Quantitative absolute continuity of planar measures with two independent Alberti representations

We study measures μ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mu $$\end{document} on the plane with two independent Alberti representations. It is known, due to Alberti, Csörnyei, and Preiss, that such measures are absolutely continuous with respect to Lebesgue measure. The purpose of this paper is to quantify the result of A–C–P. Assuming that the representations of μ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mu $$\end{document} are bounded from above, in a natural way to be defined in the introduction, we prove that μ∈L2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mu \in L^{2}$$\end{document}. If the representations are also bounded from below, we show that μ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mu $$\end{document} satisfies a reverse Hölder inequality with exponent 2, and is consequently in L2+ϵ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L^{2 + \epsilon }$$\end{document} by Gehring’s lemma. A substantial part of the paper is also devoted to showing that both results stated above are optimal.

for all Borel sets B ⊂ R d . Then, the formula ν(B) := ν ( ,P,γ ) (B) := H 1 (B ∩ γ (ω)) dP(ω) (1.5) makes sense for all Borel sets B ⊂ R d , and evidently ν(K ) < ∞ for all compact sets K ⊂ R d . We extend the definition to all sets A ⊂ R d via the usual procedure of setting first ν * (A) := inf{ν(B) : A ⊂ B Borel}. This process yields a Radon measure ν * which agrees with ν on Borel sets. In the sequel, we just write ν in place of ν * . If μ is another Radon measure on R d , we say that μ is representable by C-graphs if there is a triple ( , P, γ ) as above such that μ ν ( ,P,γ ) =: ν. In this case, the quadruple ( , P, γ, dμ dν ) is an Alberti representation of μ by C-graphs. The representation is . We also consider local versions of these properties: the representation is BoA (resp. BoB) on a Borel set B ⊂ R 2 if dμ dν ∈ L ∞ (B, ν) (resp. ( dμ dν ) −1 ∈ L ∞ (B, ν)). Two Alberti representations of μ by C 1 -and C 2 -graphs are independent, if the cones C 1 , C 2 are independent in the sense (1.2).
Representations of this kind first appeared in Alberti's paper [1] on the rank-1 theorem for BV -functions. It has been known for some time that planar measures with two independent Alberti representations are absolutely continuous with respect to Lebesgue measure; this fact is due to Alberti, Csörnyei, and Preiss, see [2,Proposition 8.6], but a closely related result is already contained in Alberti's original work, see [1,Lemma 3.3]. The argument in [2] is based on a decomposition result for null sets in the plane, [2,Theorem 3.1]. Inspecting the proof, the following statement can be easily deduced: if μ is a planar measure with two independent BoA representations, then μ ∈ L 2,∞ . The proof of [1,Lemma 3.3], however, seems to point towards μ ∈ L 2 , and the first statement of Theorem 1.6 below asserts that this is the case. Our argument is short and very elementary, see Sect. 2.1. The main work in the present paper concerns measures with two independent representations which are both BoA and BoB. In this case, Theorem 1.6 asserts an -improvement over the L 2 -integrability. Theorem 1. 6 Let μ be a Radon measure on R 2 with two independent Alberti representations. If both representations are BoA, then μ ∈ L 2 (R 2 ). If both representations are BoA and BoB on B (2), then there exists a constant C ≥ 1 such that μ satisfies the reverse Hölder inequality As a consequence, μ ∈ L 2+ (B( 1 2 )) for some > 0.

Sharpness of the main theorem
We now discuss the sharpness of Theorem 1.6. For illustrative purposes, we make one more definition. Let μ be a Radon measure on R 2 . We say that ( , P, γ, dμ dν ) is an axis-parallel representation of μ if = R, and γ : The following example shows that two independent BoA representations -even axis parallel ones-do not guarantee anything more than L 2 : Example 1.8 Fix r > 0 and consider the measure μ r = 1 r · 1 [0,r ] 2 . Note that μ r L p = r 2− p for p ≥ 1, so μ r ∈ L p uniformly in r > 0 if and only if p ≤ 2. On the other hand, consider the probability P := 1 r · L 1 | [0,r ] on = R, and the maps γ 1 := γ x and γ 2 := γ y , as above. Writing ν j := ν ( ,P,γ j ) for j ∈ {1, 2}, it is easy to check that So, μ r has two independent axis-parallel BoA representations with constants uniformly bounded in r > 0. After this, it is not difficult to produce a single measure μ with two independent axis-parallel BoA representations which is not in L p for any p > 2: simply place disjoint copies of c j μ r j along the diagonal {(x, y) : x = y}, where c j = 1 and r j → 0 rapidly. The situation where both representations are (locally) both BoA and BoB is more interesting. We start by recording the following simple proposition, which shows that Theorem 1.6 is far from sharp for axis-parallel representations: Proposition 1.9 Let μ be a finite Radon measure on R 2 with two independent axis-parallel representations, both of which are BoA and BoB on [0, 1) 2 . Then there exist constants 0 < c ≤ C < ∞, depending only on the BoA and BoB constants, such that We give the easy details in the "Appendix". In the light of the proposition, the following theorem is perhaps a little surprising: this shows that L 2+ -integrability claimed in Theorem 1.6 is sharp.

Notation 1.13
For A, B > 0, the notation A B will signify that there exists a constant C ≥ 1 such that A ≤ C B. This is typically used in a context where one or both of A, B are functions of some variable "x": then A(x) B(x) means that A(x) ≤ C B(x) for some constant C ≥ 1 independent of x. Sometimes it is worth emphasising that the constant C depends on some parameter " p", and we will signal this by writing A p B.

Higher dimensions, and connections to PDEs
The problems discussed above have natural-but harder-generalisations to higher dimen- With this definition in mind, one can discuss Radon measures on R d with d independent Alberti representations. It follows from the recent breakthrough work of De Philippis and Rindler [4] that such measures are absolutely continuous with respect to Lebesgue measure. It is tempting to ask for more quantitative statements, similar to the ones in Theorem 1.6. Such statement do not appear to easily follow from the strategy in [4].

Question 1
If μ is a Radon measure on R d with d independent BoA representations, then is μ ∈ L p for some p > 1?
In the case of independent axis-parallel representations, μ ∈ L d/(d−1) , see the next paragraph. This is the best exponent, as can be seen by a variant of Example 1.8. In general, we do not know how to prove μ ∈ L p for any p > 1. Some results of this nature will likely follow from work in progress recently announced by Csörnyei and Jones.
Question 1 is closely connected with the analogue of the multilinear Kakeya problem for thin neighbourhoods of C-graphs. A near-optimal result on this variant of the multilinear Kakeya problem is contained in the paper [6] of Guth, see [6,Theorem 7]. We discuss this connection explicitly in [3,Section 5]. It seems that the "S -factor" in [6, Theorem 7] makes it inapplicable to Question 1, and it does not even imply the qualitative absolute continuity of μ established in [4]. On the other hand, the analogue of [6,Theorem 7] without the S -factor would imply a positive answer to Question 1 with p = d/(d − 1), see the proof of [3, Lemma 5.2]. We do not know if this is a plausible strategy, but it certainly works for the axis-parallel case: the analogue of [6,Theorem 7] for neighbourhoods of axis-parallel lines is simply the classical Loomis-Whitney inequality (see [7] or [6, Theorem 3]), where no S -factor appears.
As mentioned above, the main results in this paper, and Question 1, are related to the recent work of De Philippis and Rindler [4] on A-free measures. Introducing the notation of [4] would be a long detour, but let us briefly explain some connections, assuming familiarity with the terminology of [4].
The qualitative absolute continuity result, mentioned above Question 1, follows from [4, Corollary 1.12] after realising that, for each Alberti representation of μ, (1.5) may be used to construct a normal 1-current (1.14) One may view the d-tuple of normal currents T = (T 1 , . . . , T d ) as an R d×d -valued measure T = T T , where | T| ≡ 1, and T is a finite positive measure. Since each T i is normal, div T is also a finite measure, and this is the key point relating our situation with the work of De Philippis and Rindler. If the Alberti representations of μ are BoA, then dμ/d T ∈ L ∞ ( T ), and μ ∈ L 2 (R 2 ) by Theorem 1.6. As far as we know, PDE methods do not yield the same conclusion. However, if in addition the Jacobian of T is uniformly bounded from below T almost everywhere, PDE methods look more promising. We formulate the following question, which is parallel to Question 1: Question 2 Let T = T T be a finite R d×d -valued measure, whose divergence is also a finite (signed) measure such that the Jacobian of T is uniformly bounded from below in absolute value T a.e. Is it true that T ∈ L p (R d ) for some p > 1?

Proof of the main theorem
We prove Theorem 1.6 in two parts, first considering representations which are only BoA, and then representations which are both BoA and BoB at the same time.

BoA representations
The first part of Theorem 1.6 easily follows from the next, more quantitative, statement: where the implicit constant only depends on the opening angles θ 1 , θ 2 and angular separation τ of the cones C 1 = C(e 1 , θ 1 ) and Proof It suffices to show that the restriction of μ to any dyadic square Q 0 ⊂ R 2 is in L 2 , with norm bounded (independently of Q 0 ) as in (2.2). For notational simplicity, we assume , n ∈ N, be the family of dyadic sub-squares of [0, 1) 2 of side-length 2 −n . Fix n ∈ N, pick Q ∈ D n , and write Note that {ω ∈ j : To prove (2.3), it suffices to fix a pair (γ 1 , γ 2 ) ∈ 1 × 2 , where j := C j , and show that there are τ 1 squares Q ∈ D n with γ 1 ∩ Q = ∅ = γ 2 ∩ Q. So, fix (γ 1 , γ 2 ) ∈ 1 × 2 , and assume that there is at least one square Q such that γ 1 ∩ Q = ∅ = γ 2 ∩ Q, see Fig. 1. To simplify some numerics, assume that Q = [0, 2 −n ) 2 . Pick x 1 ∈ γ 1 ∩ Q and x 2 ∈ γ 2 ∩ Q, and note that since γ 1 ∈ C 1 and γ 2 ∈ C 2 . It follows that whenever Q ∈ D n is another square with γ 1 ∩ Q = ∅ = γ 2 ∩ Q , we can find points But the independence assumption (1.2) implies that dist(y, C 1 ) τ |y| or dist(y, C 2 ) τ |y| for any y ∈ R 2 , and in particular the centre of Q . Hence, (2.4) shows that dist(Q , Q) ≤ dist(Q , 0) 2 −n , and (2.3) follows. Now, we can finish the proof of the theorem. Given any square Q ∈ D n , we note that Observe that Denoting the Lebesgue measure of Q by |Q|, and combining (2.5) with (2.3) gives This inequality shows that the L 2 -norms of the measures are uniformly bounded by the right hand side of (2.2). The proof can then be completed by standard weak convergence arguments.

Representations which are both BoA and BoB
Before finishing the proof of Theorem 1.6, we need to record a few geometric observations.
For the rest of the section, we assume that μ is a Radon measure on R 2 with μ(B(1)) > 0, and that μ has two independent Alberti representations which are both BoA and BoB on B (2). Thus, there exists a constant C ≥ 1 such that for all Borel sets A ⊂ B (2). By Theorem 2.1, we already know that μ ∈ L 2 (B(1)). We next aim to show that B(1) ⊂ spt μ, and μ is a doubling weight on B(1) in the following sense: 3 2 r )) C,τ μ (B(x, r )). (2.12) After this, it will be easy to complete the proof of the reverse Hölder inequality (1.7).

Lemma 2.13
Let μ be a measure as above. Then μ is doubling on B(1) in the sense of (2.12), where the constants only depend on C from (2.11) and τ from (1.2). In particular, B(1) ⊂ spt μ.

Proof of Lemma 2.13
Let 0 < < 1/10 be the parameter given by Corollary 2.7, applied with the angular separation constant τ > 0 of the cones C 1 , C 2 . It suffices to argue that , r )). Consequently, We observe that if ω ∈ j (B), then H 1 (G ∩ γ j (ω)) ∼ r . Indeed, if ω ∈ j (B), then certainly γ (ω) contains a point y ∈ B and then one half of the graph γ j (ω) is contained in y + C j . This half intersects A(y) in length ∼ r , and the intersection is contained in G by definition. It follows that Since G ⊂ B(x, r ), this yields (2.15) and completes the proof.
We can now complete the proof of the reverse Hölder inequality (1.7).
Concluding the proof of Theorem 1.6 Fix a ball B := B(x, r ) ⊂ B(1), and consider the restrictions of the measures P 1 , P 2 to the sets Writing P B j := (P j )| j (B) , the restriction μ B := μ| B has two independent Alberti represen- where C ≥ 1 is the constant from (2.11), so we may deduce from Theorem 2.1 that It remains to prove that since the reverse Hölder inequality (1.7) is equivalent to μ L 2 (B(x,r )) C,τ r −1 ·μ (B(x, r )).

Sharpness of the reverse Hölder exponent
The purpose of this section is to prove Theorem 1.10. The statement is repeated below: has two independent Alberti representations which are both BoA and BoB on [−1, 1] 2 .

Remark 3.3
It may be worth pointing out that, in the construction below, the BoA and BoB constants stay uniformly bounded for α ∈ (0, 1). However, the independence constant of the two representations (that is, the constant "τ " from (1.2)) tends to zero as α 1. In this section, the constants hidden in the "∼" and " " notation will not depend on α. and we let P = P j := H 1 | j . The main challenge is of course to construct the graphs γ j (ω), ω ∈ j . A key feature of f is that f (r , t) = f (t, r ) for (r , t) ∈ [−1, 1] 2 . Hence, as we will argue carefully later, it suffices to construct a single representation by C-graphs, where C is a cone around the y-axis, with opening angle strictly smaller than π/2; such a representation is depicted on the left hand side of Fig. 3. We remark that, as the picture suggests, every C-graph associated to the representation can be expressed as a countable union of line segments. The second representation is eventually acquired by rotating the first representation by π/2, see the right hand side of Fig. 4. Now we construct certain graphs γ (ω) for ω ∈ := [0, 1] × {1} ⊂ 1 . The idea is that eventually γ 1 (ω) ∩ [0, 1] 2 = γ (ω) for ω ∈ . The graphs γ (ω) will be constructed so that The right idea to keep in mind is that the graph γ (ω) "starts from ω ∈ = [0, 1]×{1}, travels downwards, and ends somewhere on [0, 1] × {0}". We will ensure that [0, 1] 2 is foliated by the graphs γ (ω), ω ∈ . Start by fixing a point p ∈ whose x-coordinate lies in (1/2, 1), see Fig. 3. The relationship between p and the exponent α in (3.2) will be specified under (3.6). Let We can now specify the graphs γ (ω) with ω ∈ I 1 . Each of them consists of two line segments: the first one connects I 1 to (( 1 2 , 1 2 ), (1, 1 2 )], and the second one is vertical, connecting (( 1 2 , 1 2 ), (1, 1 2 )] to [1/2, 1] × {0}, see Fig. 3. We also require that the graphs γ (ω) foliate the yellow pentagon R 0 in Fig.3. This description still gives some freedom on how to choose the first segments, but if the choice is done in a natural way, we will find that |{ω ∈ I 1 : γ (ω) ∩ B = ∅}| ∼ diam(B) (3.5) for all balls B ⊂ R 0 . Here, and in the sequel, |·| stands for 1-dimensional Hausdorff measure. The implicit constant of course depends on the length of I 1 (and hence p, and eventually α). We then move our attention to defining the graphs γ (ω) with ω ∈ I 0 . Look again at Fig. 3 and note the green trapezoidal regions, denoted by T j , j ≥ 0. To be precise, T 0 is the convex hull of I 0 ∪ [(0, 1 2 ), ( 1 2 , 1 2 )], and For j ≥ 0, we also define Then j is the bottom edge of the trapezoid T j−1 , and I j 0 is the top edge of the trapezoid T j for j ≥ 1. Also, We point out that 0 = , I 0 0 = I 0 and I 0 1 = I 1 . We then construct initial segments of the graphs γ (ω), ω ∈ I 0 as follows. Define the map σ 0 : I 0 → 1 by Note that as p varies in (( 1  2 , 1), (1, 1)), the number β( p) takes all values in ( 1 2 , 1). In particular, we may choose β( p) = 2 −α , where α ∈ (0, 1) is the exponent in (3.2). Now, we connect every ω ∈ I 0 to σ 0 (ω) by a line segment, see Fig. 3; this is an initial segment of γ (ω). We record that if I ⊂ 1 is any horizontal segment (or even a Borel set), then P{ω ∈ I 0 : γ (ω) ∩ I = ∅} = P{ω ∈ I 0 : σ 0 (ω) ∈ I } = β −1 · |I |. (3.7) Now, we have defined the intersections of the curves γ (ω) with T 0 ∪ R 0 . In particular, the following set families are well-defined: The graphs in (R 0 ) are already complete in the sense that they connect to [0, 1] × {0}. The graphs in (T 0 ) are evidently not complete, and they need to be extended. To do this, we define R j := 2 − j R 0 for j ≥ 1, see Fig. 3, and we define the set families for j ≥ 1. In other words, the sets in (R j ) are obtained by rescaling the graphs in (R 0 ) so they fit inside, and foliate, R j . We note that the sets in (R j ) connect points in I j 1 to Finally, we define the complete graphs γ (ω), ω ∈ I 0 as follows. Fix ω ∈ I 0 , and note that γ 0 := γ (ω) ∩ T 0 has already been defined, and the intersection γ 0 ∩ 1 contains a single point z = σ 0 (ω), which lies in either I 1 0 or I 1 1 . If z ∈ I 1 1 ⊂ R 1 , then there is a unique set γ 1 ∈ (R 1 ) with z ∈ γ 1 . Then we define In this case γ (ω) is now a complete graph, and the construction of γ (ω) terminates. Before proceeding with the case z ∈ I 1 0 , we pause for a moment to record a useful observation. If B ⊂ R 1 is a ball, consider 1 (B) := {x ∈ 1 : x ∈ γ and γ ∩ B = ∅ for some γ ∈ }.
Since all the graphs γ ∈ entering R 1 can be written as γ 0 ∪ γ 1 with γ 0 terminating at I 1 1 and γ 1 ∈ (R 1 ), the set 1 (B) can be rewritten as Then, recalling that I 1 1 = 2 −1 I 1 , and (R 1 ) = 2 −1 (R 0 ), and noting that 2B ⊂ R 0 , we see that using (3.5). The main point here is that the implicit constant is the same (absolute constant) as in (3.5). We remark that here 2B = {2x : x ∈ B} is the honest dilation of B (and not a ball with the same centre and twice the radius as B).
All the implicit constants can, again, be chosen independently of α ∈ (0, 1). Next, we fix k ≥ 0 and z ∈ int R k . Again, we choose r > 0 so small that B := B(z, r ) ⊂ R k , and we observe that (3.13) holds. The main task is again to find upper and lower bounds for P(B). Note that, by construction, every graph γ (ω) ∈ intersecting B ⊂ R k also intersects I k 1 ⊂ k (with the convention I 0 1 = I 1 and 0 = ). Hence, defining k (B) as before, in (3.14), we find that P(B) = P( k (B)) ∼ β −k · H 1 ({x ∈ k : x ∈ γ and γ ∩ B = ∅ for some γ ∈ }), using (3.11) in the last estimate. Combining this with (3.10), we find that P(B) ∼ β −k · diam(B). This implies (3.15) as before. Finally, we write z = (s, t), and observe that 2 −k+1 ≤ s ≤ 2 −k for all z ∈ R k , and also that f (s, t) ∼ s −α for all (s, t) ∈ R k (because s ≥ t/2). Consequently, for L 2 a.e. z ∈ k≥0 R k .