Regularity and modulus of continuity of space-filling curves

We study critical regularity assumptions on space-filling curves that possess certain modulus of continuity. The bounds we obtain are essentially sharp, as demonstrated by an example.


Introduction
In 1890, G. Peano [15] gave the first example of a curve whose range contains the entire unit square. More precisely, he constructed a continuous mapping g : [0, 1] → R 2 such that [0, 1] 2 ⊂ g([0, 1]). Mappings like this are now customarily called Peano curves. Peano's construction allowed also for space-filling: there exists a continuous mapping g : [0, 1] → R n , where n ∈ N, n ≥ 3, such that [0, 1] n ⊂ g ([0, 1]). In what follows, we call such a g a space-filling curve. Once these mappings are known to exist, it is natural to ask how regular they can be. Regularity properties of space-filling curves can be measured in terms of continuity and total oscillation (energy). Despite the age of the topic, the research is still active; for some recent advances, see [3], [6], [17], [18], [19]. In this paper, we consider the interplay of modulus of continuity and energy conditions, and our results are not covered by previous results.
For technical reasons, let us consider the case of closed curves g : S 1 → R n , n ≥ 2, parameterized by the unit circle S 1 . Regarding continuity, a simple dimension estimate shows that a Peano curve cannot be Hölder continuous with any exponent strictly greater that 1/2, and similarly with 1/n in the case of a spacefilling curve. Hölder continuity exponents 1/2 and 1/n are actually possible; see, e.g., [1,6,17]. Moreover, it is possible to construct an almost everywhere differentiable Peano (space-filling) curve with Hölder continuity exponent arbitrarily close to 1/2 (1/n) (see [16,Chapter 5] for one such construction).
One is then motivated to ask whether a suitable energy bound would rule out space filling. Indeed, if the component functions of g : S 1 → R n are absolutely continuous, then g(S 1 ) necessarily has finite length; hence it cannot cover a square or a cube. Thus, the case of usual Sobolev regularity is trivial, and one is led to consider fractional derivatives. The component functions of g : S 1 → R 2 belong to the fractional Sobolev space W 1 2 ,2 (S 1 ) exactly when the Poisson extension Pg : B(0, 1) → R 2 belongs to the Sobolev space W 1,2 (B(0, 1); R 2 ). Such Peano curves do exist [12,Section 5], but according to [12,Theorem D], Pg cannot be Hölder continuous (hence neither can g) under this regularity of Pg. Thus, Hölder continuity and finiteness of a suitable energy prevent g from covering a square; they actually force g(S 1 ) to be of area zero. The energy in question is the one corresponding to the Sobolev space W 1 2 ,2 (S 1 ). From the above considerations, we see that neither Hölder continuity nor finiteness of W 1 2 ,2 (S 1 )-energy is sufficient to rule out square-filling, but together they are sufficient. Let us analyze the energy assumption in more detail. Given g : S 1 → R 2 , this energy is essentially For our purposes, consider a dyadic version of this energy, given by Here, {I i, j : i ∈ N, j = 1, . . . , 2 i } is a dyadic decomposition of S 1 such that for a fixed i ∈ N, {I i, j : j = 1, . . . , 2 i } is a family of arcs of length 2π/2 i with j I i, j = S 1 . The next generation is constructed in such a way that for each j ∈ {1, . . . , 2 i+1 }, there exists a unique k ∈ {1, . . . , 2 i } satisfying I i+1, j ⊂ I i,k . We denote this parent of I i+1, j by U i+1, j , and U 1, j = S 1 for j = 1, 2. By g A , A ⊂ S 1 , we denote the mean value g A = − A g dH 1 = 1 H 1 (A) A g dH 1 . One can check that finiteness of the former energy implies the finiteness of the latter (see Theorem 3.1).
If g : S 1 → R 2 is a 1/2-Hölder continuous Peano curve, the inner sum in (1) is uniformly bounded from above. Additionally, one can rather easily obtain a positive lower bound for the inner sum in the case of a certain almost everywhere differentiable Peano curve with Hölder exponent arbitrarily close to 1/2. Thus the energy (1) does not appear critical for Hölder continuous maps. Let us consider the energy for λ ∈ R. The reader familiar with trace theorems for Sobolev spaces should consider our norm as a version of the trace norm of an Orlicz-Sobolev space corresponding to the Orlicz function ψ(t) = t 2 log λ (e + t). The value λ = −1 is critical for Hölder continuous curves by the discussion above. Our first result implies that 1/2-Hölder continuous Peano curves have finite energy E(g; 2, λ) exactly when λ < −1.
In order to deal with space-filling curves g : S 1 → R n , we modify the above definition by setting , n] and λ ∈ R. The following theorem is the space-filling analogue of Theorem 1.1.
Let us return to Peano curves in W 1 2 ,2 (S 1 ). The energy that corresponds to E(g; 2, 0) in our notation is, in principle, much larger than E(g; 2, −1). Hence one expects an optimal modulus of continuity weaker than Hölder continuity. Indeed, in [7], the optimal modulus of continuity of general mappings f ∈ W 1,2 (B(0, 1), R n ) which guarantees that the 2-dimensional Hausdorff measure of f (S) be zero has been recognized to be where C 0 , C 1 > 0 are constants. To simplify our terminology, from now on, we use the term "space-filling" even for Peano curves. Our next results deal with general energies E(g; n, λ). Theorem 1.3. Let g : S 1 → R n be a space-filling curve satisfying with some α ∈]0, 1] and C 0 , C 1 > 0, whenever x, y ∈ S 1 are close enough (|x − y| < 0 for some 0 > 0). Then E(g; n, λ) = ∞ for λ = n − 1 − αn if α ≥ n−1 n , and for λ > n − 1 − αn if α < n−1 n . Note that, when α = 1, g is assumed to be Hölder continuous. Theorem 1.4. Let g : S 1 → R n be a space-filling curve satisfying Note that by Theorem 3.1, the conclusion E(g; ν, λ) = ∞ in Theorems 1.1-1.4 implies a weaker conclusion that the integral in (7) below is infinite. Also, E(g; n, λ) = ∞ for any space-filling curve g : S 1 → R n , when λ > n − 1. This follows from the argument in [11].
In the process of verifying the above results, we actually prove a stronger statement, related to Hausdorff measures. It is given by the following theorem.
Notice that ν is not necessarily an integer. Theorems 1.1-1.5 are essentially sharp; see Sections 3 and 4.
The proof of Theorem 1.5 is based on the following idea. We consider the balls B(g i, j , r i, j ) with where j 1 and j 2 are such that I i, j = U i+1, j l , l = 1, 2. Our energy assumption gives us control over a weighted double sum of r ν i, j . Even though these balls cannot necessarily be used to cover g(S 1 ), the given modulus of continuity allows us to construct a desired cover via this sequence of balls. For related arguments in the special case of quasiconformal mappings, see [9,10]; also see [7,11] for the setting of Sobolev mappings. This paper is organized as follows. In Section 2, we introduce our basic notation and give the proof of Theorem 1.5. We discuss the connection between our energy E(g; ν, λ) and the integral energy in Section 3. Finally, in Section 4, we construct an example which shows the essential sharpness of our results.
2 Proof of Theorem 1.5 First, we introduce our basic notation. Given a > 0, we write a for the largest integer less or equal to a. If A is an arbitrary set, by A we mean the cardinality of A. For A ⊂ R n , we denote by diam(A) the diameter of A and by χ A the characteristic function of A. Given x ∈ R n and r ≥ 0, we denote by B(x, r) the open ball centered at x and having radius r; r(B) stands for the radius of the ball B. We write H ν δ (A) with ν > 0 and 0 < δ ≤ ∞ for the ν-dimensional Hausdorff content of a set A, while H ν (A) denotes its ν-dimensional Hausdorff measure. When we write L = L(·), we mean that the number L > 0 depends on the parameters listed in the parentheses. Finally, C denotes a positive constant which may depend on data and differ from occurrence to occurrence.
In addition to the standard Hausdorff content, we need a weighted Hausdorff content of a set A ⊂ R n . This is given by Before starting the proof, we state two simple lemmas concerning sequences.
Lemma 2.1. Let {a i } i∈N be a non-decreasing sequence of real numbers such that 1 ≤ a i ≤ C 0 i β for each i ≥ i 0 and some C 0 , β > 0, i 0 ∈ N. Then there exist constants c = c(β) > 0 and k 0 = k 0 (β, C 0 , i 0 ) ∈ N, such that for any k ≥ k 0 , there are at least 6k integers i in the set {k, . . . , 8k}, satisfying Proof. Let us fix an integer k ≥ i 0 and assume that there exist numbers for all of these j , provided that k > c log k. The monotonicity of the sequences On the other hand, log a 8k+1 ≤ log C 0 + β log(8k + 1), which is a contradiction when k ≥ k 0 for large enough fixed k 0 = k 0 (β, C 0 , i 0 ).

The proof of Lemma 2.2 is trivial.
Proof of Theorem 1.5. The proof is somewhat different for the cases λ ≤ 0, λ ∈]0, ν − 1[, and λ = ν − 1. We provide most of the details for all three cases. The origins of part of the proof for positive values of λ can be found in [4], while the proof for the case of negative λ adopts ideas of [7]. We may assume that the modulus of continuity (4) or (5) is satisfied globally; locality of the estimate only influences the choices of parameters. The proof is broken up into five steps.
Step 1. In this step, we build our fundamental construction. In the end of the step, we decompose S 1 into smaller sets, on which we estimate separately.
Let g i, j and r i, j be as defined in (6) for i = 1, 2, . . . and j = 1, . . . , 2 i . Since the energy E(g; ν, λ) is finite, For each pair of indices (i, j ) with r i, j > 0, we define a collection of weighted balls (8). Some balls in F may coincide; however, we treat them as if they were different. Pick x ∈ S 1 and a corresponding sequence of arcs as i → ∞. Thus, we may assume existence of at least one pair of indices (i, j ) with r i, j > 0; otherwise, g(S 1 ) would be a one-point set.
We neglect the set = x ∈ S 1 : Step 2. We fix l 1 ∈ N. The remaining steps are dedicated to the estimate for the set l 1 . In this step, for each point x ∈ l 1 we define sequences that characterize how g i (x) converges to g(x). We establish some important properties of these sequences, implied by the modulus of continuity, satisfied by g.
Let x ∈ l 1 . For each integer l ≥ l 1 , let a l (x) ∈ N be the smallest number Let l ≥ l 1 and i 0 = a l+1 (x). In the case λ = ν − 1, we have for l ≥ l 1 and a constant c ∈ N, which we specify later. We use ideas originating from [9,14] to prove that there exists an integer l ≥ 2l 1 such that for each l ≥ l . In other words, at least half of the annuli do not contain too many centers from {g i (x)} i∈N . Let us assume that (10) does not hold for some l ≥ 2l 1 . We obtain a lower bound for a l+1 (x), using the assumption that there exist at least l/2 − l 1 + 2 integers k ∈ {l 1 , . . . , l} such that θ k (x) = 0. We have which can not be true if c is fixed so that c > c 0 2 1/α /α, and l is large enough. In other words, there exists a number l = l (l 1 , α, C 0 , C 1 ) ∈ N such that (10) holds for all l ≥ l . For completeness, in the case λ = ν − 1, we put θ l (x) = 1 for each l ≥ l 1 .
Step 3. In Steps 3 and 4, we prove that for a substantial number of integers k, there exists a set of balls B ∈ F with radii in a suitable range and centers in B(g(x), 2 −k+1 ) such that r ν (B)w B ≥ const · 2 −νk k −1+δ for some δ ≥ 0. When λ = ν −1, we choose δ such that α = (ν −1 −λ)/(ν −2δ ); otherwise, we set δ = 1. Note that δ = 0 when λ ≤ 0. Without the loss of generality, we may assume that δ ≤ 1. In this Step 3, we define the desired collection of balls and provide the estimates for r ν (B)v B .
First, we concentrate on the case λ > 0. Let us take k ≥ l 1 with θ k (x) = 1. We find a set of balls M k (x) ⊂ F with radii in the range [ By the definitions of P k (x) and r i (x), we have Therefore, Then we define the collection M k (x) by We have for each B ∈ M k (x), which gives Finally, as desired. Note also that this argument works even if p k (x) = 0, because in that case the set P k (x) consists of one index i with g i (x) ∈ B(g(x), 2 −k ). The definition of r i, j implies that r i (x) > 2 −k .
Similarly, for non-positive λ, we take k ≥ l 1 with θ k (x) = 1. Starting with (12), we proceed with a similar proof with P k (x) replaced by ck 1−α α (note the case p k (x) = 0). This gives us a set of balls For completeness, for θ k (x) = 0, we put M k (x) = ∅.
Step 4. In this step, we complete the estimates announced in the beginning of the previous step by estimating the index i of each ball B i, j,q ∈ M k (x).
Finally, in the case λ = ν − 1, we apply Lemma 2.2. There exists k ≥ k 0 such that i > a k (x) ≥ C(a k+1 (x) − a k (x)) = Cp k (x) whenever i ∈ P k (x). Combining this with (11), we obtain for this k (note p k (x) = 0 again).
Step 5. In this step, we complete the estimates for l 1 and the proof. We return to the case λ = ν − 1. For k ∈ {k 0 , . . . , 8k 0 }, let us define The previous argument implies 8k 0 k =k 0 χ g(S k ) ≥ k 0 χ G l 1 . We finish the proof for λ ∈]0, ν − 1[. The techniques used are those applied in [4] in cases "C" and "D". If k 1 , k 2 ∈ {k 0 , . . . , 8k 0 } are such that thus,M k 1 ∩M k 2 = ∅ by (13). This means that there exists a constant C , independent of k 0 , such thatM k 1 ∩M k 2 = ∅ for each k 1 , k 2 ∈ {k 0 , . . . , 8k 0 } with k 1 − k 2 ≥ C log k 0 ≥ 1, once k 0 is large enough. We prove next that there are not so many numbers k ∈ {k 0 , . . . , 8k 0 }, satisfying (16). Let us assume that there are at least k 0 /2 collectionsM k , k ∈ {k 0 , . . . , 8k 0 }, with The previous conclusion implies that there are at least k 0 /2 C log k 0 collectionsM k with the property (16) consisting of different balls each. Therefore, , which contradicts (9) when k 0 is large enough. Therefore, there are at most k 0 /2 − 1 collectionsM k , satisfying (16), which yields Let us pick k ∈ {k 0 , . . . , 8k 0 } with S k = ∅ such that (16) is not true. By the Besicovitch covering theorem [13, Theorem 2.7], we can cover g(S k ) with a finite collection of balls {B k,i = B(g(x k,i ), 2 −k+1 )} i∈I k with x k,i ∈ S k such that i χ B k,i (y) ≤ N for each y ∈ R n and some N = N (n) ∈ N. We observe that, by the definition of S k , where we have used the fact that all balls in M k (x k,i ) with i ∈ I k are centered in B k,i . Since k B k,i , k 0 /2 −1 : i ∈ I k is a weighted covering of the set G l 1 , we Finally, letting k 0 tend to infinity, we obtain H ν ∞ (G l 1 ) = 0. This implies H ν (G l 1 ) = 0 and H ν (g(S 1 \ )) = 0.
In the case of non-positive λ, the proof is more trivial. We simply notice that M k 1 ∩M k 2 = ∅ for each k 1 = k 2 . As in (17), using the Besicovitch covering theorem and the definition of S k , we find a covering {B k,i } i∈I k of the set g(S k ) such that We conclude that Finally, in the case λ = ν − 1, we apply the Besicovitch covering theorem to obtain a collection of balls {B i = B(g(x i ), 2 −k i +1 )} i∈I with x i ∈ l 1 and k i ≥ k 0 covering the set G l 1 such that i χ B i (y) ≤ N and (15) holds with x = x i and k = k i for each i ∈ I . We estimate

How to compute the energy
In this section, we use a fixed dyadic decomposition of S 1 again. We denote by D i the set of 2 i dyadic intervals of S 1 of generation i having length 2 1−i π. We denote the parent interval of I ∈ D i by U(I ) ∈ D i−1 . Recall that the parent interval of I is the dyadic interval of previous generation which contains I . Theorem 3.1. Let g : S 1 → R n , n ≥ 2, be continuous. Let λ ∈ R and ν > 1. If Proof. We may assume that |g(x)| ≤ 1 for all x ∈ S 1 . Let for each k ∈ N, where |·| S 1 is the length of the shortest arc of S 1 connecting x and y. In the following, where I is some dyadic interval and k ∈ N.
Here and in what follows, we use the convention that for any function ψ : S 1 × S 1 → R and all y ∈ S 1 .
for t > t 0 and y > ϕ(t 0 ), where t 0 is a suitably fixed number. In addition, we require thatφ(ϕ(t)) ≥ Ct for all t ≥ t 0 and some constant C > 0. Let I ∈ D i . Using these facts and Jensen's inequality, we obtain Raising the last term on the right hand side of (18) to the power of ν, multiplying it by i λ , and summing over all I ∈ D i and over all i ∈ N, we get some finite number. Now we concentrate on the first term on the right hand side of (18). For simplicity, we set We start with Hölder's inequality: From here on we split the argument into two cases, according to the value of λ. Assume first that λ ≤ 0. Since |g| ≤ 1 and ϕ(t) ≤ ϕ(t 0 ) + t ν log λ (e + t) , we have, for k ≥ 1, where C 1 depends on ϕ(t 0 ) and ν only. Therefore, We estimate now the contribution of the first term on the right hand side of (18) to the sum (2). Using estimates (18), (19), (20), and the definitions of I and A k (y), we obtain the estimate This finishes the proof in the case λ ≤ 0.
We may now assume that λ > 0. In order to estimate the logarithmic term from above, we define We start with the estimate (19) as in previous calculation to obtain the estimate To estimate P 1 , we may proceed in the same way as in the case λ ≤ 0. For P 2 , we use the definition of χ as follows: Therefore,

Example
This section is dedicated to the following example. Note that ν in it is not necessarily an integer.
There exists a mapping f : S 1 → R n with (21) and H ν ( f (S 1 )) > 0 which is continuous with modulus of continuity where C 0 > 0 and C 1 > 0 depend on the given parameters.
In the following subsections, we concentrate on the case λ = ν − 1. The remaining case is addressed in Subsection 4.9.

Construction.
The following construction is a modification of the example given in [8]. We define the map f : I → R n , where I ∈ R is an interval. It is easy to modify the construction to a map f : S 1 → R n .
Put R j = exp − j β and r j = 2 n exp −( j + 1) β . Let n 0 be a large fixed integer, to be determined later. We utilize the numbers R j and r j , with j ≥ n 0 in a Cantor-type construction used in defining the mapping. We start with an interval I n 0 of length 2r n 0 . Divide I n 0 into 2 n intervals of length 2R n 0 +1 . We denote these intervals by I i,n 0 +1 , where i = 1, 2, . . . , 2 n , and denote the corresponding midpoints of these intervals by a i,n 0 +1 . Define also intervals I i,n 0 +1 with the same centers and lengths 2r n 0 +1 . We choose n 0 large enough to ensure that R j > r j . We also use the notation A i,n 0 +1 : = I i,n 0 +1 \ I i,n 0 +1 . We continue inductively.
Assuming that for i = 1, 2, . . . , 2 n( j −n 0 ) we have defined the intervals I i, j of lengths 2r j , we divide each interval I i, j into 2 n intervals of length 2R j +1 . We denote these intervals by I i , j +1 and their midpoints by a i , j +1 , where i = 1, 2, . . . , 2 n( j +1−n 0 ) . Again, we choose the intervals I i , j +1 with midpoints a i , j +1 and lengths 2r j +1 , and define A i , j +1 : = I i , j +1 \ I i , j +1 .
Let d > 0 satisfy 2 n d ν = 1. For each j ∈ N, we define ϕ j : For the derivative of ϕ j , j ∈ N and j ≥ n 0 , we have the estimate for large j . Let e 1 , e 2 , . . . , e n be the standard basis of R n . Let v i , i = 1, 2, . . . , 2 n , be the elements of the set n k =1 a k e k : a k = ±1 in some fixed order. For each integer k > 2 n , we define v k = v k−2 n .
For each pair (i, j ) of indices, j = n 0 + 1, n 0 + 2, . . ., i = 1, 2, . . . , 2 n( j −n 0 ) , and each x ∈ I n 0 , we defineφ i j ( In addition, we set It is clear that the above sum converges at every point. In the following subsections, we prove (21) and establish the correct modulus of continuity. But first, we explain why H ν ( f (I n 0 )) > 0. More precisely, we show that f (I n 0 ) contains a Cantor set with positive ν-dimensional Hausdorff measure. Define Q i j , j = n 0 + 1, n 0 + 2, . . ., i = 1, . . . , 2 n( j −n 0 ) , to be the closed cube of side length d j centered at f (a i, j ), with sides parallel to the coordinate axes. The set [2,Theorem 8.6] or [13, 4.13]. Since f (I n 0 ) contains the centers of each Q i j , it must also contain the set C, by continuity.

Finiteness of energy.
Next, we show that for every x ∈ A i, j with fixed j ≥ n 0 and i ∈ {1, . . . , 2 n( j −n 0 ) }, the mapping f satisfies the estimate with some positive constants C, C n 0 , independent of x, i, and j . To prove this, we split the interval I n 0 into a finite number of pieces and show the estimate for each of these pieces separately. This is done in the following subsections. The estimate (24) then follows by summing all the obtained estimates. We first take care of the case λ ≥ 0 and then indicate the required changes necessary for the case λ < 0. Fix x ∈ A i, j . We may assume that a i, j = 0 and that x ≥ 0. We also choose n 0 large enough to guarantee r j < R j /2. Additionally, notice that Both of these estimates are elementary, and we use them without explicit mention. where γ = (2 n+1 − 1)/2 n+1 and ζ is the right endpoint of I n 0 . We assume that n 0 is so large that ϕ k+1 (R k+1 ) ≥ ϕ k (r k ) for all k = n 0 + 1, . . . . If x/2 < r j , we estimate, using (23),

Points close to
for each y ∈ [a, b] and large enough j . Otherwise, Therefore, as desired. For the remaining points y ∈ I n 0 \ [a, b], we have |x − y| > 2 −n−1 x.

Parent set and its descendants.
In this subsection, we integrate over the set contained in I i , j −1 , and of 2 n intervals I i , j and I i, j \ [a, b] of length 2r j , embraced by those A i , j and A i, j . In the case j = n 0 + 1, the only difference is the absence of the set A i , j −1 \ [a, b]. For a point y in any of these sets, we have Thus it remains to estimate the integral of the quotient | f (x) − f (y)| ν /|x −y| 2 . This is done differently for each of the mentioned sets.

Interval containing x. For y ∈
Furthermore, log |y| Combining this estimate with (26) and (27), we obtain the desired estimate.

Siblings of
Let us fix an index i such that both A i, j and A i , j are contained in the same interval I i , j −1 of generation j − 1. Then Note also that |x − y| ≥ R j /4 for each of these y. Hence, substituting t = x|y − a i , j |/R 2 j into each of the two intervals, we obtain The desired estimate follows from the last inequality, (26), and (28). Note that there are 2 n − 1 suitable i .

Children of A i, j .
We consider one of 2 n sets A i 0 , j +1 , such that Using the inequality |x − y| ≥ 2 −n−1 x for y / ∈ [a, b], we further estimate, with the substitution t = 2 n |y − a i 0 , j +1 |/x, Together with (26) and (29), this gives us (24).

Parent of A i, j .
We fix the unique i such that A i, j ⊂ I i , j −1 . Similarly to (29), interchanging the roles of y and x and shifting the index, we obtain , since |a i , j −1 | ≤ r j −1 and |y| ≥ R j . We also have y ≥ 3x/2, when y > 0. Therefore, log 2|y| The conclusion is as in the previous subsections.

Grandchildren of A i, j , nephews of A i, j and their descendants.
As before, we fix i 0 with A i 0 , j +1 ⊂ I i, j . Let us integrate over the set I i 0 , j +1 . Here we use the fast decay of the numbers r j . The estimate is the following: If j = n 0 + 1, there are no intervals of earlier generations; if j = n 0 + 2, the only intervals of earlier generations are the intervals of the generation n 0 + 1, which are treated in (34) and in Section 4.4.4. Using (31), we get the following estimate for all j ≥ n 0 + 3: Next, we show that for all large k, This is rather easy: 2 n exp β(k − 1) β−1 ≥ 2, when k is large enough. Now the sum on the right hand side of (32) is seen to be smaller than Besides, similarly to (30), we observe that R j R j−1 log ν R j r j ≤ C, which implies the desired estimate (33) On the other hand, on the set of points y such that | f (x) − f (y)| / |x − y| > R