Restricted families of projections onto planes: The general case of nonvanishing geodesic curvature

It is shown that if $\gamma: [a,b] \to S^2$ is $C^3$ with $\det(\gamma, \gamma', \gamma'') \neq 0$, and if $A \subseteq \mathbb{R}^3$ is a Borel set, then $\dim \pi_{\theta} (A) \geq \min\left\{ 2,\dim A, \frac{ \dim A}{2} + \frac{3}{4} \right\}$ for a.e. $\theta \in [a,b]$, where $\pi_{\theta}$ denotes projection onto the orthogonal complement of $\gamma(\theta)$ and ``$\dim$'' refers to Hausdorff dimension. This partially resolves a conjecture of F\"assler and Orponen in the range $1<\dim A \leq 3/2$, which was previously known only for non-great circles. For $3/2<\dim A<5/2$ this improves the known lower bound for this problem.


Let dim
Denote the s-dimensional Hausdorff measure in Euclidean space by H s .In R 2 the classical Marstrand projection theorem [11] states that if A ⊆ R 2 is a Borel set and P e denotes orthogonal projection onto the 1-dimensional subspace through e ∈ S 1 , then for dim A ≤ 1, dim P e (A) = dim A, H 1 -a.e. e ∈ S 1 , and for dim A > 1, H 1 (P e (A)) > 0, H 1 -a.e. e ∈ S 1 .
This was generalised to higher dimensions by Mattila.In R 3 there are two versions of the Martrand-Mattila projection theorem; one for lines and one for planes.The version for lines is analogous to the above with S 1 replaced by S 2 .For planes, it states that if A ⊆ R 3 is a Borel set, then if dim A ≤ 2, (1.1) dim π v (A) = dim A, H 2 -a.e.v ∈ S 2 , and if dim A > 2, then where π v denotes projection onto v ⊥ .Restricted projection families can be formed by constraining v to move along a one-dimensional curve γ : [a, b] → S 2 , and the restricted projection problem asks whether (1.1) still holds with a natural 1dimensional measure replacing the surface measure H 2 on S 2 .Without the assumption that det(γ, γ ′ , γ ′′ ) is nonvanishing, the equality can only hold (in general) for dim A ≤ 1, and was proved in this range by Järvenpää, Järvenpää, Ledrappier, and Leikas [6] using the energy method of Kaufman [8].
Considering the counterexample where γ is a great circle contained in a plane P and A ⊆ P shows that some extra assumption on γ is necessary for (1.2) to hold in general for 1 < dim A ≤ 2. The following conjecture is due to Fässler and Orponen; ρ θ denotes projection onto the 1-dimensional subspace through γ(θ).For projections onto planes, progress was made by Fässler-Orponen [3], Oberlin-Oberlin [13], Orponen [14], Orponen-Venieri [15], and by the author in [5].The condition dim A > 5/2 remains the best known sufficient condition that ensures H 2 (π θ (A)) > 0 for a.e.θ ∈ [a, b]; this is due to Oberlin-Oberlin [13].A comparison with some of the previous bounds is shown in Figure 1.
The improvement over [5] in Theorem 1.1 stems from Definition 4.1, the application of which is the main novelty of this work.Definition 4.1 reformulates the projection problem as an averaged inequality over collections of "bad" tubes.The proof of Theorem 1.1 follows a similar approach to [5] (which used ideas from [13,15,4,9]), and proceeds by splitting an integral into a "good" and "bad" part.A self-similarity in the proof allows the "bad" part to be bounded by re-using Definition 4.1, which circumvents the appeal to the Orponen-Venieri lemma ([15, Lemma 2.3]).Since the use of this lemma was the only step in [5] specific to nongreat circles, this allows the proof to be generalised to curves in S 2 of nonvanishing geodesic curvature.The bound on the "good" part uses the decoupling theorem for the cone in R 3 , from [1].
Section 4 contains the proof of Theorem 1.1.Section 2 is a proof of the refined Strichartz inequality, needed in Section 4. Section 3 contains a derivation of the wave packet decomposition needed in Section 4, and is independent of Section 2. Section 5 is a discussion of some related problems.The current and some of the previous a.e.lower bounds for dim π θ (A), in the range dim A ∈ (1, 5/2).The Orponen-Venieri theorem and the bound from [5] were for the special case of nongreat circles.
Acknowledgements.Alex Barron helped find the counterexample at the end of Section 5. I also thank Shaoming Guo for some comments on an earlier version of this article.

Refined Strichartz inequality
where B ≥ 1 is the smallest constant such that Given δ > 0 and τ ∈ P R −1 (Γ(γ)), let Then for any A ≥ 1 and ǫ > 0, there exists δ 0 > 0 such that the following holds for all 0 < δ < δ 0 .Let R ≥ 1 and suppose that T τ , and each T τ is an A-overlapping set of translates of T τ,0 intersecting B(0, R).Assume that for all T ∈ W, with f T 2 constant over T ∈ W up to a factor of 2. Let Y be a disjoint union of R 1/2 -balls in B(0, R), each of which intersects at most M sets 2T with T ∈ W.

and
(2.5) It follows that for each T ∈ T τ , there are ∼ 1 sets ∈ P κ(τ ) with T ∩ 10 = ∅, and moreover T ⊆ 100 whenever T ∩ 10 = ∅.For each such T let = (T ) ∈ P κ be some choice such that T ∩ 10 = ∅.For each τ , let and For each T ∈ W and S ∈ S τ (T ) , let where {η S } S∈Sτ is a smooth partition of unity, such that where the implicit constant is absolute, with η S supported in This partition can be constructed using the Poisson summation formula.By dyadic pigeonholing and by (2.3), there are dyadic numbers µ and ν such that , where, for each , W is a subset of (S, T ) : such that for any (S 0 , T 0 ) ∈ W , and with the property that The dyadic range of ν was constrained; relying on the tail term in (2.6) to handle the contribution from those f S,T with f S,T 2 ≤ R −2000 f T 2 .Hence for all and (S, T ) ∈ W , where the implicit constant is absolute.
For each κ and ∈ P κ , let {Q } Q be a finitely overlapping cover of 100 by translates of the ellipsoid Using Poisson summation again, let {η Q } Q ∈Q be a smooth partition of unity such that on 10 3 , and such that each η Q satisfies By dyadic pigeonholing, , where, for each , Y is a union over a subset of the sets Q , and η Y is the corresponding sum over η Q , such that each Q ⊆ Y intersects a number # ∈ [M ′ ( ), 2M ′ ( )) different sets 3S with (S, T ) ∈ W , up to a factor of 2. By pigeonholing again, , where |W | and M ′ = M ′ ( ) are constant over ∈ B up to a factor of 2. By one final pigeonholing step, , where Y ′ is a union over R This will be bounded using the inductive assumption, following a Lorentz rescaling.
The inequalities (2.4) and (2.5) imply that for each (S, T ) ∈ W , the set L −1 (S) is a equivalent (up to a factor 1.01) to a box of length R 1/2+δ/2 in its longest direction parallel to Moreover, it will be shown that To prove this, let where Moreover, (2.10) Combining (2.9), (2.10) and (2.11) gives (2.8).
Inductively applying the theorem at scale R 1/2 therefore gives The second bracketed term is 1, since By definition of M ′ and M ′′ , The inequality (2.12) above follows from the observation that if To prove this, recall that L −1 (S) is equivalent (up to a factor 1.01) to a box of length R 1/2+δ/2 in its longest direction and of length R 1/4+δ/2 in its other two directions.The sets

Wave packet decomposition
Throughout this section, assume that γ : and that To simplify notation, the convention N = {0, 1, 2, . . .} will be assumed. and The parameter J should be thought of as morally equal to 1, and is only used to exclude low frequency pieces.For any τ ∈ Λ and constant C ≥ 0, let Cτ be the box with the same centre as τ but with side lengths scaled by C.
, and and such that if Moreover, there exists a constant K, depending only on γ and ε, such that if k 1 , k 2 ≥ K and (3.9) holds, then be a point in the intersection, where each side satisfies the conditions in any of (3.5),(3.6),(3.7)or (3.8), multiplied by the factor 1.01.Then (3.12) by comparing norms on either side.By symmetry it may be assumed that and therefore k 2 ≤ 100 by (3.11) and the triangle inequality.This shows that all conclusions of the lemma hold if sgn λ 1 = sgn µ 1 (assuming that K ≥ 1000 and C ≥ 2 100 max{1, b − a}), so assume that sgn λ 1 = sgn µ 1 .By (3.11) and (3.12), together with (3.1), (3.13) and the assumption sgn λ 1 = sgn µ 1 , gives By (3.2), (3.14), the mean value theorem and the identity .
A similar argument gives that provided k 1 and k 2 are sufficiently large depending on ε and γ.
If k 1 = j 1 then the lemma follows, so assume that k 1 < j 1 .By the generalised mean value theorem and the assumption that where the rate of decay to zero in the error term is uniform in φ and θ.Hence Using (3.11), (3.15), letting θ = θ τ1 and φ = θ τ2 , and taking the dot product of both sides of (3.11) with γ (θ τ1 ), gives for some constant C γ which may depend on γ.Since |λ 3 | 2 j1−k1 and sgn λ 3 = − sgn λ 1 = − sgn µ 1 , and since the decay to zero in the error term is uniform, the left-hand side is 2 j1−k1 provided k 1 and k 2 are sufficiently large depending only on γ.This implies that k 2 k 1 , and therefore |k 1 − k 2 | ≤ C provided that C is sufficiently large (depending on γ).Lemma 3.3.There exist constants ̺ > 0, ε ∈ (0, 1] and J 1 ≥ 0 such that for all J ≥ J 1 , there is a partition of unity {ψ τ } τ ∈Λ J subordinate to the cover 1.01τ : τ ∈ Λ J of τ ∈Λ J,o τ, such that for each τ ∈ Λ J,o ∩ Λ j,k , the function ψ τ is smooth and satisfies for all l ∈ N, t ∈ R and x, v ∈ R 3 .
Given any δ > 0, for every τ ∈ Λ there exists a smooth partition of unity {η T } T ∈Tτ subordinate to the cover Proof.For τ ∈ Λ + , let and for τ ∈ Λ − , let where g 0 is a smooth function on R with 0 ≤ g 0 ≤ 1, g 0 (x) = 1 for x ≥ 1/1000 and g(x) = 0 for x ≤ 0. Choose J 1 large enough and ε small enough to ensure that if such a choice of J 1 and ε exists by the angle condition (3.10) in Lemma 3.2, and by (3.16).
By translating and rescaling a fixed bump function on the unit cube, for each τ ∈ Λ let f τ be a smooth bump function which is equal to 1 on τ , nonzero in the interior of 1.01τ , with f τ ≥ 1/100 on 1.009τ and with For τ ∈ Λ j,j ∩ Λ J , let Then τ ∈Λ J ψ τ (x) = 1 for x ∈ τ ∈Λ J τ , by the choice of J 1 and ε.
It will be shown that if the constant ̺ in (3.4) is small enough, and if J 1 is large enough, then for any τ ∈ Λ j,k ∩ Λ J,o with k < j, If τ ∈ Λ + (which can be assumed; the argument for τ ∈ Λ − being similar), this follows from the following argument.Given , and define k ′ by If J 1 is chosen sufficiently large and ε is sufficiently small, then (by (3.16)) the parameter k ′ is well-defined and satisfies |k − k ′ | 1, k ′ ≥ 0, and moreover Then (by (3.16)) if J 1 is sufficiently large and then ̺ is chosen sufficiently small (depending on ε): By letting τ ′ ∈ Λ j ′ ,k ′ ∩ Λ + be the box corresponding to the angle θ τ ′ , this proves (3.20).If k ′ ≥ j ′ , the above argument still works by taking θ τ ′ ∈ Θ j ′ instead.The covering property in (3.20) implies that the denominator in the definition of ψ τ in (3.18) is bounded away from zero on the support of the numerator, and therefore (by Lemma 3.2) ψ τ is smooth and satisfies the inequalities in (3.17) whenever τ ∈ Λ J,o ∩ Λ j,k with k < j.For the first part of the lemma, it remains to prove (3.17) in the case k = j.
For the case k = j, it will be shown that if the constant ̺ in (3.4) is small enough, and if J 1 is large enough, then for any τ ∈ Λ j,j ∩ Λ J,o , To see this, given x ∈ (1.01τ ) \ {x : x, γ(θ τ ) > 4.005}, write Then (by (3.16)) if J 1 is sufficiently large and then ̺ is chosen sufficiently small (depending on ε): In this case, let k ′ = j ′ − 1, and choose Then (by (3.16)) if J 1 is sufficiently large and then ̺ is chosen sufficiently small (depending on ε): In either case, by letting τ ′ ∈ Λ j ′ ,k ′ ∩ Λ + be the cap corresponding to angle θ τ ′ , this proves (3.21).This implies that the denominator in the definition of ψ τ in (3.19) is bounded away from zero on the support of the numerator, and therefore (by Lemma 3.2) ψ τ is smooth and satisfies (3.17) whenever τ ∈ Λ j,j ∩ Λ J,o .This proves the first part of the lemma.
The second part of the lemma is straightforward.
In the proof of the main theorem, only the behaviour of µ on tubes of radius at least 2 −j0 is considered, so there is no loss in convolving µ with the bump function above, and this (crucially) localises the frequencies to the ball of radius ≈ 2 j0 .
The precise exponent in the "bad" part µ b is defined as above in such a way that the average L 1 norm of the measures π θ# µ b can be controlled by re-using Definition 4.1, using the strategy below.
Proof of Lemma 3.5.Most of the lemma follows by defining µ b as in (3.23) and by defining µ g = µ * φ j0 − µ b ; the only nontrivial thing to check is that the sum in (3.23) converges in L ∞ .For this it suffices to show that for any T with τ (T ) ∈ Λ j,k and j ≥ j 0 , for any N ≥ 0. By Hausdorff-Young, the definition of M T , and the assumption that µ is finite, it suffices to show that for j ≥ 2j 0 , (3.24) By the Schwartz property of φ, where m(τ ) denotes the Lebesgue measure of τ .By replacing N with 3N , this gives (3.24) and proves the lemma.Lemma 3.6.Let δ > 0. Let ε, ̺ > 0 and J 1 ≥ 0 be parameters ensuring the existence of the partition of unity in Lemma 3.3.If J ≥ J 1 is sufficiently large, and if τ ∈ Λ j,k ∩ Λ J,o , then for any T ∈ T τ and any N ≥ 1.
Proof.If j > j 0 + J/10, the inequality follows by Cauchy-Schwarz, Plancherel and the Schwartz decay of φ.Assume then that j ≤ j 0 + J/10.By the definition of M T (µ * φ j0 ), where τ is the "dual" box to τ centred at the origin; with axes parallel to τ but reciprocal side lengths.The first integral is 2 3jδ (µ * φ j0 )(1.5T ), which is smaller than 2 3jδ µ(2T ) since k > J and j ≤ j 0 + J/10.The second integral is N 2 −jN (µ * φ j0 )(1.5T ) by Lemma 3.3 and repeated integration by parts.Similarly, by Lemma 3.3 and repeated integration by parts, the third integral is if N ′ is chosen large enough.This proves the lemma.
Lemma 3.7.Let δ > 0. Let ε, ̺ > 0 and J 1 ≥ 0 be parameters ensuring the existence of the partition of unity in Lemma 3.3, and let J ≥ J 1 .Then there exists Proof.By identifying the complex measure π θ# M T f with its Radon-Nikodym derivative with respect to H 2 , By repeated integration by parts, it suffices to show that for all t ∈ R, l ≥ 1, ξ ∈ τ and The assumed lower bound (3.25) on ε, together with (3.16) and the assumption that γ is C 2 with det(γ, γ ′ , γ ′′ ) nonvanishing, yields provided k is sufficiently large.Hence to prove (3.26) it suffices to show that By Lemma 3.3 and (3.16), where the last line follows from the assumed lower bound (3.25) on ε.

Proof of the main theorem
for all Borel measures µ on the unit ball with c α (µ) ≤ 1, for any R ≥ 1, and for any collection of sets {D θ : θ ∈ [a, b]} such that the integrand of (4.1) is measurable, where, for each θ The proof of the main theorem will be broken up into several separate lemmas.Lemma 4.2 deals with the contribution from the "bad" part of the measure, whilst Lemmas 4.3, 4.4, 4.5, 4.6 and 4.7 deal with the "good" part.Lemma 4.8 converts everything into a lower bound for α 0 in Definition 4.1, which is then used to obtain the main theorem.
)), then there exists δ ′ > 0 such that for any δ ∈ (0, δ ′ ], there is a positive integer J 0 such that for all j 0 ≥ J ≥ J 0 with J ∈ [(j 0 ǫ)/1000, 1000j 0 ǫ] and for all Borel measures µ on the unit ball with Proof.Let δ ′′ = δ ′′ (α, α ′ 0 ) ∈ (0, 1/100) be an exponent that works in (4.1) with α * replaced by α ′ 0 + 100δ ′ and with A = 1, for some positive δ ′ , and after taking δ ′ smaller if necessary assume that δ ′ ≤ (δ ′′ ǫ 2 )/100.Let δ ∈ (0, δ ′ ] be given.Let j 0 and J be such that j 0 ≥ J ≥ J 0 , where J 0 is implicity chosen sufficiently large (depending on δ) so that the argument below holds.By Lemma 3.5, By Lemma 3.7, the contribution from (4.4) is If J 0 is sufficiently large, then by Lemma 3.6 the contribution from (4.3) is where For fixed j and k, let {B l } l be a finitely overlapping cover of B 3 (0, 1) by balls of radius 2 −(j−k) .For each θ and l let Let µ j,k be the pushforward of µ under x → 2 j−k−2kδ x.Then where with b l the centre of B l .Up to translation and finite overlaps, B ′ j,k,l and µ j,k,l satisfy the conditions of Definition 4.1; for each θ the set 2 j−k−2kδ B j,k,l (θ) is contained in a union of tubes of radius 2 −k/2 parallel to γ(θ), with the number of tubes 2 k(α ′ 0 +100δ ′ )/2 µ j,k,l (R 3 ), such that each tube overlaps 2 10kδ of the others.Moreover c α ( µ j,k,l ) ≤ 1 and µ j,k,l is supported in a ball of radius 1. Hence , Putting this into (4.6)yields Substituting this into (4. provided J 0 is sufficiently large.This proves the lemma.Then there exists a constant σ > 0 depending on γ, and for each ǫ ∈ (0, 1) an integer J 0 ≥ 0 depending on γ and ǫ, such that for any τ ∈ Λ J ∩ Λ j,k with J ≥ J 0 and k ∈ [jǫ, j], Proof.If either |η| < 2 j−10 or |η| > 2 j+10 this is immediate, so it may be assumed that Then there exists a positive integer J 0 such that for all j 0 ≥ J ≥ J 0 and for all finite Borel measures µ on the unit ball, b a where µ g = µ g,j0,J,α,ǫ,δ,α ′ 0 is defined by Lemma 3.5.Proof.This follows from the triangle inequality, Lemma 4.3, the definition of µ b , and the rapid decay of F (M T (µ * φ j0 )) outside 1.01τ (T ).
A similar inequality will be shown in the case k = j.By the wave packet decomposition and Lemma 3.2, (4.9) Let µ be defined by setting µ equal to the function inside the modulus signs above.By the finite overlapping property of the sets τ , where the sets T ′ cover T with planks of dimensions ≈ 1 × 2 −j/2 × 2 −j , with long direction parallel to γ(θ τ (T ) ), medium direction parallel to γ ′ (θ τ (T ) ) and short direction parallel to (γ × γ ′ ) (θ τ (T ) ), and M T ′ µ = η T ′ M T µ, where {η T ′ } T ′ is a smooth partition of unity subordinate to the cover {T ′ } T ′ .By the 2-dimensional Plancherel theorem followed by the uncertainty principle (bounding the L 2 norm by the L ∞ norm, followed by Hausdorff-Young and Cauchy-Schwarz), This shows that (4.11)-(4.12)holds also in the case k = j, although possibly with a 2 10jδ loss, and with {B m } m equal to the cover of B(0, 1) by the single ball B(0, 10) in that case.The remainder of the proof will therefore cover both cases simultaneously (k ≤ j).
Applying Plancherel to the non-negligible term in the right hand side of (4.12) gives (4.14) Let ν be the restriction of µ * φ j0 to 2 10kδ B m .By Cauchy-Schwarz, The integral in (4.15) satisfies 1+2 jN |x| N for some very large N .This follows from the uncertainty principle since f is supported in a ball of radius 2 j .By dyadic pigeonholing, there is a subset such that f T 2 is constant up to a factor of 2 as T varies over W, and .
By pigeonholing again and by Hölder's inequality, there is a disjoint union Y of balls Q of radius 2 −j+k/2 , such that (4.16) , and such that each Q ⊆ Y intersects a number # ∈ [M, 2M ) boxes 3T as T varies over W, for some dyadic number M .By rescaling and then applying the refined Strichartz inequality (Theorem 2.2) with p = 6, the first factor in (4.16) satisfies For the second factor in (4.16), the assumed inequality By Plancherel, f T 2 ≤ M T (µ * φ j0 ) 2 for every T .Assuming the tail terms are not dominant, substituting into (4.17) and then into (4.14)yields .
By cancelling the common factors, this yields .
By substituting back into (4.11)-(4.12)and summing over m, this proves the lemma (the m for which the tail terms dominate make a negligible contribution to the sum).
) , then there exists δ 0 > 0, and for any δ ∈ (0, δ 0 ] a J 0 ≥ 0, such that for all j 0 ≥ 3J and J ≥ J 0 with J ∈ [(j 0 ǫ)/1000, 1000j 0 ǫ] and for all Borel measures µ on the unit ball with c α (µ) ≤ 1, follows straightforwardly from the rapid decay of φ j0 outside B 0, 2 j0 ; see [5, pp. 13-14] for a more detailed calculation of a similar inequality.The other term in (4.19) can be written as By Lemma 4.4, the first term satisfies The change of variables .
The last part of the lemma follows directly from the first part, and from Definition 4.1 which defines α 0 .
1 Issues of measurability will be ignored since they can be easily adjusted for.
Figure1.The current and some of the previous a.e.lower bounds for dim π θ (A), in the range dim A ∈ (1, 5/2).The Orponen-Venieri theorem and the bound from[5] were for the special case of nongreat circles.