacy given by Poincaré's Theorem. By its construction, this semi-conjugacy cannot be a (full) conjugacy. Nevertheless, h −1 (z) is a singleton whenever z does not belong to the orbit of 0 under the rotation R α , i.e. whenever Otherwise, h −1 (z) is a doubleton. Furthermore, for any two points x, y ∈ S 1 \ O Rα , we also have that where |J n | denotes the length of interval J n and (x, y) is the shorter of the two arcs connecting x and y; if the intervals have equal length, take the one in positive orientation.
In general, i.e. for any points x, y ∈ S 1 , the above formula actually still holds. More precisely, it holds for those respective preimages of x and y for which the arc connecting them is the shortest.

Hausdorff Measure on Denjoy Minimal Sets.
As we are working with subtle measure estimates we deemed it prudent to put here the precise definitions in use in this paper.
First, recall our circle has length 1. We will use the standard version of the definition of the Hausdorff measure.
Definition 5. The outer measure is the following where the infimum is take over all countable covers of Y satisfying the conditions as stated. By Carathéodory's extension this gives the (typical) Hausdorff measure.
The next definition is also standard.
Definition 6. The Hausdorff dimension of the set Y is given by the formula

Diophantine Properties.
We denote the standard continued fraction expansion of an irrational number α ∈ (0, 1) as [a 1 , a 2 , . . .]. The corresponding convergents are denoted as pn qn . Recall that the convergents are the subsequent closest approximations of α. Also, the following properties hold: where q −1 = 0 and q 0 = 1, and We now bring up the following classical definition.
has infinitely many positive integral solutions q for every µ < ν and at only finitely many for every µ > ν. A number of Diophantine class ν = ∞ is commonly called a Liouville number.
Recall that the set of irrational numbers of Diophantine class ν = 1 is of full Lebesgue measure. However, for any ν ∈ [1, +∞] there exist α of Diophantine class ν.
Remark 8. Observe that if ν > 1, then for infinitely many n we have a very fast growth of q n 's. Combining (2.8) and (2.9) gives q n+1 ≈ q ν n for infinitely many ns.

Recurrence Results.
To estimate the Hausdorff measure and Hausdorff dimension of the minimal set Ω, we will apply the method described in a paper by the first named author, [3].
The result we need is this.
Theorem 9. If (X, d) is a metric space, µ is a Borel probability measure on X, and T : X → X is a Borel measurable map preserving measure µ, then for every β > 0 we have that This result is applied in the following way. For µ-almost, but see the comment in the next paragraph, every x ∈ X find a lower bound of the recurrence speed, i.e. the left hand side of the inequality above. This gives a lower bound on the above function g, giving in turn a lower bound on the Hausdorff measure of X by using the formula and the fact that µ(X) = 1.
Note that we do not have to really care about what the invariant measure actually is X is compact and the map T : X → X is continuous. Such measure then exists because of Bogolubov-Krylov Theorem. Indeed, ff we do know what the invariant measure is, then we may do the calculations for almost every point with respect to this invariant. If not, then we need to do the calculations for all points; except countably many of them, if the measure is non-atomic.

The Main Results.
As the general statement of our estimating method is rather complicated, we will state a few simpler versions. Also, our method can be applied under weaker assumptions, ex. for some sequences ( n ) n∈Z not satisfying the requirements of Definition 1; see Example 16.
Theorem 10. Fix δ ∈ (0, 1) and irrational rotation number α. Consider the corresponding Denjoy map f : S 1 → S 1 of the circle S 1 that is semi-conjugate to the rotation by R α with gaps J n , n ∈ Z whose sizes |J n | = n satisfy formulas (2.1) and (2.2) of Definition 1. For every n ∈ N, set N n := q n + q n+1 2 .
Then for any β > 0 we have that In fact, the following more complicated but stronger, estimate holds. (2.12) Th strongest estimate we prove is the following. For every n ∈ N, set Q n := q n + q n+1 and denote by P n the set of all sequences (P A direct straightforward application of the first estimate in the above Theorem 10 gives the following.
Corollary 11. With the notation and hypotheses of Theorem 10, we have that Lastly, with some additional work, as a consequence of the above corollary, we will prove (in Section 4) the lower bound on the Hausdorff dimension of Ω f which was obtained first by Kra and Schmelling in [2] with an entirely different method.
Proposition 12. Under assumptions as above (2.14) Let us also observe that if the Diophantine class ν of the rotation number α is equal to 1, then we may easily get estimates from above on the Haudorff dimension (and in some cases the Hausdorff measure).
Theorem 13. For ν = 1 and the classic Denjoy sequence (see Example 2) the Hausdorff dimension equals δ and the relevant Hausdorff measure is positive and finite.

Proof of Theorem 10
Fix an irrational number α ∈ (0, 1). Take a point x 0 in Ω f . For convenience assume that x 0 / ∈ O Rα . As we have already remarked, we may ignore all points in O Rα since this set is countable. In fact, our estimates also work for such points, but then one needs to be wary of the somewhat annoying double preimages.
We are to estimate from below the number lim inf n→∞ n (d(T n (x 0 ), x 0 )) β . Using (2.6), we see that where the sum is taken over all n, for which nα ∈ (h(x 0 ), h(f n (x 0 ))).
To find the lower limit we are only interested in the sequence of subsequent closest returns. Since the map h semi-conjugates the dynamics on Ω f with the rotation by α on the circle, the closest returns on Ω f are the closest returns for the rotation. They in turn are given by the reducts of the continued fraction expansion of α. Thus, we only need to estimate the numbers Let us denote the n-th closest return of x 0 as x n = f qn (x 0 ). First, we estimate the sum of lengths (3.1) from below by taking only the biggest element.
where k n ∈ Z is such that k n α ∈ h(x 0 ), h(x n ) and the number |J kn | is biggest possible (see Fig. 2). Now, we need to find bounds on k n . By the semi-conjugacy h(x n ) = h(x 0 ) + q n α, and the search for k n may by restated as follows.
Since we are assuming that h(x 0 ) is not in orbit of 0 under R α , the gap between h(x n ) and h(x 0 ) cannot be equal to any of the above gaps. The answer to our above question is given in the following lemma. Proof. We use the three gaps theorem as stated in [5]. Actually, we only need Lemma 2.1. from this paper, which (with the notation of this paper) states the following. Take a forward orbit of a point under irrational rotation R α . For K = q n +q n+1 −1 the set {αk : k = 0, . . . , K} partitions the circle into q n gaps of length q n+1 α and q n+1 gaps of length q n α . The figure below illustrates this phenomenon. We took α = √ 3 − 1 and drew the first 25 iterates. Note that all the gaps are only of two lengths. In our setting we are working with (tα) N t=−N . Rotating the circle (and all our points) by N α, gives the set (tα) 2N t=0 and does not change the gap sizes, leading straight to the required result. The next figure shows the forward and backward orbit for the same α.  Note that there is one gap on the left of the circle that is of a different size. This is the one resulting from the fact that such type of orbit (forward and backward) always has an odd number of points necessitating the rounding in our result. Removing 13 gives the same picture as before but rotated in such a way that −13 becomes 0.
Recall that we defined N n = qn+q n+1 2 . By Lemma 14, we know that |k n | ≤ N n , and we may estimate the size of J kn from below as Inserting this into our estimate yields This ends the proof of (2.11).
To get better bounds we need to utilize all the elements of the sum in (3.1). Recall that the finite sequence kα, k ∈ [−N n , . . . , N n ], hits every gap of size q n α . Similarly, the sequence kα, k ∈ [N n , . . . , 3N n ], hits every gap of the same size, and so does every sequence of type kα, [(2l − 1)N n , . . . , (2l + 1)N n ], for every integer l. Summing the lengths of the smallest intervals J i (on every sequence), leads straightforward to the estimate (2.12).
It is now easy to observe that our finite sequence may be moved around, i.e. for any P ∈ Z the sequence kα, for k ∈ [P, P + Q n ) hits every gap of size q n α . We used Q n instead of N n = [Q n /2] since there is no need for the finite sequences to be symmetric w.r.t. zero. Also, removing the 'integer part function' improves the estimate a little bit.
It remains to divide for every n ∈ N the set of all integers into sets of length Q n ; in the statement of (2.13), we called this 'dividing' sequence P and consider the classical Denjoy sequence where, we recall, c δ is the normalizing constant. Then the minimum in (2.11) is attained at the largest possible i, i.e. at i = N n , and, putting α * := 1+ √ 5 2 our simpler estimate gives Therefore, H β (Ω) > 0 for β ∈ (0, δ], thus proving that dim H (Ω) ≥ δ, and we get the following estimate on the Hausdorff measure where c δ = n∈Z (|n| + 1) −1/δ . We may easily improve the estimate by using the more complicated inequality (2.12). The minima in the sum are attained at the endpoints of the intervals, i.e. at (2k ± 1)N n (where the sign depends on whether k is positive or not). Let us first estimate the sum that occurs in the lower limit where we utilised the Riemann ζ function to simplify the harmonic sums. We used the following trivial equality: Finally, we see that, the only difference in comparison to the previous estimate is the last (rather complicated) constant. Thus we may repeat the same calculation, getting at the end β = δ. Let us also use the 'closed form' of c δ (2.4) arriving at but change the sequence |J n |, n ∈ Z, slightly. Start with n = (|n| + 1) −1/δ but change the elements ±4 k := 7 −k , k ∈ N. Normalize the sum by c and put |J n | := c n .
Important note: this does not satisfy (2.2)! That is, our construction does not give a diffeomorphism on the entire circle -only a homeomorphism. However, the method is still applicable. Now, using our estimate in the straightforward manner by taking tiny gaps as the minimum 'pollutes' the gap sizes, gives a weak result.
However, recall that the sequences kα, k ∈ [−N n , . . . , N n ] and kα, k ∈ [N n , . . . , 3N n ], hit every gap of size q n α . So, the sequence kα, k ∈ [N n , . . . , 5N n ], hits every gap at least twice. Now observe that the growth of the sequence N n ≈ α n * ≤ 1.7 n , n ∈ N, is much slower than the growth of the distances between numbers of gaps with bad sizes, i.e. 4 k , k ∈ N. Thus our set [N n , . . . , 5N n ] for n large enough may hit at most one element of the sequence 4 k , k ∈ N. Since every gap is hit at least twice, we may ignore that one, and recover the estimate as before, only slightly worsened.
Therefore, H β (Ω) > 0 for β ∈ (0, δ], thus proving that dim H (Ω) ≥ δ, and we get the following estimate on the Hausdorff measure As above one could improve this estimate by using the whole sum, i.e. applying (2.13).
Example 17. Take α ∈ (0, 1) whose continued fraction expansion satisfies the relation q n+1 = q 2 n for every n ∈ N. Then a n ≈ c 2 n for some integer c ≥ 2. α is then of Diophantine class ν = 2. Set where c is the corresponding normalizing constant. We know that for such quickly growing numbers q n we have for any ε > 0 and all integers n ≥ 1 large enough that Using this formula in our estimate, i.e. the computation as in the previous example, leads in the current case to the following.
Proof of Proposition 12. To get the desired lower bound on the Hausdorff dimension of Ω f , we need two ingredients: some kind of a universal bound on the lengths |J n |, n ∈ Z, and some bound on the growth of the denominators of the continued fraction expansion.
The assumption (2.2) imposed on the sequence n = |J n |, n ∈ Z, easily leads to the following result which was stated and proved as Lemma 2.3 in [2].
For all 0 < θ < δ < 1 there exists s θ ≥ 1 such that for all integers n > s θ . Now we shall deal with the growth of the q n s, n ∈ N. Combining the continued fraction property (2.8) and the Diophantine class definition (2.9) we get for every ε > 0 and all n ≥ s ε that 1 q n+1 > q n α > 1 q ν+ε n . This gives the desired estimate q n+1 < q ν+ε n . To do this, first fix θ ∈ (0, δ) and also any ε > 0. Then fix any integer n > max(n ε , n θ ). We will obtain the desired lower estimate by using first (4.2), then the definition of N n together with the fact N n < q n+1 , and finally (4.3). With any β > 0, all of these lead to lim inf n→∞ q n min β |J i | : −N n ≤ i ≤ N n ≥ lim inf So, this lower limit is positive for β = θ ν+ε . Therefore, by applying Corollary 11, we get that dim H (Ω) ≥ θ ν+ε . Hence, by letting first ε go to 0 and then θ to δ, we obtain that dim H (Ω) ≥ δ ν .
Proof of Theorem 13. By the previous results, we only need to prove that H δ (Ω) < +∞. Observe the set S 1 \ n k=−n J k ⊃ Ω.
It consists of 2n + 1 disjoint intervals, let us call them I p , whose total length equals For the classic sequence both of the sums on the RHS are equal and may be trivially bounded by and integral giving |k|>n |J k | ≤ 2 δ δ − 1 c δ (n − 1) 1−1/δ .
As δ < 1, function x δ is concave and application of Jensen's inequality leads to the following bound on the covering sum (for clarity, put N = 2n + 1 until the last equality).
Clearly, the lengths (diameters) of the intervals |I p | shrink to 0 as n → +∞. Thus, we arrive at Remark 18. Note that the calculation above works for all irrational α, but whenever ν > 1 it does not give the proper upper estimate on the dimension. It should not come as a surprise as in the proof above we essentially treat the cover as its elements were of comparable sizes.
On the other hand, if ν > 1, then looking at the rotation by α sometimes the orbit returns very close to the starting point. So in the proof above some of the sets I p will have very small length in comparison with others. Thus, a cover were the sizes are similar may not be optimal.
Following the proof of the previous result we get the following generalisation.