Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 27, 2002, 325–339

ON THE COMPLEXIFICATION OF THE WEIERSTRASS NON-DIFFERENTIABLE FUNCTION

Krzysztof Barański

Warsaw University, Institute of Mathematics ul. Banacha 2, PL-02-097 Warsaw, Poland; baranski@mimuw.edu.pl

Abstract. It is shown that for the Weierstrass nowhere differentiable functions X_{a,b}(t) = ∑_{n=0}^{∞} a^{n} cos(b^{n}t) and Y_{a,b}(t) = ∑_{n=0}^{∞} a^{n} sin(b^{n}t) the set (X_{a,b}, Y_{a,b})([0, 2π]) has a non-empty interior in R^{2} , provided b ∈ N, b ≥ 2 and a < 1 is sufficiently close to 1. It follows that the box dimension of graph(X_{a,b}, Y_{a,b}) is equal to 3 − 2α where α = − log a/ log b and its Hausdorff dimension is at least 2. Moreover, the level sets L(s) for X_{a,b} and Y_{a,b} have Hausdorff dimension at least α for open sets of s ∈ R, so the Hausdorff dimension of graph X_{a,b} and graph Y_{a,b} is at least 1 + α.

1. Introduction

This paper concerns the famous functions

X_{a,b}(t) = ∑_{n=0}^{∞} a^{n} cos(b^{n}t), Y_{a,b}(t) = ∑_{n=0}^{∞} a^{n} sin(b^{n}t)

for t ∈ [0, 2π] and 0 < a < 1, b > 1, ab ≥ 1. The first one was introduced by Weierstrass in 1872 as an example of a continuous, nowhere differentiable function. In fact, the non-differentiability for all given above parameters a, b was proved by Hardy in [Ha]. Later, the graphs of these and related functions were studied as fractal curves. A basic question which arises in this context is computing the Hausdorff dimension (HD) of these curves. However, this problem is still unsolved for the classical functions X_{a,b} and Y_{a,b}.

For ab = 1, the graphs of X_{a,b} and Y_{a,b} have Hausdorff dimension 1 and σ - finite 1-dimensional Hausdorff measure, as was proved by Mauldin and Williams in [MW]. For ab > 1, it is easy to check that the functions X_{a,b} and Y_{a,b} are Hölder continuous with exponent α for

α = −\frac{log a}{log b}

2000 Mathematics Subject Classification: Primary 28A80, 28A75, 28A78.

Research supported by Polish KBN Grant 2 P03A 009 17.

(i.e. a = b^{−α}). Consequently, the box dimension (BD) of their graphs is at most 2 − α (see Lemma 2.2). In fact, it is equal to 2 − α, as was proved in [KMY]. Hence, HD(graph X_{a,b}), HD(graph Y_{a,b}) cannot exceed 2 − α. Moreover, the packing dimension of these graphs is 2 − α (see [R]).

It is believed that the Hausdorff dimension of these graphs should also be equal to 2 − α. Note that

X_{a,b}(t) = aX_{a,b}(bt) + cos t, Y_{a,b}(t) = aY_{a,b}(bt) + sin t,

which means that the graphs are roughly self-similar for the scaling with horizontal factor b and vertical factor a.

The difficulties lie in the lower estimates of the Hausdorff dimension. There are not too many results in this direction. Mauldin and Williams in [MW] gave the lower bound of the form 2 − α − C/ log b for a constant C > 0 independent of b, which approaches the upper bound as b → ∞. Przytycki and Urbański proved in [PU] that the Hausdorff dimension of graph X_{a,b}, graph Y_{a,b} is greater than 1 for b ∈ N, b ≥ 2. In [Hu], Hunt showed that the Hausdorff dimension of the graph of the function

∑_{n=0}^{∞} a^{n} cos(b^{n}t + θ_{n})

with θ_{n} chosen independently with respect to the uniform probability measure on [0, 2π], is almost surely equal to 2 − α.

It turns out that it is easier to consider the problem for the Weierstrass function with cosine replaced by some other continuous periodic function g: R → R. For instance, take g(t) = dist(t, Z) (the sawtooth function), which was studied by Besicovitch and Ursell in [BU]. For a = b, we obtain the van der Waerden–Tagaki function, which has Hausdorff dimension 1 and σ -finite 1-dimensional Hausdorff measure. This was proved by Anderson and Pitt in [AP]. Moreover, by the work of Ledrappier [L], the case b = 2, a > \frac{1}{2} can be brought to the case of the Bernoulli convolutions ∑ ±a^{n}, where the signs are chosen independently with probability \frac{1}{2} on [0, 1]. Then the work of Solomyak [S] implies that the Hausdorff dimension of the graph is 2 − α for almost all a ∈ (\frac{1}{2}, 1).

Another interesting problem is studying various measures related to these graphs. For a function f: [t_{0}, t_{1}] → R^{m} denote by μ_{f} the image under f of the uniform probability measure on [t_{0}, t_{1}]. Little is known about the measures μ_{X_{a,b}} and μ_{Y_{a,b}}. It is conjectured that the Hausdorff dimension of graph X_{a,b} (or graph Y_{a,b}) is 2−α if and only if the measure μ_{X_{a,b}} (or μ_{Y_{a,b}}) is absolutely continuous with respect to the Lebesgue measure on R. This holds for the Bernoulli convolutions, as was proved in [PU]. Kôno showed in [K] that if b ∈ N and ab is sufficiently large and the suitable measure μ_{X_{a,b}} or μ_{Y_{a,b}} has a bounded density function with respect to the Lebesgue measure, then the Hausdorff dimension of the graph is 2 − α.

On the complexification of the Weierstrass non-differentiable function 327

In this paper we consider the complexification of the functions X_{a,b} and Y_{a,b}, i.e.

F_{a,b}(z) = ∑_{n=0}^{∞} a^{n}z^{b^{n}}, z ∈ C, |z| ≤ 1,

for a ∈ (0, 1) and b ∈ N, b ≥ 2. Then F_{a,b} is holomorphic in the open unit disc, continuous in the closed unit disc and

Re(F_{a,b}(e^{it})) = X_{a,b}(t), Im(F_{a,b}(e^{it})) = Y_{a,b}(t).

We prove that if a is sufficiently close to 1, then the image of the unit circle S^{1} under F_{a,b} (i.e. the image of the segment [0, 2π] under the map (X_{a,b}, Y_{a,b})) is a curve which has non-empty interior in the topology of the plane. More precisely, we show

Theorem 1.1. There exist a_{0} < 1 and c > 0, such that for every a ∈ [a_{0}, 1) and every b ∈ N, b ≥ 2, the set F_{a,b}(S^{1}) contains a disc of radius c/(1 − a). Moreover, F_{a,b}(S^{1}) is the closure of its interior in the topology of the plane.

Figure 1. The curve F_{a,2}(S^{1}) for a = 0.7 (left) and a = 0.8 (right).

The idea of complexifying the Weierstrass function is not new. In [Ha] Hardy used its harmonic extension to prove non-differentiability. Our approach, however, is not analytical but relies on some elementary geometric facts (Lemma 3.3).

Apart from presenting an interesting example of a “plane-filling” curve, Theorem 1.1 has some consequences concerning the graphs of the functions X_{a,b}, Y_{a,b}. First, we can compute the exact value of the box dimension of graph(X_{a,b}, Y_{a,b}) (as a subset of R^{3} ), which is equal to 3 − 2α. The Hausdorff dimension of this graph is at least 2. These results are shown in Corollary 4.1.

For s ∈ R define the level sets of X_{a,b}, Y_{a,b} as

L_{X_{a,b}}(s) = {t ∈ R : X_{a,b}(t) = s}, L_{Y_{a,b}}(s) = {t ∈ R : Y_{a,b}(t) = s}.

In Corollary 4.3 we show that the Hausdorff dimension of L_{X_{a,b}}(s) and L_{Y_{a,b}}(s) is at least α for some open sets of s ∈ R. This implies (Corollary 4.4) that

HD(graph X_{a,b}), HD(graph Y_{a,b}) ≥ 1 + α.

Theorem 1.1 and the corollaries are true for a close to 1, i.e. for α close to 0. The functions X_{a,b}, Y_{a,b} are Hölder continuous with exponent α, so the map (X_{a,b}, Y_{a,b}) is also Hölder continuous with the same exponent. This implies (see Lemma 2.2) that

HD(F_{a,b}(S^{1})) = HD((X_{a,b}, Y_{a,b})([0, 2π])) ≤ \underline{BD}((X_{a,b}, Y_{a,b})([0, 2π])) ≤ 3 − 2α,

so for α > \frac{1}{2} we have HD(F_{a,b}(S^{1})) < 2. In particular, F_{a,b}(S^{1}) has 2-dimensional Lebesgue measure 0, μ_{(X_{a,b},Y_{a,b})} is singular with respect to this measure and Theorem 1.1 cannot be true. It would be of interest to check whether Theorem 1.1 holds for every α ≤ \frac{1}{2} . (See Figure 1, where the left picture shows the curve for α = 0.5145 … and the right one for α = 0.3219 . . ..) The most interesting case is α = \frac{1}{2} . Indeed, we have the following:

Fact. Suppose α = \frac{1}{2} and F_{a,b}(S^{1}) has positive 2-dimensional Lebesgue measure. Then μ_{(X_{a,b},Y_{a,b})} is not singular with respect to this measure and the measures μ_{X_{a,b}}, μ_{Y_{a,b}} are not singular with respect to 1-dimensional Lebesgue measure. Moreover, HD(graph X_{a,b}) = HD(graph Y_{a,b}) = 2 − α = \frac{3}{2} .

Proof. Let Leb_{m} be the m-dimensional Lebesgue measure. By Lemma 2.2, we have Leb_{2} |_{F_{a,b}(S^{1})} ≤ Cμ_{(X_{a,b},Y_{a,b})} for a constant C. Suppose μ_{(X_{a,b},Y_{a,b})} is singular with respect to Leb_{2} and take a set A ⊂ F_{a,b}(S^{1}) such that μ_{(X_{a,b},Y_{a,b})}(A) = 1 and Leb_{2}(A) = 0. Then

Leb_{2}(F_{a,b}(S^{1})) = Leb_{2}(F_{a,b}(S^{1}) \ A) ≤ Cμ_{(X_{a,b},Y_{a,b})}(F_{a,b}(S^{1}) \ A) = 0,

which contradicts the assumption. Hence, μ_{(X_{a,b},Y_{a,b})} is not singular with respect to Leb_{2} . Since μ_{X_{a,b}}, μ_{Y_{a,b}} are orthogonal projections of μ_{(X_{a,b},Y_{a,b})} on the coordinate axes, they are not singular with respect to Leb_{1} . The last part follows from Corollary 4.4, because 1 + α = \frac{3}{2} = 2 − α.

2. Preliminaries

We recall some basic definitions and facts concerning the Hausdorff and box dimension.

Definition 2.1. For A ⊂ R^{n} and δ > 0 the (outer) δ -Hausdorff measure of A is defined as

ℋ ^{δ}(A) = lim_{ε→0} inf ∑_{U∈𝒰} (diam U)^{δ},

where infimum is taken over all countable coverings 𝒰 of A by open sets of diameters less than ε.

The Hausdorff dimension of A is defined as

HD(A) = sup{δ > 0 : ℋ ^{δ}(A) = +∞} = inf{δ > 0 : ℋ ^{δ}(A) = 0}.

Let N_{ε}(A) be the minimal number of balls of diameter ε needed to cover A. Define the lower and upper box dimension as

\underline{BD}(A) = lim inf_{ε→0} \frac{log N_{ε}(A)}{− log ε}, \overline{BD}(A) = lim sup_{ε→0} \frac{log N_{ε}(A)}{− log ε}.

The box dimension is also called the box-counting or Minkowski dimension. It is easy to check that

HD(A) ≤ \underline{BD}(A).

The definitions of the Hausdorff and box dimension easily imply

Lemma 2.2. Let A ⊂ R^{n} and let f: A → R^{m} be a map such that

∥f(x) − f(y)∥ ≤ c∥x − y∥^{β}

for every x, y ∈ A and constants c > 0, 0 < β ≤ 1. Then for every δ > 0,

ℋ ^{δ/β}(f(A)) ≤ c^{δ/β}ℋ ^{δ}(A),  so  HD(f(A)) ≤ HD(A)/β.

Moreover,

\underline{BD}(graph f) ≤ \underline{BD}(A) + m(1 − β),
\overline{BD}(graph f) ≤ \overline{BD}(A) + m(1 − β).

We shall use the following theorem estimating the Hausdorff dimension of a planar set by the dimensions of its level sets (for the proof see e.g. [F]).

Theorem 2.3. Let E ⊂ R^{2} and A ⊂ R. Suppose that there exists β > 0, such that if x ∈ A, then ℋ ^{β}({y ∈ R : (x, y) ∈ E}) > c, for some constant c. Then for every δ > 0,

ℋ ^{δ+β}(E) ≥ bcℋ ^{δ}(A),

where b depends only on β and δ. In particular,

HD(E) ≥ HD(A) + inf_{x∈A} HD({y ∈ R : (x, y) ∈ E}).

Notation. The symbols cl, int and ∂ denote respectively the closure, interior and boundary in the topology of the plane. The euclidean distance is denoted by dist. \overline{AB} is the segment with endpoints A, B and |\overline{AB}| is its length. We write D_{r}(x) for the open disc centred at x ∈ C of radius r. For t ∈ R we denote by [t] the integer part of t, i.e. the largest integer not greater than t.

330 Krzysztof Barański

3. Proof of Theorem 1.1

Let 0 < a < 1, b ∈ N, b ≥ 2. By the definition of F_{a,b}, we have

F_{a,b}(e^{2πit_{1}}) − F_{a,b}(e^{2πit_{2}}) = 2i ∑_{n=0}^{∞} a^{n} sin(πb^{n}(t_{1} − t_{2}))e^{πib^{n}(t_{1}+t_{2})}

for every t_{1}, t_{2} ∈ [0, 1]. Let

z_{n,k} = e^{2πik/b^{n}} for n ≥ 0, k = 1, …, b^{n}

and fix j ∈ Z. Then

F_{a,b}(z_{n,k+j}) − F_{a,b}(z_{n,k}) = 2i ∑_{l=0}^{n−1} a^{l} sin(πj/b^{n−l})e^{πi(2k+j)/b^{n−l}}

= 2ia^{n} ∑_{m=1}^{n} a^{−m} sin(πj/b^{m})e^{πi(2k+j)/b^{m}}

= 2ia^{n} ∑_{m=1}^{n} u_{m}^{(j)}(a, b)ζ_{m,k}^{(j)}(b),

where

u_{m}^{(j)}(a, b) = a^{−m} sin(πj/b^{m}), ζ_{m,k}^{(j)}(b) = e^{πi(2k+j)/b^{m}}.

Note that u_{m}^{(j)}(a, b) ∈ R, ζ_{m,k}^{(j)}(b) ∈ S^{1} . Moreover,

z_{n,k} = z_{n+n_{0},b^{n_{0}} k}

and

ζ_{m,b^{n_{0}} k}^{(j)}(b) = e^{πi(2b^{n_{0}} k+j)/b^{m}} = e^{πij/b^{m}}e^{2πikb^{n_{0}−m}}

for every n_{0} ≥ 0. Thus,

ζ_{m,b^{n_{0}} k}^{(j)}(b) = e^{πij/b^{m}} for m ≤ n_{0}

and

(1) \begin{matrix}F_{a,b}(z_{n+n_{0},b^{n_{0}} k+j}) − F_{a,b}(z_{n,k}) = 2ia^{n+n_{0}} ∑_{m=1}^{n+n_{0}} u_{m}^{(j)}(a, b)ζ_{m,b^{n_{0}} k}^{(j)}(b) \\ = 2ia^{n+n_{0}} ∑_{m=1}^{n_{0}} u_{m}^{(j)}(a, b)e^{πij/b^{m}} + 2ia^{n+n_{0}} ∑_{m=n_{0}+1}^{n+n_{0}} u_{m}^{(j)}(a, b)ζ_{m,b^{n_{0}} k}^{(j)}(b) \\ = 2ia^{n+n_{0}}( ∑_{m=1}^{∞} u_{m}^{(j)}(a, b)e^{πij/b^{m}} − ∑_{m=n_{0}+1}^{∞} u_{m}^{(j)}(a, b)e^{πij/b^{m}} \\ + ∑_{m=n_{0}+1}^{n+n_{0}} u_{m}^{(j)}(a, b)ζ_{m,b^{n_{0}} k}^{(j)}) \\ = 2ia^{n+n_{0}}(U^{(j)}(a, b) + Δ_{n,k,n_{0}}^{(j)}(a, b)), \end{matrix}

where

U^{(j)}(a, b) = ∑_{m=1}^{∞} u_{m}^{(j)}(a, b)e^{πij/b^{m}}.

Note that

(2) |Δ_{n,k,n_{0}}^{(j)}(a, b)| ≤ 2 ∑_{m=n_{0}+1}^{∞} |u_{m}^{(j)}(a, b)| ≤ 2πj ∑_{m=n_{0}+1}^{∞}(ab)^{−m} = \frac{2πj}{ab − 1}(ab)^{−n_{0}}.

Let

U^{(j)}(b) = ∑_{m=1}^{∞} sin(πj/b^{m})e^{πij/b^{m}}.

Lemma 3.1. For every b ∈ N, b ≥ 2, there exists an integer j_{0} such that (1) U^{(j_{0})}(b), U^{(−j_{0})}(b) ≠ 0 and Arg(U^{(j_{0})}(b)) ≠ Arg(U^{(−j_{0})}(b)),

(2) if b tends to ∞, then U^{(±j_{0})}(b) tend respectively to U^{(±j_{0})} ≠ 0 such that Arg(U^{(j_{0})}) ≠ Arg(U^{(−j_{0})}).

Proof. By definition,

Re(U^{(j)}(b)) = \frac{1}{2} ∑_{m=1}^{∞} sin(2πj/b^{m}), Im(U^{(j)}(b)) = ∑_{m=1}^{∞} sin^{2}(πj/b^{m}).

Note that for every j ≠ 0 we have Im(U^{(j)}(b)) > 0, so U^{(j)}(b) ≠ 0 and Arg(U^{(j)}(b)) ∈ (0, π). Moreover,

Re(U^{(−j)}(b)) = − Re(U^{(j)}(b)), Im(U^{(−j)}(b)) = Im(U^{(j)}(b)),

so Arg(U^{(j)}(b)) ≠ Arg(U^{(−j)}(b)) if and only if Re(U^{(j)}(b)) ≠ 0.

Let

j_{0} = \{ \begin{matrix}1 & for b < 4, \\ [\frac{1}{4}b] & for b ≥ 4. \end{matrix}

Then 0 < 2πj_{0}/b^{m} ≤ π for all b ≥ 2, m ≥ 1 and the equality holds only if b = 2, m = 1. This implies Re(U^{(j_{0})}(b)) > 0, so Arg(U^{(j_{0})}(b)) ≠ Arg(U^{(−j_{0})}(b)).

Note that

(3)  0 < j_{0} ≤ \frac{1}{2}b.

Using this, we obtain

|U^{(±j_{0})}(b) − sin(±πj_{0}/b)e^{±πij_{0}/b}| ≤ ∑_{m=2}^{∞} \frac{πj_{0}}{b^{m}} = \frac{πj_{0}}{b(b − 1)} ≤ \frac{π}{2(b − 1)},

which tends to 0 as b tends to ∞, so U^{(±j_{0})}(b) tends to

U^{(±j_{0})} = lim_{b→∞} sin(±π[\frac{1}{4}b]/b)e^{±πi[b/4]/b} = \frac{1}{2}(±1 + i).

Lemma 3.2. Let j_{0} = j_{0}(b) be the number defined in the proof of Lemma 3.1. If a tends to 1, then U^{(±j_{0})}(a, b) tend respectively to U^{(±j_{0})}(b) uniformly with respect to b ≥ 2.

Proof. Recall that

u_{m}^{(±j_{0})}(a, b) = a^{−m} sin(±πj_{0}/b^{m})⟶_{a→1} ± sin(πj_{0}/b^{m}),

so it is sufficient to show that the series

∑_{m=1}^{∞} u_{m}^{(±j_{0})}(a, b)e^{±πij_{0}/b^{m}}

are convergent uniformly with respect to b ≥ 2 and a ∈ [a_{1}, 1) for some a_{1} < 1. To check this, it is enough to notice that by (3), we have

|u_{m}^{(±j_{0})}(a, b)e^{±πij_{0}/b^{m}}| = a^{−m} sin(\frac{πj_{0}}{b^{m}} ) ≤ \frac{π}{2a}(ab)^{1−m} ≤ π(\frac{2}{3})^{m}

for every a ∈ [\frac{3}{4}, 1).

The proof of Theorem 1.1 is based on the following elementary planar geometric property.

Lemma 3.3. Let A, B, C be three non-collinear points in the plane. Then there exist a point P in the interior of the triangle ABC and constants ε, c > 0, such that for every q < 1 sufficiently close to 1 there exists r > c/(1 − q) such that

D_{r}(P) ⊂ D_{qr}(Ã) ∪ D_{qr}(B̃) ∪ D_{qr}(C̃)

for every Ã ∈ D_{ε}(A), B̃ ∈ D_{ε}(B), C̃ ∈ D_{ε}(C).

Proof. Let P be the unique point in the interior of the triangle ABC, such that ∡AP B = ∡BP C = ∡CP A = \frac{2}{3}π. For Z = A, B, C denote by S_{Z} the closed angle of measure \frac{2}{3}π and vertex P, symmetric with respect to the line P Z and containing Z. Then

(4)  D_{r}(P) = (D_{r}(P) ∩ S_{A}) ∪ (D_{r}(P) ∩ S_{B}) ∪ (D_{r}(P) ∩ S_{C}).

Take r > |\overline{AP} |. Let Q, Q′ be the two points in ∂D_{r}(P), such that ∡AP Q = ∡AP Q′ = \frac{1}{3}π and let R be the point of intersection of the line AP with ∂D_{r}(P)∩ S_{A} (see Figure 2).

Then

(5)  max\{dist(A, Z) : Z ∈ ∂(D_{r}(P) ∩ S_{A})\} = |\overline{AQ}|.

To see this, observe that max\{dist(A, Z) : Z ∈ \overline{P Q}\} is achieved for Z ∈ {P, Q}. Moreover, it is easy to check that dist(A, Z) decreases as Z goes along

Figure 2. The set D_{r}(P) ∩ S_{A}.

∂D_{r}(P) ∩ S_{A} from Q to R. Since ∡P QA < \frac{1}{3}π = ∡AP Q, we have |\overline{AQ}| > |\overline{AP} |. This shows (5). By (5) and the triangle inequality, if

(6)  qr > |\overline{AQ}| + ε,

then

D_{r}(P) ∩ S_{A} ⊂ D_{qr}(Ã)

for every Ã ∈ D_{ε}(A). Since

|\overline{AQ}| = \sqrt{r^{2} − |\overline{AP} |r + |\overline{AP} |^{2}} ,

the condition (6) is equivalent to

(7)  (1 − q^{2})r^{2} − (|\overline{AP} | − 2εq)r + |\overline{AP} |^{2} − ε^{2} < 0.

Solving the quadratic inequality, it is easy to check that if ε > 0 is sufficiently small and q is sufficiently close to 1, then (7) holds for r ∈ [c′_{A}/(1 − q), c_{A}/(1 − q)], where c_{A} > 0 depends only on |\overline{AP} | and c′_{A} > 0 is arbitrarily small if ε and 1 − q are sufficiently small. Replacing A by B and C and repeating the above arguments, we obtain by (4)

D_{r}(P) ⊂ D_{qr}(Ã) ∪ D_{qr}(B̃) ∪ D_{qr}(C̃)

for every Ã ∈ D_{ε}(A), B̃ ∈ D_{ε}(B), C̃ ∈ D_{ε}(C) and

r ∈ [max(c′_{A}, c′_{B}, c′_{C})/(1 − q), min(c_{A}, c_{B}, c_{C})/(1 − q)]

(if ε is sufficiently small and q is sufficiently close to 1). Hence, the lemma holds for c = min(c_{A}, c_{B}, c_{C})/2 and r = 2c/(1 − q).

Now we can prove the main lemma which is used in the proof of Theorem 1.1.

Lemma 3.4. There exist a_{0} < 1, n_{0} > 0 and c > 0, such that for every a ∈ [a_{0}, 1) and every b ∈ N, b ≥ 2 there exist z_{0} ∈ C and ϱ > c/(1 − a), such that for every n ≥ 0 and k ∈ {1, …, b^{n}},

D_{ϱa^{n}}(F_{a,b}(z_{n,k}) + z_{0}a^{n}) ⊂ ⋃_{l=b^{n_{0}} (k−1)}^{b^{n_{0}} (k+1)} D_{ϱa^{n+n_{0}}} (F_{a,b}(z_{n+n_{0},l})).

Proof. Let j_{0} be the number defined in the proof of Lemma 3.1. Take b ≥ 2 and define A, B, C ∈ C setting

A = 0,  B = U^{(j_{0})}(b),  C = U^{(−j_{0})}(b).

By Lemma 3.1, the points A, B, C are not collinear. For a < 1, n ≥ 0, k ∈ {1, …, b^{n}} and n_{0} > 0 let

\begin{aligned}Ã & = 0, \\ B̃ & = a^{n_{0}}(U^{(j_{0})}(a, b) + Δ_{n,k,n_{0}}^{(j_{0})}(a, b)), \\ C̃ & = a^{n_{0}}(U^{(−j_{0})}(a, b) + Δ_{n,k,n_{0}}^{(−j_{0})}(a, b)). \end{aligned}

Take a small ε > 0. By (2) and (3) we obtain

a^{n_{0}}|Δ_{n,k,n_{0}}^{(±j_{0})}(a, b)| < π \frac{1}{a − 1/b}(ab)^{−n_{0}} ≤ 4π(\frac{2}{3})^{n_{0}}

for every a ∈ [\frac{3}{4}, 1), every b ≥ 2, every n_{0} and every k. Fix n_{0} such that

4π(\frac{2}{3})^{n_{0}} < \frac{1}{3}ε.

By Lemma 3.2, there exists \frac{3}{4} < ã_{0} < 1 such that for every a ∈ [ã_{0}, 1) and every b ≥ 2,

(8)  |a^{n_{0}}U^{(±j_{0})}(a, b) − U^{(±j_{0})}(b)| < \frac{1}{3}ε.

This implies |B̃ − B|, |C̃ − C| < ε. Apply Lemma 3.3 for the points A, B, C, Ã, B̃, C̃. By this lemma, there exist a point P ∈ C and constants c_{b} > 0, q_{0} < 1 (depending only on b), such that for every q ∈ [q_{0}, 1) there exists r > c_{b}/(1 − q) such that

(9)  D_{r}(P) ⊂ D_{qr}(Ã) ∪ D_{qr}(B̃) ∪ D_{qr}(C̃)

Take a_{0}(b) < 1 such that a_{0}(b) > ã_{0} and (a_{0}(b))^{n_{0}} > q_{0} . Let

z_{0} = 2iP.

Take a ∈ [a_{0}(b), 1) and q = a^{n_{0}} . Then q ≥ (a_{0}(b))^{n_{0}} > q_{0} , so by (9) and (1), we have

D_{r}(z_{0}/(2i)) ⊂ D_{ra^{n_{0}}} (0)

∪ D_{ra^{n_{0}}} (\frac{F_{a,b}(z_{n+n_{0},b^{n_{0}} k+j_{0}}) − F_{a,b}(z_{n,k})}{2ia^{n}})

∪ D_{ra^{n_{0}}} (\frac{F_{a,b}(z_{n+n_{0},b^{n_{0}} k−j_{0}}) − F_{a,b}(z_{n,k})}{2ia^{n}}),

where

(10)  r > \frac{c_{b}}{1 − a^{n_{0}}} > \frac{c_{b}}{n_{0}(1 − a)}.

Multiplying by 2ia^{n} and adding F_{a,b}(z_{n,k}) we obtain

(11)  \begin{aligned}D_{2ra^{n}}(F_{a,b}(z_{n,k}) + z_{0}a^{n}) & ⊂ D_{2ra^{n+n_{0}}} (F_{a,b}(z_{n,k})) \\ & ∪ D_{2ra^{n+n_{0}}} (F_{a,b}(z_{n+n_{0},b^{n_{0}} k+j_{0}})) \\ & ∪ D_{2ra^{n+n_{0}}} (F_{a,b}(z_{n+n_{0},b^{n_{0}} k−j_{0}})). \end{aligned}

In this way we have proved the lemma with a_{0} , c and r depending on b. To get the independence of b, let

A = 0,  B = U^{(j_{0})},  C = U^{(−j_{0})}

and define Ã, B̃, C̃, n_{0} and ã_{0} as previously. Note that by Lemma 3.1, for given ε > 0 there exists b_{0} such that for b > b_{0} we have

|U^{(±j_{0})}(b) − U^{(±j_{0})}| < \frac{1}{3}ε.

Using this together with (8), we have

|a^{n_{0}}U^{(±j_{0})}(a, b) − U^{(±j_{0})}| < \frac{2}{3}ε

for a ∈ [ã_{0}, 1) and b > b_{0} . Repeating the previous arguments, we show that there exist a_{0}(∞) < 1, c_{∞} > 0, z_{0} ∈ C and r > 0 such that

(12)  r > \frac{c_{∞}}{n_{0}(1 − a)}

and (11) holds for every a ∈ [a_{0}(∞), 1) and every b > b_{0} .

Define

a_{0} = max(a_{0}(2), …, a_{0}(b_{0}), a_{0}(∞)), c = \frac{min(c_{2}, …, c_{b_{0}}, c_{∞})}{2n_{0}}.

Take a ∈ [a_{0}, 1), b ≥ 2 and let ϱ = 2r for r from (11). Then by (10) and (12) we have ϱ > c/(1 − a) and

D_{ϱa^{n}}(F_{a,b}(z_{n,k}) + z_{0}a^{n}) ⊂ D_{ϱa^{n+n_{0}}} (F_{a,b}(z_{n,k}))
∪ D_{ϱa^{n+n_{0}}} (F_{a,b}(z_{n+n_{0},b^{n_{0}} k+j_{0}}))
∪ D_{ϱa^{n+n_{0}}} (F_{a,b}(z_{n+n_{0},b^{n_{0}} k−j_{0}})).

By (3), this implies

D_{ϱa^{n}}(F_{a,b}(z_{n,k}) + z_{0}a^{n}) ⊂ ⋃_{l=b^{n_{0}} (k−1)}^{b^{n_{0}} (k+1)} D_{ϱa^{n+n_{0}}} (F_{a,b}(z_{n+n_{0},l})).

Remark. In fact, the symmetry between B and C gives z_{0} ∈ R, z_{0} < 0. Proof of Theorem 1.1. Let

A_{n}(ϱ, p, q) = ⋃_{l=p}^{q} D_{ϱa^{n}}(F_{a,b}(z_{n,l})).

Take n ≥ 0, k ∈ {1, …, b^{n}} and m > 0. Applying Lemma 3.4 a number of times we obtain

D_{ϱa^{n}}(F_{a,b}(z_{n,k}) + z_{0}a^{n}) ⊂ A_{n+n_{0}}(ϱ, b^{n_{0}}k − b^{n_{0}}, b^{n_{0}}k + b^{n_{0}}),

D_{ϱa^{n}}(F_{a,b}(z_{n,k}) + z_{0}a^{n} + z_{0}a^{n+n_{0}})
⊂ A_{n+2n_{0}}(ϱ, b^{2n_{0}}k − b^{2n_{0}} − b^{n_{0}}, b^{2n_{0}}k + b^{2n_{0}} + b^{n_{0}}),
⋯

D_{ϱa^{n}}(F_{a,b}(z_{n,k}) + z_{0}a^{n} \frac{1 − a^{mn_{0}}}{1 − a^{n_{0}}} )

⊂ A_{n+mn_{0}}(ϱ, b^{mn_{0}}k − b^{n_{0}} \frac{b^{mn_{0}} − 1}{b^{n_{0}} − 1}, b^{mn_{0}}k + b^{n_{0}} \frac{b^{mn_{0}} − 1}{b^{n_{0}} − 1})

⊂ A_{n+mn_{0}}(ϱ, b^{mn_{0}}(k − 2), b^{mn_{0}}(k + 2))

for every a ∈ [a_{0}, 1), b ≥ 2 and suitable z_{0} ∈ C, ϱ > c/(1 − a). Hence,

D_{ϱa^{n}/2}(F_{a,b}(z_{n,k}) + \frac{z_{0}a^{n}}{1 − a^{n_{0}}} ) ⊂ ⋂_{m=m_{0}}^{∞} A_{n+mn_{0}}(ϱ, b^{mn_{0}}(k − 2), b^{mn_{0}}(k + 2))

for sufficiently large m_{0} . This means

D_{ϱa^{n}/2}(F_{a,b}(z_{n,k}) + \frac{z_{0}a^{n}}{1 − a^{n_{0}}} ) ⊂ ⋂_{m=m_{0}}^{∞} ⋃_{t∈[(k−2)/b^{n},(k+2)/b^{n}]} D_{ϱa^{mn_{0}}} (F_{a,b}(e^{2πit})).

By the compactness of F_{a,b}(S^{1}), we have

(13) D_{ϱa^{n}/2}(F_{a,b}(z_{n,k})+ \frac{z_{0}a^{n}}{1 − a^{n_{0}}} ) ⊂ F_{a,b}(\{e^{2πit} : t ∈ [(k −2)/b^{n}, (k +2)/b^{n}]\}),

which easily implies both parts of the theorem.

4. Corollaries

Corollary 4.1. For every a ∈ [a_{0}, 1) and every b ∈ N, b ≥ 2,

\underline{BD}(graph(X_{a,b}, Y_{a,b})) = \overline{BD}(graph(X_{a,b}, Y_{a,b})) = 3 − 2α.

Moreover, ℋ ^{2}(graph(X_{a,b}, Y_{a,b})) > 0, so HD(graph(X_{a,b}, Y_{a,b})) ≥ 2.

Proof. Consider the first part of the corollary. Since the map (X_{a,b}, Y_{a,b}) is Hölder continuous with exponent α, we have by Lemma 2.2,

\overline{BD}(graph(X_{a,b}, Y_{a,b})) ≤ 3 − 2α,

so it is sufficient to show the opposite inequality. Take ε > 0. Let n be the maximal number for which 2π/b^{n} > ε and let

I_{k} =[\frac{2π2k}{b^{n}} , \frac{2π(2k + 1)}{b^{n}}] for k = 0, …, [\frac{b^{n}}{2} − 1].

Then |I_{k}| > ε and dist(I_{k_{1}}, I_{k_{2}}) > ε for k_{1} ≠ k_{2} . By (13), the set (X_{a,b}, Y_{a,b})(I_{k}) contains a disc of diameter ca^{n} for a constant c > 0 independent of k, n. Hence, to cover graph(X_{a,b}, Y_{a,b})|_{I_{k}} we need at least c^{2}a^{2n}ε^{−2} balls of diameter ε with non-empty intersections with graph(X_{a,b}, Y_{a,b})|_{I_{k}}. Since for k_{1} ≠ k_{2} we have dist(I_{k_{1}}, I_{k_{2}}) > ε, such balls for k_{1} and k_{2} are disjoint. Hence,

N_{ε}(graph(X_{a,b}, Y_{a,b})) ≥ c_{1}a^{2n}ε^{−3} = c_{1}b^{−2αn}ε^{−3} ≥ c_{2}ε^{2α−3}

for some constants c_{1}, c_{2} > 0. This implies \underline{BD}(graph(X_{a,b}, Y_{a,b})) ≥ 3 − 2α.

To prove the second part, note that F_{a,b}(S^{1}) is the orthogonal projection of graph(X_{a,b}, Y_{a,b}). Since the projection is a Lipschitz map, it is enough to use Theorem 1.1 and Lemma 2.2 for β = 1.

The next corollary shows that for the functions X_{a,b}, Y_{a,b} we can improve the general estimates from Lemma 2.2.

Corollary 4.2. For every a ∈ [a_{0}, 1) and every b ∈ N, b ≥ 2 there exist U_{X_{a,b}}, U_{Y_{a,b}} ⊂ R, such that U_{X_{a,b}} (or U_{Y_{a,b}}) is open and dense in X_{a,b}([0, 2π]) (or Y_{a,b}([0, 2π])) and for every δ > 0,

if ℋ ^{δ}(A) > 0, then ℋ ^{α(δ+1)}(X_{a,b}^{−1}(A)) > 0 for every set A ⊂ U_{X_{a,b}},
if ℋ ^{δ}(A) > 0, then ℋ ^{α(δ+1)}(Y_{a,b}^{−1}(A)) > 0 for every set A ⊂ U_{Y_{a,b}}.

In particular,

HD(X_{a,b}^{−1}(A)) ≥ α(HD(A) + 1) for every set A ⊂ U_{X_{a,b}},
HD(Y_{a,b}^{−1}(A)) ≥ α(HD(A) + 1) for every set A ⊂ U_{Y_{a,b}}.

Proof. Let U_{X_{a,b}} (or U_{Y_{a,b}}) be the orthogonal projection of int F_{a,b}(S^{1}) on the real (or imaginary) axis. By Theorem 1.1, U_{X_{a,b}} (or U_{Y_{a,b}}) is open and dense in X_{a,b}([0, 2π]) (or Y_{a,b}([0, 2π])). Take A ⊂ U_{X_{a,b}} such that ℋ ^{δ}(A) > 0. By definition, for every s ∈ A the set

({s} × R) ∩ (X_{a,b}, Y_{a,b})([0, 2π])

contains a non-trivial interval. Hence, by Theorem 2.3, we have

ℋ ^{δ+1}((A × R) ∩ (X_{a,b}, Y_{a,b})([0, 2π])) > 0.

Since the map (X_{a,b}, Y_{a,b}) is Hölder with exponent α, we have by Lemma 2.2,

ℋ ^{α(δ+1)}((X_{a,b})^{−1}(A)) = ℋ ^{α(δ+1)}((X_{a,b}, Y_{a,b})^{−1}(A × R)) > 0.

The case A ⊂ U_{Y_{a,b}} is symmetric.

Taking A = {s} in Corollary 4.2, we obtain the following result on level sets L_{X_{a,b}}(s), L_{Y_{a,b}}(s).

Corollary 4.3. For every a ∈ [a_{0}, 1) and every b ∈ N, b ≥ 2,

ℋ ^{α}(L_{X_{a,b}}(s)) > 0 for every s ∈ U_{X_{a,b}},
ℋ ^{α}(L_{Y_{a,b}}(s)) > 0 for every s ∈ U_{Y_{a,b}}.

In particular,

HD(L_{X_{a,b}}(s)) ≥ α for every s ∈ U_{X_{a,b}},
HD(L_{Y_{a,b}}(s)) ≥ α for every s ∈ U_{Y_{a,b}}.

Moreover,

int_{R}(Y_{a,b}(L_{X_{a,b}}(s))) ≠ ∅ for every s ∈ U_{X_{a,b}},
int_{R}(X_{a,b}(L_{Y_{a,b}}(s))) ≠ ∅ for every s ∈ U_{Y_{a,b}}.

By Corollary 4.3 and Theorem 2.3, we get Corollary 4.4. For every a ∈ [a_{0}, 1) and every b ∈ N, b ≥ 2,

ℋ ^{1+α}(graph X_{a,b}), ℋ ^{1+α}(graph Y_{a,b}) > 0.

In particular, HD(graph X_{a,b}), HD(graph Y_{a,b}) ≥ 1 + α.

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Received 16 July 2001