Weak tangent and level sets of Takagi functions

In this paper, we study some properties of Takagi functions and their level sets. We show that for Takagi functions Ta,b\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$T_{a,b}$$\end{document} with parameters a, b such that ab is a root of a Littlewood polynomial, there exist large level sets. As a consequence, we show that for some parameters a, b, the Assouad dimension of graphs of Ta,b\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$T_{a,b}$$\end{document} is strictly larger than their upper box dimension. In particular, we can find weak tangents of those graphs with large Hausdorff dimension, larger than the upper box dimension of the graphs.


Introduction
In this paper, we study graphs of the following functions, where a, b are real parameters a, b such that a < 1, b > 1, ab ≥ 1 and T : R → R is the tent map which has period 1 and defined on the unit interval as follows, x ∈ 0, 1 Such functions T a,b are called Takagi functions. Originally the Takagi function was referred to as T 1/2,2 but no confusion should appear if we also call T a,b to be a Takagi function. There has been a lot of interest in the Hausdorff and box counting dimensions of the graphs of such functions. For the box dimensions we know from [14,Section 2] and [4,Theorem 2.4] that the upper box dimension of graphs of these functions T a,b can be computed by the following formula The Hausdorff dimensions of graphs of these functions are harder to obtain, see [19], [5] and the references therein for more recent results on related questions. One of the results of this paper is about the Assouad dimension of some Takagi functions. In what follows, for a function f : R → R, we denote the following set A is the Assouad spectrum with parameter 1/B. In this paper, we focus only on the Assouad dimension. For more details on Assouad spectrum, see [10].
Notice that by keeping the product ab unchanged and making b larger, this lower bound can be larger than the upper box dimension ln ab 2 ln b for large b. For example, when we choose parameters such that ab = √ 5+1 The rightmost inequality can be strict. Theorem 1.1 follows from the existence of a large level set of graphs T a,b and we think this result is interesting on its own. Theorem 1.4 Let T a,b be as in the statement of Theorem 1.1 but we also allow ab = 1. For each y ∈ R we define the following level set The restriction of the product ab as a certain type of algebraic integers seems to be strong. However, with some effort, we can show that those algebraic integers are dense in [ 1 2 , 2]. Theorem Let L be the set of algebraic integers which are roots of Littlewood polynomials namely, x ∈ C and there exist an integer k ≥ 1 and a finite sequence 2]. Proofs of the above result and its generalizations can be found in [2,3,18].

Discussions and future work
In this section, we give some backgrounds of Theorems 1.1 and 1.4. We also pose some questions which are related to the results in this paper.  We have not completely determined the Assouad dimension of any Takagi function yet. We only showed that the Assouad dimension could be strictly larger than the upper box dimension for graphs of Takagi functions. Question 2.2 For a, b ∈ R + and ab > 1, determine the Assouad dimension for the graph of T a,b .

Level sets of Takagi functions
For more details about the level sets of Takagi functions, see [1,16]. Notice that if we set a = 0.5, b = 2 then we can find level set of T a,b with Hausdorff dimension at least 0.5. This is sharp, see [6]. For other values of a, b for example a = ( √ 5+1)/16, b = 8 we see that we can find a level set with Hausdorff dimension at least 1/3 and we do not know whether this is sharp.

Notation
1. For a real number x ∈ R we use the symbol x + to denote a number x + where > 0 is some fixed positive number whose value can be chosen freely and we will point out the specific value of when necessary. Similarly, we use x − for a number smaller but close to x. 2. For a function f : R → R, the following set is called the graph of f over the interval [0, 1]. 3. For a real number x, we use x to denote the greatest integer that is not strictly larger than x.

Preliminaries
We will now introduce some notions of dimensions which will be used in this paper. We use N r (F) for the minimal covering number of a bounded set F in R n with balls of side length r > 0.

Hausdorff dimension
The Hausdorff dimension of F is defined to be

Upper box dimension
The upper box dimension of F is

Assouad dimension and weak tangents
The where B(x, R) denotes the closed ball of centre x and radius R. An important tool for studying the Assouad dimension is weak tangents introduced in [17] and microsets in [11]. The next definition appeared in [9, Definition 1.1]. Definition 4.1 Let X ∈ K(R n ) be a fixed reference set (usually the closed unit ball or cube) and let E, F ⊂ R n be compact sets. Suppose there exists a sequence of similarity maps T k : is a complete metric space with the Hausdorff metric, namely, for two compact subsets A, B ⊂ R n is defined by where for any compact set C ⊂ R n C δ = {x ∈ R n : |x − y| < δ for some y ∈ C}.

Theorem 4.2 (KOR) Let F be a compact set with dim A F = s. Then there exist a weak tangent E of F such that
In other words, we have

Covering by disjoint cubes
For convenience, in this paper we will count covering number with disjoint squares rather than balls. We denote S(a, R) for a ∈ R 2 , R > 0 as the square centred at a with side length 2R whose sides are parallel to the coordinate axis. Since we are dealing with graph of functions, the choice of axis is natural. We denote the following covering number, This is equivalent to N r (F ∩ B(a, R)) in the sense that there exists a constant C > 0 such that for all a ∈ F, 0 < r < R < 1 we have the following inequality, B(a, R)).

Some properties of Takagi functions
In this paper, we will use the following result whose proof can be found in [13], and we use the version presented in [4,Theorem 2.4].
Notice that if a < 1, ab > 1 then − ln a ln b ∈ (0, 1) and we see that if |J | < 1 then

Remark 4.4
When T is the tent map defined in the beginning of the first section, it is known that when a < 1, ab ≥ 1, the function T a,b is nowhere differentiable therefore only the second property of Lemma 4.3 can be true.

Large level sets, proof of Theorem 1.4
We show that there exist large level sets for function T a,b with certain parameters a, b.
Since ab is a root of a Littlewood polynomial we see that for an integer k ≥ 1 and some choice of i ∈ {±1}. Next we consider the first k terms partial sum The derivative of the above function is not continuous at x = mb −k for integers m.
Let us now assume that x is an irrational number then the derivative is Therefore we can find at least 2 disjoint intervals of length 1 2b k−1 where F 1 (x) = 0 and F 1 (x) = a 1 is for a constant a 1 ≥ 0 on those two intervals. Indeed, when So there are at least two intervals we can find and the union is symmetric with respect to the line {x = 0.5}. Then because F 1 is also symmetric with respect to the line {x = 0.5} we see that F 1 (.) takes the same value on those two intervals, say, I 1 and I 2 .
We consider the next k terms sum Then we can find b many intervals of length 1/2b 2k−1 in I 1 , I 2 such that the above sum stays constant a 2 ≥ 0 on those intervals. To see this, consider I 1 , which is an interval of length 1/2b k . Now observe the following Therefore the graph of F 2 is an affine copy, or intuitively speaking, a narrowed version of the graph of F 1 . Then we see that there are exactly b many intervals in I 1 of length 1/2b 2k−1 such that F 2 equals to a 2 on all those intervals. Indeed, over any interval the form [l/b k−1 , (l + 1)/b k−1 ] the graph of F 2 has 2b many plateaux of the same level. That is to say, we can find 2b many 1/2b 2k−1 length intervals on which F 2 (x) = a 2 .
Since I 1 is only a half 1/b k−1 length interval, therefore we can find b many plateaux over I 1 . Here we used the mirror symmetry of F 2 . We can apply the above argument to j-th k-terms partial sums for each j ≥ 2 and as a result we can find a Cantor set C such that T a,b (C) = {c} for a constant c. By construction, this Cantor set is obtained by first taking two intervals I 1 , I 2 of length 1/2b k−1 then placing b-many intervals of length 1/2b 2k−1 inside of each I 1 , I 2 . Then, inside of those intervals of length 1/2b 2k−1 , we place b many intervals of length 1/2b 3k−1 . This procedure goes on indefinitely and self-similarly (just like the construction of the middle third Cantor set). As a result, the obtained Cantor set is a self-similar set satisfying the open set condition and its Hausdorff dimension is 1/k (the contraction ratio is 1/b k and branching number is b, see for example [7, Theorem 9.3]). Thus we have proved Theorem 1.4.

Squashing and counting, proof of Theorem 1.1
In order to deal with the Assouad dimension of graphs of Takagi functions we need to handle the following quantity The situation is not too bad when we want to deal with the above quantity for f +g when one of the functions, say f , is Lipschitz continuous. The following result is a localized and quantitative version of [

Remark 6.2
For the case when R/r is not an integer, we can replace r with a larger value As we will eventually choose R/r to be arbitrarily large, r and r are essentially the same. For any δ > 0, if R/r is large enough the following relation holds Then the inequality of this theorem holds with 1/(M + 2) being replaced by some other constant which depends only on M.
Proof of Lemma 6.1 Let a ∈ R 2 and consider the square S(a, R/2). For any r < R we consider the following rectangles for i = 0, 1, . . . , R r − 1

× [a − R/2, a + R/2] ⊂ S(a, R/2).
Each rectangle S i contains the following squares for j = 0, 1, . . . , R r − 1 Now if S i j ∩ g = ∅ we colour it black, otherwise we colour it white. Let n i ≥ 0 denote the number of black squares among S i j , j = 0, 1, 2, . . . , R r − 1. By continuity of g, the fact that we have n i black squares implies the following inequality By Lipschitz property of function f we see that Then we see that So we see that to cover the set we need at least n i − 2 − M many squares of side length r . Summing over all i we see that to cover the set R r many squares with side length r . The next fact to notice is that the following set is contained in a R × (M + 1)R rectangle. This rectangle can be covered by M + 2 squares with side length R, so for at least one of the M + 2 squares need at least many squares with side length r to cover. It is then easy to see that This implies that and since R r is integer we see that .
We can take the supreme of a on the right hand side of the above inequality and the lemma concludes.
Now we can move on to dealing with the Assouad dimension of Takagi functions. We shall need the following lemma.
Then we have the following result, Proof For any positive integer M > 0, we can find an integer k and and on this subset we have the following condition for the oscillation of the following M-th partial sum M n=0 a n T (b n x 1 )) − M n=0 a n T (b n x 2 )) < Then we can write that where the functions F M , G M are the first sum and second sum in the third expression.
Then we see that the graph G M is actually a 'squashed' version of T a,b , namely we have the following relation where the linear transformation X M : R 2 → R 2 is defined to be as follows We see that since ab > 1 this linear transformation squashes a square to a very thin rectangle for large enough M. Now we concentrate on the strip The graph G M over this strip is the squashed version of the graph T a,b over [0, 1]. We want to find a 1 b M+1 -square contained in this strip such that we need a reasonably large amount of r -squares to cover G M , where r > 0 is number that will be specified later. Now because of the bijective linear map X M , covering a 1 b M+1 -square with rsquare in G M is the same thing as covering the original graph T a,b over [0, 1] inside a 1 × 1 (ab) M+1 -rectangle with rb M+1 × r a m+1 -rectangles. Now consider the level set L with lower box dimension D mentioned in the second condition. For some real number y we have that then we do box counting in the 1× 1 (ab) M+1 -rectangle containing L. For any (x, y ) ∈ L the graph T a,b intersects the middle axis of the rectangle at (x, y ), so by Lemma 4.3 we see that there exists a positive constant C 3 > 0 such that the projection of graph T a,b inside any 1 (ab) M+1 -square centred in L to the vertical axis has length at least If M is large enough we need at least 1 ab −D − (M+1) many 1 (ab) M+1 -square to cover this 1 × 1 (ab) M+1 -rectangle because for covering the level set L we already need that many squares. Since each square is sufficiently occupied by T a,b in the sense that the curve occupies at least min(C 3 , 1) portion of the vertical length. This means that for each such square we need some constant times 1/(ab) M+1 r /a M+1 = 1 rb M+1 many rb M+1 × r a m+1 -rectangle to cover. Now we choose the following value for r where B = 2 + ln a ln b is the upper box dimension of T a,b . We denote R = 1 b M+1 and note that r = R 1 θ with θ = 1 B and we get the following relation The above inequality holds for arbitrarily large M and therefore it holds also for arbitrarily small R, R 1 θ . This is a covering property for G M and we can translate it to a covering property for T a,b . By using Lemma 6.1 and Remark 6.
Then by definition of the Assouad dimension, we see that This concludes the proof.
We can now finish the proof of Theorem 1.1. By Lemma 6.4 and Theorem 1.4 we see that it is enough to show that the Takagi functions satisfy condition (1) in the statement of Lemma 6.4. In fact, condition (1) is satisfied by T a,b whenever a < 1, b ∈ {2} ∪ [2, ∞], ab ∈ (1, 2). We put the last step of proving Theorem 1.1 in the following lemma. l, l + 1 is a pair of integers, one of them is odd and the other is even. It is easy to see that this is also true for b = 2. Then we see that given any sequence ω i ∈ {±1}, i = 0, 1, 2 . . . it is possible to choose a sequence of integers j i , i = 0, 1, . . . such that Then by Lemma 6.3 we see that the result follows.