1 Introduction

Let \(Ef=E_{\mathcal P} f\) be the extension operator associated with the unit paraboloid \({\mathcal P} = \{ \xi \in \mathbb R^n : \xi _n = \xi _1^2 + \ldots + \xi _{n-1}^2 \le 1 \}\) in \(\mathbb R^n\):

$$\begin{aligned} Ef(x) = \int _{\mathbb B^{n-1}} e^{-2 \pi i x \cdot (\omega , |\omega |^2)} f(\omega ) d\omega , \end{aligned}$$

where \(\mathbb B^{n-1}\) is the unit ball in \(\mathbb R^{n-1}\).

Our starting point is the following fractal restriction theorem of Du and Zhang [4]. (Throughout this paper, we denote a cube in \(\mathbb R^n\) of center x and side-length r by \(\widetilde{B}(x,r)\).)

Theorem1-A

(Du and Zhang [4, Corollary 1.6]) Suppose \(n \ge 2\), \(1 \le \alpha \le n\), \(R \ge 1\), \(X= \cup _k \widetilde{B}_k\) is a union of lattice unit cubes in \(\widetilde{B}(0,R) \subset \mathbb R^n\), and

$$\begin{aligned} \gamma = \sup \frac{\#\{ \widetilde{B}_k : \widetilde{B}_k \subset \widetilde{B}(x',r) \}}{r^\alpha }, \end{aligned}$$

where the sup is taken over all pairs \((x',r) \in \mathbb R^n \times [1,\infty )\) satisfying \(\widetilde{B}(x',r) \subset \widetilde{B}(0,R)\). Then to every \(\epsilon > 0\) there is a constant \(C_\epsilon \) such that

$$\begin{aligned} \int _X |Ef(x)|^2 dx \le C_\epsilon R^\epsilon \, \gamma ^{2/n} R^{\alpha /n} \Vert f \Vert _{L^2(\mathbb B^{n-1})}^2 \end{aligned}$$
(1)

for all \(f \in L^2(\mathbb B^{n-1})\).

In [4], Theorem 1 was used to derive the sharp \(L^2\) estimate on the Schrödinger maximal function (see [4, Theorem 1.3] and the paragraph following the statement of [4, Corollary 1.6]). The authors of [4] also used Theorem 1 to obtain new results on the Hausdorff dimension of the sets where Schrödinger solutions diverge (see [11]), achieve progress on Falconer’s distance set conjecture in geometric measure theory (see [6]), and improve on the decay estimates of spherical means of Fourier transforms of measures (see [16]).

The purpose of this paper is threefold:

  • Show that Theorem 1 is a borderline sharp Kakeya result in the sense that (1) is the endpoint of a family of estimates (see (2) in the statement of Conjecture 1.1) such that each member of the family (other than (1)) implies a certain sharp Kakeya result that we will formulate in §3 below.

  • Show that the sharp Kakeya result is true in certain cases in \(\mathbb R^3\); see Theorem 4.1.

  • Prove Conjecture 1.1 in \(\mathbb R^2\) (see Theorem 5.1) in the hope that this will shed some light on whether it would be possible to modify the Du-Zhang argument to also prove it in higher dimensions and consequently obtain the Kakeya result without having to pass through the restriction conjecture.

Conjectute 1.1

(when \(\beta =2/n\) or \(n=2\), this is a theorem) Suppose n, \(\alpha \), R, X, and \(\gamma \) are as in the statement of Theorem 1.

Let \(\beta \) be a parameter satisfying \(1/n \le \beta \le 2/n\), and define the exponent p by

$$\begin{aligned} p = 2 + \frac{n-\alpha }{n-1} \Big ( \frac{2}{n} - \beta \Big ). \end{aligned}$$

Then to every \(\epsilon > 0\) there is a constant \(C_\epsilon \) such that

$$\begin{aligned} \int _X |Ef(x)|^p dx \le C_\epsilon R^\epsilon \gamma ^\beta R^{\alpha /n} \Vert f \Vert _{L^p(\mathbb B^{n-1})}^p \end{aligned}$$
(2)

for all \(f \in L^p(\mathbb B^{n-1})\).

We note that when \(\beta = 2/n\), (2) becomes (1), so, to prove Conjecture 1.1 we need to perform the following trade: lower the power of \(\gamma \) in (1) from 2/n to \(\beta \) in return for raising the Lebesgue space exponent from 2 to p.

We will show below that if (2) holds for any \(\beta < 2/n\), then we obtain the sharp Kakeya result of §3.

As noted above, in dimension \(n=2\), (2) is true for all \(1/2 \le \beta \le 1\) (and hence Conjecture 1.1 is a theorem in the plane). We will prove this in the last three sections of the paper by using weighted bilinear restriction estimates and the broad-narrow strategy of [1].

Before we discuss the implications of Conjecture 1.1 to the Kakeya problem, it will be convenient to write (2) in an equivalent form, which is, perhaps, more user-friendly. This is the purpose of the next section.

2 Writing (2) in an Equivalent Form

Suppose \(n \ge 1\) and \(0 < \alpha \le n\). Following [12] (see also [3] and [13]), for Lebesgue measurable functions \(H: \mathbb R^n \rightarrow [0,1]\), we define

$$\begin{aligned} A_\alpha (H)= \inf \Big \{ C : \int _{B(x_0,R)} H(x) dx \le C R^\alpha \text{ for } \text{ all } x_0 \in \mathbb R^n \text{ and } R \ge 1 \Big \}, \end{aligned}$$

where \(B(x_0,R)\) denotes the ball in \(\mathbb R^n\) of center \(x_0\) and radius R. We say H is a weight of fractal dimension \(\alpha \) if \(A_\alpha (H) < \infty \). We note that \(A_\beta (H) \le A_\alpha (H)\) if \(\beta \ge \alpha \), so we are not really assigning a dimension to the function H; the phrase “H is a weight of dimension \(\alpha \)” is merely another way for us to say that \(A_\alpha (H) < \infty \).

Proposition 2.1

Suppose n, \(\alpha \), R, X, \(\gamma \), \(\beta \), and p are as in the statement of Conjecture 1.1. Then the estimate (2) holds if and only if to every \(\epsilon > 0\) there is a constant \(C_\epsilon \) such that

$$\begin{aligned} \int _{B(0,R)} |Ef(x)|^p H(x) dx \le C_\epsilon R^\epsilon A_\alpha (H)^\beta R^{\alpha /n} \Vert f \Vert _{L^p(\mathbb B^{n-1})}^p \end{aligned}$$
(3)

for all functions \(f \in L^p(\mathbb B^{n-1})\) and weights H of fractal dimension \(\alpha \).

Proof

Let H be the characteristic function of X. By the definition of \(\gamma \), we have

$$\begin{aligned} \int _{\widetilde{B}(x_0,r)} H(x) dx \le \gamma \, (r+2)^\alpha \le \gamma \, (3r)^\alpha \end{aligned}$$

for all \(x_0 \in \mathbb R^n\) and \(r \ge 1\). Thus H is a weight on \(\mathbb R^n\) of fractal dimension \(\alpha \), and \(A_\alpha (H) \le 3^\alpha \gamma \). This immediately shows that (3) implies (2).

To prove the reverse implication, we follow [4, Proof of Theorem 2.2].

We consider a covering \(\{ \widetilde{B} \}\) of B(0, R) by unit lattice cubes. Since every unit cube is contained in a ball of radius \(\sqrt{n}\), we have \(\int _{\widetilde{B}} H(x) dx \le A_\alpha (H) n^{\alpha /2}\), so, if we define \(v(\widetilde{B})=A_\alpha (H)^{-1} \int _{\widetilde{B}} H(x) dx\) and \(V_k = \{ \widetilde{B} : 2^{k-1} < n^{-\alpha /2} v(\widetilde{B}) \le 2^k \}\), then

$$\begin{aligned} B(0,R) \subset \cup \, \widetilde{B} \subset \cup _{k=-\infty }^0 V_k. \end{aligned}$$

We note that

$$\begin{aligned}{} & {} {\int _{\widetilde{B}} H(x) dx \le \Big ( \int _{\widetilde{B}} H(x)^{1/\beta } dx \Big )^\beta \le \Big ( \int _{\widetilde{B}} H(x) dx \Big )^\beta } \nonumber \\{} & {} \quad = \Big ( A_\alpha (H) v(\widetilde{B}) \Big )^\beta \le n^{\alpha \beta /2} A_\alpha (H)^\beta 2^{k \beta } \end{aligned}$$
(4)

for all \(\widetilde{B} \in V_k\), where we have used the assumptions \(\beta \le 2/n \le 1\) and \(\Vert H \Vert _{L^\infty } \le 1\).

The vast majority of the sets \(V_k\) are negligible for us. In fact, letting \(k_1\) be the sup of the set \(\{ k \in \mathbb Z :2^k \le R^{-1000n/\beta } \}\), we see that

$$\begin{aligned} \int _{\cup _{k=-\infty }^{k_1} \cup _{\widetilde{B} \in V_k}} |Ef(x)|^p H(x) dx\le & {} \Vert f \Vert _{L^1(\mathbb B^{n-1})}^p \sum _{k=-\infty }^{k_1} \sum _{\widetilde{B} \in V_k} \int _{\widetilde{B}} H(x) dx \\\le & {} C A_\alpha (H)^\beta \Vert f \Vert _{L^1(\mathbb B^{n-1})}^p \sum _{k=-\infty }^{k_1} R^n 2^{k \beta } \\\le & {} C R^{-999n} A_\alpha (H)^\beta \Vert f \Vert _{L^1(\mathbb B^{n-1})}^p, \end{aligned}$$

where we used (4) on the line before the last, and the fact that \(2^{k_1} \le R^{-1000n/\beta }\) on the last line. Therefore, we only need to estimate

$$\begin{aligned} \int _{\cup _{k=k_1+1}^0 \cup _{\widetilde{B} \in V_k}} |Ef(x)|^p H(x) dx = \sum _{k=k_1+1}^0 \sum _{\widetilde{B} \in V_k} \int _{\widetilde{B}} |Ef(x)|^p H(x) dx. \end{aligned}$$

Letting \(k_0 \in \{ k_1+1, k_1+2, \ldots , 0 \}\) be the integer satisfying

$$\begin{aligned} \sum _{\widetilde{B} \in V_{k_0}} \int _{\widetilde{B}} |Ef(x)|^p H(x) dx = \max _{k_1+1 \le k \le 0} \Big [ \sum _{\widetilde{B} \in V_k} \int _{\widetilde{B}}|Ef(x)|^p H(x) dx \Big ], \end{aligned}$$

we see that

$$\begin{aligned}{} & {} {\int _{B(0,R)} |Ef(x)|^p H(x) dx} \nonumber \\{} & {} \quad \le (-k_1) \sum _{\widetilde{B} \in V_{k_0}} \int _{\widetilde{B}} |Ef(x)|^p H(x) dx + C R^{-999n} A_\alpha (H)^\beta \Vert f \Vert _{L^1(\mathbb B^{n-1})}^p. \end{aligned}$$
(5)

Since \(-k_1 {\mathop {\sim }\limits ^{\textstyle <}}\log (2R)\), it follows that we only need to estimate

$$\begin{aligned} \sum _{\widetilde{B} \in V_{k_0}} \int _{\widetilde{B}} |Ef(x)|^p H(x) dx. \end{aligned}$$

We start by using the uncertainty principle in the following form. Let \(d\sigma \) be the pushforward of the \((n-1)\)-dimensional Lebesgue measure under the map \(T : \mathbb B^{n-1} \rightarrow {\mathcal P}\) given by \(T(\omega )= (\omega , |\omega |^2)\). Since the measure \(d\sigma \) is compactly supported and \(Ef=\widehat{gd\sigma }\), where g is the function on \({\mathcal P}\) defined by the equation \(f= g \circ T\), it follows that there is a non-negative rapidly decaying function \(\psi \) on \(\mathbb R^n\) such that

$$\begin{aligned} \sup _{\widetilde{B}} |Ef|^p {\mathop {\sim }\limits ^{\textstyle <}}|Ef|^p *\psi (c(\widetilde{B})), \end{aligned}$$

where \(c(\widetilde{B})\) is the center of \(\widetilde{B}\). Thus

$$\begin{aligned} \int _{\widetilde{B}} |Ef(x)|^p H(x) dx {\mathop {\sim }\limits ^{\textstyle <}}\Big ( \int _{\widetilde{B}} H(x) dx \Big ) |Ef|^p *\psi (c(\widetilde{B})). \end{aligned}$$

From (4) we know that \(\int _{\widetilde{B}} H(x) dx {\mathop {\sim }\limits ^{\textstyle <}}A_\alpha (H)^\beta 2^{k_0 \beta }\) for all \(\widetilde{B} \in V_{k_0}\). Also,

$$\begin{aligned} |Ef|^p *\psi (c(\widetilde{B}))&= \int _{B(c(\widetilde{B}),R^\epsilon )} |Ef(x)|^p \psi (c(\widetilde{B})-x) dx \\&\quad \ + \int _{B(c(\widetilde{B}),R^\epsilon )^c} |Ef(x)|^p \psi (c(\widetilde{B})-x) dx \\&{\mathop {\sim }\limits ^{\textstyle <}}\int _{B(c(\widetilde{B}),R^\epsilon )} |Ef(x)|^p dx + R^{-1000n} \Vert f \Vert _{L^1(\mathbb B^{n-1})}^p \end{aligned}$$

and

$$\begin{aligned} \sum _{\widetilde{B} \in V_{k_0}} \chi _{B(c(\widetilde{B}),R^\epsilon )} {\mathop {\sim }\limits ^{\textstyle <}}R^{n\epsilon }, \end{aligned}$$
(6)

so

$$\begin{aligned}{} & {} {\sum _{\widetilde{B} \in V_{k_0}} \int _{\widetilde{B}} |Ef(x)|^p H(x) dx} \nonumber \\{} & {} \quad {\mathop {\sim }\limits ^{\textstyle <}}R^{n\epsilon } A_\alpha (H)^\beta 2^{k_0 \beta } \int _V |Ef(x)|^p dx + A_\alpha (H)^\beta R^{-999n} \Vert f \Vert _{L^1(\mathbb B^{n-1})}^p, \end{aligned}$$
(7)

where \(V= \cup _{\widetilde{B} \in V_{k_0}} B(c(\widetilde{B}),R^\epsilon )\).

We now let \(\{ \widetilde{B}^* \}\) be the set of all the unit lattice cubes that intersect V, and \(X= \cup \, \widetilde{B}^*\). We plan to apply (2) on this set X, but we first need to estimate \(\gamma \).

Let \(B_r\) be a ball in \(\mathbb R^n\) of radius \(r \ge R^\epsilon \) (if \(1 \le r \le R^\epsilon \), then, clearly, \(\# \{ \widetilde{B}^* : \widetilde{B}^* \subset B_r \} {\mathop {\sim }\limits ^{\textstyle <}}R^{n \epsilon }\)), and \(V_r\) the subset of \(V_{k_0}\) that consists of all unit cubes \(\widetilde{B}\) such that \(B(c(\widetilde{B}),2R^\epsilon ) \cap B_r \not = \emptyset \). If \(B_r\) intersects any of the cubes \(\widetilde{B}^*\) that make up X, then \(B_r\) intersects \(B(c(\widetilde{B}),2R^\epsilon )\) for some \(\widetilde{B} \in V_r\). Therefore,

$$\begin{aligned} \# \{ \widetilde{B}^* : \widetilde{B}^* \subset B_r \} {\mathop {\sim }\limits ^{\textstyle <}}R^{n \epsilon } \#(V_r). \end{aligned}$$

Our assumption \(r \ge R^\epsilon \), tells us that

$$\begin{aligned} \cup _{\widetilde{B} \in V_r} B(c(\widetilde{B}),2R^\epsilon ) \subset B_{5r}, \end{aligned}$$

so (using (6))

$$\begin{aligned}{} & {} {R^{n \epsilon } \int _{B_{5r}} H(x) dx {\mathop {\sim }\limits ^{\textstyle >}}\sum _{\widetilde{B} \in V_r} \int _{B(c(\widetilde{B}),2R^\epsilon )} H(x) dx} \\{} & {} \quad \ge \sum _{\widetilde{B} \in V_r} \int _{\widetilde{B}} H(x) dx = \sum _{\widetilde{B} \in V_r} v(\widetilde{B}) A_\alpha (H) \ge \#(V_r) \, n^{\alpha /2} 2^{k_0-1} A_\alpha (H). \end{aligned}$$

On the other hand,

$$\begin{aligned} \int _{B_{5r}} H(x) dx \le A_\alpha (H) (5r)^\alpha , \end{aligned}$$

so \(\#(V_r) {\mathop {\sim }\limits ^{\textstyle <}}R^{n \epsilon } 2^{-k_0} r^\alpha \), and so

$$\begin{aligned} \# \{ \widetilde{B}^* : \widetilde{B}^* \subset B_r \} {\mathop {\sim }\limits ^{\textstyle <}}R^{2n \epsilon } 2^{-k_0} r^\alpha . \end{aligned}$$

Therefore, \(\gamma {\mathop {\sim }\limits ^{\textstyle <}}R^{2n \epsilon } 2^{-k_0}\).

Applying (2), we now obtain

$$\begin{aligned} \int _V |Ef(x)|^p dx \le \int _X |Ef(x)|^p dx {\mathop {\sim }\limits ^{\textstyle <}}R^{5\epsilon } 2^{-k_0 \beta } R^{\alpha /n} \Vert f \Vert _{L^p(\mathbb B^{n-1})}^p, \end{aligned}$$

which, combined with (5) and (7), implies that

$$\begin{aligned} \int _{B(0,R)} |Ef(x)|^p H(x) dx&{\mathop {\sim }\limits ^{\textstyle <}} R^{(n+6)\epsilon } (2^{k_0})^{\beta -\beta } A_\alpha (H)^\beta R^{\alpha /n} \Vert f \Vert _{L^p(\mathbb B^{n-1})}^p \\&= R^{(n+6)\epsilon } A_\alpha (H)^\beta R^{\alpha /n} \Vert f \Vert _{L^p(\mathbb B^{n-1})}^p, \end{aligned}$$

which is our desired estimate (3).

3 Conjecture 1.1 Implies a Sharp Kakeya Result

Let \(\Omega \) be a subset of \(\mathbb R^n\) that obeys the following property: there is a number \(\alpha \) between 1 and n such that

$$\begin{aligned} |\Omega \cap B_R| \le C R^\alpha \end{aligned}$$
(8)

for all balls \(B_R\) in \(\mathbb R^n\) of radius \(R \ge 1\). (Given \(E \subset \mathbb R^n\) a Lebesgue measurable set, we let |E| denote its Lebesgue measure.)

For large L, we divide the unit paraboloid \({\mathcal P}\) into finitely overlapping caps \(\theta _j\) each of radius \(L^{-1}\), and we associate with each \(\theta _j\) a family \(\mathbb T_j\) of parallel \(1 \times L\) tubes that tile \(\mathbb R^n\) and point in the direction normal to \(\theta _j\) at its center. We let N be the cardinality of the set

$$\begin{aligned} J = \{ j : \text{ there } \text{ is } \text{ a } \text{ tube } \text{ of }\,\mathbb T_j\hbox { that lies in}\,\Omega \cap B(0,5L) \}. \end{aligned}$$
(9)

It is easy to see that the Kakeya conjecture (in its maximal operator form) implies the following bound on N: to every \(\epsilon > 0\) there is a constant \(C_\epsilon \) such that

$$\begin{aligned} N \le C_\epsilon L^\epsilon L^{\alpha -1} \end{aligned}$$
(10)

for all \(L \ge 1\). In fact, [2, Proposition 2.2] presents a proof of the fact that the Kakeya conjecture implies (10) in the case when \(\Omega \) is a neighborhood of an algebraic variety. This proof easily extends to general sets \(\Omega \) satisfying (8). (For the connection between neighborhoods of algebraic varieties and the condition (8), we refer the reader to [14].)

We note that (10) implies that if \(\Omega \cap B(0,5L)\) contains at least one tube from each direction (i.e. at least one tube from each of the \(\sim L^{n-1}\) families \(\mathbb T_j\)), then \(\alpha = n\).

In the special case when \(\Omega \) is a neighborhood of an algebraic variety, this bound on N was proved by Guth [7] in \(\mathbb R^3\), conjectured by Guth [8] to be true in \(\mathbb R^n\) for all \(n \ge 3\), and proved by Zahl [17] in \(\mathbb R^4\); see also [9]. The conjecture of [8] was then settled in all dimensions by Katz and Rogers in [10].

In this section we prove that Conjecture 1.1 about the extension operator implies that all sets \(\Omega \subset \mathbb R^n\) that satisfy the dimensionality condition (8) will also possess the Kakeya property (10). Here is the precise statement.

Theorem 3.1

Suppose (3) (or equivalently (2)) holds for some \(1/n \le \beta < 2/n\). Then (10) holds for all Lebesgue measurable sets \(\Omega \subset \mathbb R^n\) that obey (8).

Proof

We first write the set J as \(\{ j_1, j_2, \ldots , j_N \}\), and for each \(1 \le l \le N\), we let \(T_l\) be a tube from \(\mathbb T_{j_l}\) that lies in \(\Omega \cap B(0,5L) = \Omega \cap B_{5L}\). Then

$$\begin{aligned}{} & {} {NL = \sum _{l=1}^N |T_l| = \sum _{l=1}^N \int _{B_{5L} \cap \, \Omega } \chi _{T_l}(x) \, dx = \int _{B_{5L} \cap \, \Omega } \sum _{l=1}^N \chi _{T_l}(x) \, dx} \nonumber \\{} & {} \;\;\;\;\;\;\;\;\;\; = L^{2(n-1)} \int _{B_{5L} \cap \, \Omega } \sum _{l=1}^N \Big ( \frac{1}{L^{n-1}} \, \chi _{T_l}(x) \Big )^2 dx. \end{aligned}$$
(11)

Recall that \(\mathbb T_{j_l}\) is a family of parallel \(1 \times L\) tubes that tile \(\mathbb R^n\) and point in the direction normal to the \(L^{-1}\)-cap \(\theta _{j_l}\). The projection of \(\theta _{j_l}\) into \(\mathbb B^{n-1}\) is an \(L^{-1}\)-ball. We denote this ball by \(B_l\) and let \(\omega _l\) be its center and \(\chi _l\) its characteristic function. Then

$$\begin{aligned} |E \chi _l(x)| = \Big | \int _{B_l} e^{-2 \pi i x \cdot (\omega , |\omega |^2)} d\omega \Big | = \Big | \int _{B_l} e^{-2 \pi i \big ( (x_1 + 2x_2 \omega _l)(\omega -\omega _l) + x_2|\omega -\omega _l|^2 \big )} d\omega \Big | \end{aligned}$$

for all \(x=(x_1,x_2) \in \mathbb R^{n-1} \times \mathbb R\). Since \(|\omega -\omega _l| \le L^{-1}\) for all \(\omega \in B_l\), it follows that \(|E \chi _l(x)| {\mathop {\sim }\limits ^{\textstyle >}}|B_l| \sim L^{-(n-1)}\) on the set \(\{ x \in \mathbb R^n : |x_1 + 2x_2 \omega _l| {\mathop {\sim }\limits ^{\textstyle<}}L \text{ and } |x_2| {\mathop {\sim }\limits ^{\textstyle <}}L^2 \}\), and hence \(|E \chi _l(Lx)| {\mathop {\sim }\limits ^{\textstyle >}}L^{-(n-1)}\) on the set \(\{ x \in \mathbb R^n : |x_1 + 2x_2 \omega _l| {\mathop {\sim }\limits ^{\textstyle<}}1 \text{ and } |x_2| {\mathop {\sim }\limits ^{\textstyle <}}L \}\). Since \(|\omega _l| \le 1\), this last set contains a \(1 \times L\) tube \(\widetilde{T}_l\) that is parallel to the normal vector of the cap \(\theta _{j_l}\) at its center \((\omega _l, |\omega _l|^2)\). Moreover,

$$\begin{aligned} |E\chi _l(L x)| {\mathop {\sim }\limits ^{\textstyle >}}\frac{1}{L^{n-1}} \, \chi _{\widetilde{T}_l}(x) \end{aligned}$$

for all \(x \in \mathbb R^n\).

The tube \(\widetilde{T}_l\) is parallel to the tube \(T_l\) that we chose at the beginning of the proof and has the same dimensions, so \(T_l = v + \widetilde{T}_l\) for some vector \(v \in \mathbb R^n\), and so

$$\begin{aligned} |E\chi _l(L x)| {\mathop {\sim }\limits ^{\textstyle >}}\frac{1}{L^{n-1}} \, \chi _{T_l}(x+v) \end{aligned}$$

for all \(x \in \mathbb R^n\). Defining the function \(f_l\) on \(\mathbb R^{n-1}\) by

$$\begin{aligned} f_l(\omega ) = e^{2 \pi i L v \cdot (\omega , |\omega |^2)} \chi _l(\omega ), \end{aligned}$$

we see that \(Ef_l(x) = E\chi _l(x-Lv)\), so that

$$\begin{aligned} |Ef_l(Lx)| = |E\chi _l(L x - Lv)| = |E\chi _l(L(x - v))| {\mathop {\sim }\limits ^{\textstyle >}}\frac{1}{L^{n-1}} \, \chi _{T_l}(x) \end{aligned}$$

for all \(x \in \mathbb R^n\). Returning to (11) and letting \(H = \chi _\Omega \), we arrive at

$$\begin{aligned} NL {\mathop {\sim }\limits ^{\textstyle <}}L^{2(n-1)} \int _{B_{5L}} \sum _{l=1}^N |Ef_l(Lx)|^2 H(x) dx. \end{aligned}$$

Next, we let \(\epsilon _l = \pm 1\) be random signs, define the function \(f: \mathbb B^{n-1} \rightarrow \mathbb C\) by \(f = \sum _{l=1}^N \epsilon _l f_l\), and use Khintchin’s inequality to get

$$\begin{aligned} NL {\mathop {\sim }\limits ^{\textstyle <}}L^{2(n-1)} \, \mathbb E \Big ( \int _{B_{5L}} |Ef(Lx)|^2 H(x) dx \Big ), \end{aligned}$$

where \(\mathbb E\) is the expectation sign. Since \(p \ge 2\), we can apply Hölder’s inequality in the inner integral to get

$$\begin{aligned} N L&{\mathop {\sim }\limits ^{\textstyle<}} L^{2(n-1)} \Big ( \int _{B_{5L}} H(x) dx \Big )^{1-(2/p)} \, \mathbb E \Big ( \int _{B_{5L}} |Ef(Lx)|^p H(x) dx \Big )^{2/p} \\&{\mathop {\sim }\limits ^{\textstyle <}} L^{2(n-1)} L^{\alpha (1-(2/p))} \, \mathbb E \Big ( \int _{B_{5L}} |Ef(Lx)|^p H(x) dx \Big )^{2/p}. \end{aligned}$$

Applying the change of variables \(u=Lx\) and defining the weight \(H^*\) by \(H^*(u)= H(x)= H(u/L)\), this becomes

$$\begin{aligned} N L {\mathop {\sim }\limits ^{\textstyle <}}L^{2(n-1)} L^{\alpha (1-(2/p))} L^{-2n/p} \, \mathbb E \Big ( \int _{B_{5L^2}} |Ef(u)|^p H^*(u) du \Big )^{2/p}, \end{aligned}$$

so that

$$\begin{aligned} N L^{3-n} {\mathop {\sim }\limits ^{\textstyle <}}L^{(n+\alpha )(1-(2/p))} \, \mathbb E \Big ( \int _{B_{5L^2}} |Ef(u)|^p H^*(u) du \Big )^{2/p}. \end{aligned}$$
(12)

We note that

$$\begin{aligned}{} & {} {\int _{B(u_0,R)} H^*(u) du = L^n \int _{B(u_0/L,R/L)} H(x) dx} \\{} & {} \quad \le L^n A_\alpha (H) \big ( \frac{R}{L} \big )^\alpha = L^{n-\alpha } A_\alpha (H) R^\alpha \end{aligned}$$

if \(R \ge L\). On the other hand, if \(R \le L\), then

$$\begin{aligned} \int _{B(u_0,R)} H^*(u) du {\mathop {\sim }\limits ^{\textstyle <}}R^n = R^{n-\alpha } R^\alpha \le L^{n-\alpha } R^\alpha . \end{aligned}$$

Therefore,

$$\begin{aligned} A_\alpha (H^*) {\mathop {\sim }\limits ^{\textstyle <}}L^{n-\alpha }. \end{aligned}$$

We are now in a good shape to apply (3), which tells us that

$$\begin{aligned} \int _{B_{5L^2}} |Ef(u)|^p H^*(u) du&{\mathop {\sim }\limits ^{\textstyle<}} (L^2)^\epsilon A_\alpha (H^*)^\beta (L^2)^{\alpha /n} \Vert f \Vert _{L^p(\mathbb B^{n-1})}^p \\&{\mathop {\sim }\limits ^{\textstyle <}} L^{2 \epsilon } L^{(n-\alpha )\beta } L^{2\alpha /n} \frac{N}{L^{n-1}}. \end{aligned}$$

Inserting this back in (12), we get

$$\begin{aligned} N L^{3-n} {\mathop {\sim }\limits ^{\textstyle <}}L^{2 \epsilon } L^{(n+\alpha )(1-(2/p))} \Big ( L^{(n-\alpha )\beta } L^{2\alpha /n} L^{1-n} N \Big )^{2/p}, \end{aligned}$$

so that

$$\begin{aligned} N^{1-(2/p)} L^{3-n} {\mathop {\sim }\limits ^{\textstyle <}}L^{2 \epsilon } L^{(n+\alpha )(1-(2/p))} \Big ( L^{(n-\alpha )(\beta -\frac{2}{n}) + 2 - \frac{2\alpha }{n}} L^{\frac{2\alpha }{n}} \frac{L^{-2}}{L^{n-3}} \Big )^{2/p}, \end{aligned}$$

so that

$$\begin{aligned} N^{1-(2/p)} {\mathop {\sim }\limits ^{\textstyle <}}L^{2 \epsilon } L^{(n-3)(1-(2/p))} L^{(n+\alpha )(1-(2/p))} L^{(n-\alpha )(\beta -\frac{2}{n})(\frac{2}{p})}. \end{aligned}$$
(13)

Therefore,

$$\begin{aligned} N {\mathop {\sim }\limits ^{\textstyle <}}L^{O(\epsilon )} L^{n-3} L^{\alpha } L^{n} L^{\frac{(n-\alpha )(\beta -\frac{2}{n})(\frac{2}{p})}{1- \frac{2}{p}}} = L^{O(\epsilon )} L^{2n-3+\alpha } L^{\frac{(n-\alpha )(\beta -\frac{2}{n})}{\frac{p}{2}-1}}. \end{aligned}$$

But

$$\begin{aligned} \frac{(n-\alpha )(\beta -\frac{2}{n})}{\frac{p}{2}-1} = (n-\alpha )\left( \beta -\frac{2}{n}\right) \frac{2(n-1)}{(n-\alpha )(\frac{2}{n}-\beta )} = - 2(n-1) = 2 - 2n, \end{aligned}$$

so

$$\begin{aligned} N {\mathop {\sim }\limits ^{\textstyle <}}L^{O(\epsilon )} L^{2n-3+\alpha +2-2n} = L^{O(\epsilon )} L^{\alpha - 1}. \end{aligned}$$

At this point, it might be helpful for the reader to observe how the above argument breaks down in the \(p=2\) case: recalling that

$$\begin{aligned} p = 2 + \frac{n-\alpha }{n-1} \Big ( \frac{2}{n} - \beta \Big ), \end{aligned}$$

we see that \(\beta = 2/n\) and (13) becomes \(1 {\mathop {\sim }\limits ^{\textstyle <}}L^{2\epsilon }\), which tells us nothing.

4 Proof of (10) in the Regime \(1 \le \alpha \le 2\) in \(\mathbb R^3\)

The fact that the Kakeya conjecture is true in \(\mathbb R^2\) tells us that (10) is also true there. In this section, we use Wolff’s hairbrush argument from [15], as adapted by Guth in [7], to prove the following bound on N.

Theorem 4.1

In \(\mathbb R^3\), we have

$$\begin{aligned} N {\mathop {\sim }\limits ^{\textstyle <}}\left\{ \begin{array}{ll} (\log L)^2 L^{\alpha -1} &{} \text{ if }\,1 \le \alpha \le 2, \\ (\log L)^2 L^{2\alpha -3} &{} \text{ if }\,2 \le \alpha \le 3. \end{array} \right. \end{aligned}$$

Proof

Let \(\Omega \) be a subset of \(\mathbb R^3\) that obeys (8). As we did in the previous section, for large L, we consider a decomposition \(\{ \theta _j \}\) of \({\mathcal P}\) into finitely overlapping caps each of radius \(L^{-1}\), and we associate with each \(\theta _j\) a family \(\mathbb T_j\) of parallel \(1 \times L\) tubes that tile \(\mathbb R^3\) and point in the direction of the normal vector \(v_j\) of \({\mathcal P}\) at the center of \(\theta _j\). The quantity N that we need to estimate is the cardinality of the set J as defined in (9).

For each \(j \in J\), we let \(T_j\) be a member of \(\mathbb T_j\) that lies in \(\Omega \cap B(0,5L)\), and \(S= \{ T_j \}\). Of course, \(N = \#(S)\).

We tile \(\Omega \cap B(0,5L)\) by unit lattice cubes \(\widetilde{B}\). Then (8) tells us that

$$\begin{aligned} \#(\{ \widetilde{B} \}) {\mathop {\sim }\limits ^{\textstyle <}}L^\alpha . \end{aligned}$$
(14)

Also, each tube \(T_j\) intersects \(\sim L\) of the cubes \(\widetilde{B}\).

We now define the function \(f : \{ \widetilde{B} \} \rightarrow \mathbb Z\) by

$$\begin{aligned} f(\widetilde{B}) = \# \{ T_j \in S : T_j \cap \widetilde{B} \not = \emptyset \}. \end{aligned}$$

Then

$$\begin{aligned} \sum _{\widetilde{B}} f(\widetilde{B}) \sim N L. \end{aligned}$$

So, by Cauchy–Schwarz and (14),

$$\begin{aligned} N L {\mathop {\sim }\limits ^{\textstyle<}}\Big ( \sum _{\widetilde{B}} f(\widetilde{B})^2 \Big )^{1/2} \Big ( \#(\{ \widetilde{B} \}) \Big )^{1/2} {\mathop {\sim }\limits ^{\textstyle <}}\Big ( \sum _{\widetilde{B}} f(\widetilde{B})^2 \Big )^{1/2} L^{\alpha /2}, \end{aligned}$$

and so

$$\begin{aligned} \sum _{\widetilde{B}} f(\widetilde{B})^2 {\mathop {\sim }\limits ^{\textstyle >}}N^2 L^{2 - \alpha }, \end{aligned}$$

which means that the set

$$\begin{aligned} \{ (\widetilde{B}, T_i, T_j) : T_i, T_j \in S, \, T_i \cap \widetilde{B} \not = \emptyset , \text{ and } T_j \cap \widetilde{B} \not = \emptyset \} \end{aligned}$$

has cardinality \({\mathop {\sim }\limits ^{\textstyle >}}N^2 L^{2 - \alpha }\). Therefore, the set

$$\begin{aligned} X = \{ (\widetilde{B}, T_i, T_j) : T_i, T_j \in S, \, T_i \cap \widetilde{B} \not = \emptyset , \, T_j \cap \widetilde{B} \not = \emptyset \text{ and } i \not = j \} \end{aligned}$$

has cardinality

$$\begin{aligned} \ge C_1 N^2 L^{2 - \alpha } - \sum _{\widetilde{B}} f(\widetilde{B}) \ge C_1 N^2 L^{2 - \alpha } - C_2 N L. \end{aligned}$$

If \(C_1 N^2 L^{2 - \alpha } \le 5 C_2 N L\), then \(N \le (5C_2/C_1) L^{\alpha -1}\) and the theorem will be proved. So, we may assume that \(N \ge C_3 L^{\alpha -1}\) for some large constant \(C_3\). Therefore, \(\#(X) {\mathop {\sim }\limits ^{\textstyle >}}N^2 L^{2 - \alpha }\).

For \(l \in \mathbb N\), we define \(X_l\) to be the subset of X for which

$$\begin{aligned} \frac{2^{l-1}}{L} \le \text{ Angle }(v_i,v_j) \le \frac{2^l}{L}. \end{aligned}$$

Since the angle between any two tubes in our set S ranges between \(L^{-1}\) and 1, it follows by the pigeonhole principle that \(\#(X) {\mathop {\sim }\limits ^{\textstyle <}}(\log L) \#(X_{l_0})\) for some \(l_0 \in \mathbb N\). Denoting \(2^{l_0} L^{-1}\) by \(\theta \), and \(X_{l_0}\) by \(X'\), we have \(L^{-1} \le \theta \le 1\) and \(\#(X') {\mathop {\sim }\limits ^{\textstyle >}}N^2 L^{2 - \alpha } (\log L)^{-1}\).

There are N tubes in S. By the pigeonhole principle, one of the tubes must appear in \({\mathop {\sim }\limits ^{\textstyle >}}N^2 L^{2 - \alpha } (\log L)^{-1}/N = N L^{2 - \alpha } (\log L)^{-1}\) of the elements of \(X'\). We call this tube T, and we define

$$\begin{aligned} \mathbb H = \{ T_j \in S : (\widetilde{B}, T, T_j) \in X' \}. \end{aligned}$$

Let v be the direction of the tube T. Since the angle between v and \(v_j\) is \(\sim \theta \), it follows that \(|T \cap T_j| {\mathop {\sim }\limits ^{\textstyle <}}\theta ^{-1}\). So, the set \(\{ \widetilde{B} : (\widetilde{B}, T, T_j) \in X' \}\) has cardinality \({\mathop {\sim }\limits ^{\textstyle <}}\theta ^{-1}\), and so

$$\begin{aligned} \#(\mathbb H) {\mathop {\sim }\limits ^{\textstyle >}}\frac{N L^{2 - \alpha } (\log L)^{-1}}{\theta ^{-1}} = \theta N L^{2 - \alpha } (\log L)^{-1}. \end{aligned}$$

To finish the proof, we need to also have an upper bound on \(\#(\mathbb H)\). We first observe that

$$\begin{aligned} \bigcup _{T_j \in \mathbb H} T_j \subset \Omega \cap {\textbf{B}}, \end{aligned}$$

where B is a box in \(\mathbb R^3\) of dimensions \(L \times \theta L \times \theta L\). Since B can be covered by \(\sim L/(\theta L)\) balls of radius \(\theta L\), and since \(\theta L \ge 1\), the dimensionality property (8) tells us that

$$\begin{aligned} \Big | \bigcup _{T_j \in \mathbb H} T_j \Big | {\mathop {\sim }\limits ^{\textstyle <}}\theta ^{-1} (\theta L)^\alpha . \end{aligned}$$

Next, we use the (by now) standard fact that the tubes \(T_j\) in \(\mathbb H\) are morally disjoint (see [7, Lemma 4.9] for a very nice explanation of this idea) to see that

$$\begin{aligned} \Big | \bigcup _{T_j \in \mathbb H} T_j \Big | {\mathop {\sim }\limits ^{\textstyle >}}\frac{\#(\mathbb H) \, |T_j|}{\log L} = \frac{\#(\mathbb H) \, L}{\log L}. \end{aligned}$$

Therefore,

$$\begin{aligned} \#(\mathbb H) {\mathop {\sim }\limits ^{\textstyle <}}(\log L) \theta ^{-1} L^{-1} (\theta L)^\alpha = (\log L) (\theta L)^{\alpha -1}. \end{aligned}$$

Comparing the lower and upper bounds we now have on the cardinality of \(\mathbb H\), we conclude that

$$\begin{aligned} \theta N L^{2 - \alpha } (\log L)^{-1} {\mathop {\sim }\limits ^{\textstyle <}}(\log L) (\theta L)^{\alpha -1}. \end{aligned}$$

Therefore,

$$\begin{aligned} N {\mathop {\sim }\limits ^{\textstyle <}}(\log L)^2 \theta ^{\alpha - 2} L^{2 \alpha - 3}. \end{aligned}$$

If \(\alpha \ge 2\), then the fact that \(\theta \le 1\) tells us that

$$\begin{aligned} N \le (\log L)^2 L^{2 \alpha - 3}. \end{aligned}$$

If \(1 \le \alpha < 2\), then the fact that \(\theta \ge 1/L\) tells us that

$$\begin{aligned} N {\mathop {\sim }\limits ^{\textstyle <}}(\log L)^2 L^{2 - \alpha } L^{2 \alpha - 3} = (\log L)^2 L^{\alpha - 1}. \end{aligned}$$

It might be interesting for the reader to observe that the sharp result that we get in the case \(1 \le \alpha < 2\) is due to the fact that we are using ‘substantial’ information about \(\theta \) (namely, \(\theta \ge 1/L\)), whereas in the case \(2 \le \alpha \le 3\) we only can use the relatively ‘unsubstantial’ information that \(\theta \le 1\).

We note that if \(\Omega \subset \mathbb R^3\) obeys (8) and \(\Omega \cap B(0,5L)\) contains at least one tube from each direction (i.e. at least one tube from each of the \(\sim L^2\) families \(\mathbb T_j\)), then Theorem 4.1 implies that \(\alpha \ge 5/2\) (cf. [15]).

5 Proof of Conjecture 1.1 in the Plane

The rest of the paper is concerned in proving that Conjecture 2.1 is true in \(\mathbb R^2\). In view of Proposition 2.1, this task will be accomplished as soon as we prove Theorem 5.1 below.

We alert the reader that the extension operator in Theorem 5.1 is the one associated with the unit circle \(\mathbb S^1 \subset \mathbb R^2\) and is given by

$$\begin{aligned} Ef(x) = \int _{\mathbb S^1} e^{-2 \pi i x \cdot \xi } f(\xi ) d\sigma (\xi ) \end{aligned}$$

for \(f \in L^1(\sigma )\), where \(\sigma \) is induced Lebesgue measure on \(\mathbb S^1\). The proof for the extension operator associated with the unit parabola is similar (and a little easier).

Theorem 5.1

Suppose \(1 \le \alpha \le 2\) and \(R \ge 1\). Let \(\beta \) be a parameter satisfying \(1/2 \le \beta \le 1\), and define the exponent p by

$$\begin{aligned} p = 2 + (2 - \alpha )(1 - \beta ). \end{aligned}$$

Then to every \(\epsilon > 0\) there is a constant \(C_\epsilon \) such that

$$\begin{aligned} \int _{B(0,R)} |Ef(x)|^p H(x) dx \le C_\epsilon R^\epsilon A_\alpha (H)^\beta R^{\alpha /2} \Vert f \Vert _{L^p(\sigma )}^p \end{aligned}$$
(15)

for all functions \(f \in L^p(\sigma )\) and weights H of fractal dimension \(\alpha \).

The proof of Theorem 5.1 will use ideas from [5, 12, 16], and [4]. The overarching idea, however, is the broad-narrow strategy of [1]. Implementing this strategy involves

  • proving a bilinear estimate (see (22) in Subsection 7.1 below) that will be used to control Ef on the broad set

  • proving a linear estimate (see (28) in Subsection 7.2 below) that will be used to establish (15) when the function f is supported on an arc of small size (i.e. \(\sigma \)-measure), which will provide the base of a recursive process

  • carrying out a recursive process on the size of the function’s support that will establish (15) for general f.

The main new idea in the proof of Theorem 5.1 is a localization of the weight argument that will help us in deriving the bilinear estimate (22). We use this argument to take advantage of the locally constant property of the Fourier transform, and we will end this section by formulating the intuition that lies behind it in a lemma.

Given a function \(f : \mathbb R^n \rightarrow \mathbb C\) and a number \(K > 0\), we say that f is essentially constant at scale K if there is a constant C such that

$$\begin{aligned} \sup _{Q_K} |f| \le C \inf _{Q_K} |f| \end{aligned}$$
(16)

for all cubes \(Q_K \subset \mathbb R^n\) of side-length K.

Lemma 5.1

Suppose \(1 \le \alpha \le 2\), \(1/2 \le \beta \le 1\), \(R > K^2 \ge 1\), and Q is a box in \(\mathbb R^2\) of dimensionsFootnote 1\(R/K \times R\). Also, suppose that f is a non-negative function on \(\mathbb R^2\) that is essentially constant at scale K, and H is a weight on \(\mathbb R^2\) of fractal dimension \(\alpha \). Then

$$\begin{aligned} \int _Q f(x) H(x) dx {\mathop {\sim }\limits ^{\textstyle <}}K^{-m} A_\alpha (H)^\beta R^{\alpha /2} \Vert f \Vert _{L^2(\widetilde{Q})} \end{aligned}$$

for some \(m \ge 0\) (in fact, \(m=\beta -(1/2)+(1-\beta )(\alpha -1)\)), where \(\widetilde{Q}\) is a box of dimensions \(2R/K \times 2R\) that has the same center as Q, and the implicit constant depends only on \(\alpha \) and \(\beta \) and the constant C from (16).

Proof

We tile \(\mathbb R^2\) by cubes \(\widetilde{B}_l\) of side-length K. If \(\widetilde{B}_l \cap Q \not = \emptyset \), we let \(c_l\) be the center of \(\widetilde{B}_l\) and write

$$\begin{aligned}{} & {} {\int _Q f(x) H(x) dx = \sum _l \int _{\widetilde{B}_l \cap Q} f(x) H(x) dx {\mathop {\sim }\limits ^{\textstyle<}}\sum _l f(c_l) \int _{\widetilde{B}_l} H(x) dx} \\{} & {} \quad = \sum _l K^{-2} \int _{\widetilde{B}_l} f(c_l) H'(y) dy {\mathop {\sim }\limits ^{\textstyle <}}K^{-2} \int _{\widetilde{Q}} f(y) H'(y) dy, \end{aligned}$$

where \(H': \mathbb R^2 \rightarrow [0,\infty )\) is given by

$$\begin{aligned} H'(y) = \int _{\widetilde{B}_l} H(x) dx \hspace{0.25in} \text{ for } \hspace{0.25in} y \in \widetilde{B}_l. \end{aligned}$$

For \(y \in \widetilde{B}_l\), we have

$$\begin{aligned} H'(y)= & {} \Big ( \int _{\widetilde{B}_l} H(x) dx \Big )^{1-\theta } \Big ( \int _{\widetilde{B}_l} H(x) dx \Big )^\theta \\\le & {} K^{2(1-\theta )} A_\alpha (H)^\theta (\sqrt{2} K)^{\alpha \theta }, \end{aligned}$$

where \(0 \le \theta \le 1\) is a parameter that will be determined later in the argument.

Next, we define the function \({\mathcal H} : \mathbb R^2 \rightarrow [0,1]\) by

$$\begin{aligned} {\mathcal H}(y) = 2^{-\alpha \theta /2} A_\alpha (H)^{-\theta } K^{-2(1-\theta )-\alpha \theta } H'(y) \end{aligned}$$

and observe that

$$\begin{aligned}{} & {} {\int _{B(x_0,r)} {\mathcal H}(y) dy \le K^2 A_\alpha (H)^{-\theta } K^{-2(1-\theta )-\alpha \theta } \int _{B(x_0,3r)} H(y) dy} \\{} & {} \quad \le K^2 A_\alpha (H)^{-\theta } K^{-2(1-\theta )-\alpha \theta } A_\alpha (H) (3 r)^\alpha = 3^\alpha A_\alpha (H)^{1-\theta } K^{\theta (2-\alpha )} r^\alpha \end{aligned}$$

for all \(x_0 \in \mathbb R^2\) and \(r \ge K\). On the other hand, when \(1 \le r \le K\) we use the fact that

$$\begin{aligned}{} & {} {{\mathcal H}(y) \le A_\alpha (H)^{-\theta } K^{-2(1-\theta )-\alpha \theta } H'(y) \le A_\alpha (H)^{-\theta } K^{-2(1-\theta )-\alpha \theta } \sup _l \int _{\widetilde{B}_l} H(x) dx} \\{} & {} \;\;\;\;\;\;\;\;\;\;\;\;\;\; {\mathop {\sim }\limits ^{\textstyle <}}A_\alpha (H)^{-\theta } K^{-2(1-\theta )-\alpha \theta } A_\alpha (H) K^\alpha = A_\alpha (H)^{1-\theta } K^{\theta (2-\alpha )} K^{\alpha -2} \end{aligned}$$

for all \(y \in \mathbb R^2\) to see that

$$\begin{aligned} \int _{B(x_0,r)} {\mathcal H}(y) dy {\mathop {\sim }\limits ^{\textstyle <}}A_\alpha (H)^{1-\theta } K^{\theta (2-\alpha )} K^{\alpha -2} r^2 \le A_\alpha (H)^{1-\theta } K^{\theta (2-\alpha )} r^\alpha \end{aligned}$$

(because \(K^{\alpha -2} \le r^{\alpha -2}\)). Therefore, \({\mathcal H}\) is a weight on \(\mathbb R^2\) of fractal dimension \(\alpha \) with

$$\begin{aligned} A_\alpha ({\mathcal H}) {\mathop {\sim }\limits ^{\textstyle <}}A_\alpha (H)^{1-\theta } K^{\theta (2-\alpha )}. \end{aligned}$$

Going back to our integral, we now have

$$\begin{aligned} \int _Q f(x) H(x) dx {\mathop {\sim }\limits ^{\textstyle <}}A_\alpha (H)^\theta K^{\theta (\alpha -2)} \int _{\widetilde{Q}} f(y) {\mathcal H}(y) dy. \end{aligned}$$

Bounding the integral on the right-hand side by Cauchy–Schwarz, this becomes

$$\begin{aligned} \int _Q f(x) H(x) dx {\mathop {\sim }\limits ^{\textstyle <}}A_\alpha (H)^\theta K^{\theta (\alpha -2)} \Big ( \int _{\widetilde{Q}} {\mathcal H}(y) dy \Big )^{1/2} \Vert f \Vert _{L^2(\widetilde{Q})}. \end{aligned}$$

But \(\widetilde{Q}\) can be covered by \(\sim K\) balls of radius R/K, so

$$\begin{aligned} \int _{\widetilde{Q}} {\mathcal H}(y) dy&{\mathop {\sim }\limits ^{\textstyle<}} K A_\alpha ({\mathcal H}) (K^{-1} R)^\alpha \nonumber \\&{\mathop {\sim }\limits ^{\textstyle <}} K A_\alpha (H)^{1-\theta } K^{\theta (2-\alpha )} (K^{-1} R)^\alpha , \end{aligned}$$
(17)

and so

$$\begin{aligned} \int _Q f(x) H(x) dx {\mathop {\sim }\limits ^{\textstyle <}}K^{1/2} A_\alpha (H)^{(1+\theta )/2} K^{\theta (\alpha -2)/2} (K^{-1} R)^{\alpha /2} \Vert f \Vert _{L^2(\widetilde{Q})}. \end{aligned}$$

We now determine \(\theta \) by solving the equation \((1+\theta )/2=\beta \), which gives \(\theta =2\beta -1\), and we arrive at

$$\begin{aligned} \int _Q f(x) H(x) dx {\mathop {\sim }\limits ^{\textstyle <}}K^{-m} A_\alpha (H)^\beta R^{\alpha /2} \Vert f \Vert _{L^2(\widetilde{Q})} \end{aligned}$$

with \(m=\beta -(1/2)+(1-\beta )(\alpha -1)\).

6 Preliminaries for the Proof of Theorem 5.1

This section contains basic facts that we need to prove Theorem 5.1 that we include to make the paper as self-contained as possible.

6.1 The \(L^1\) Norm of a Rapidly Decaying Function over a Box

In the rigorous version of the localization argument that we described in the previous section, instead of integrating over a proper \(R/K \times R\) box, we will be integrating against a Schwartz function that is essentially supported on such a box. It is easy to see that (17) continues to be true in this case. Here are the details.

Lemma 6.1

Suppose \(0 < \alpha \le n\), \(R \ge K^2 \ge 1\), and \(\Psi \) is a non-negative Schwartz function on \(\mathbb R^n\). Then

$$\begin{aligned} \int \Psi \Big ( \frac{x_1-\nu _1}{R K^{-1}}, \ldots , \frac{x_{n-1}-\nu _{n-1}}{R K^{-1}} \frac{x_n-\nu _n}{R} \Big ) H(x) dx {\mathop {\sim }\limits ^{\textstyle <}}K A_\alpha (H) (K^{-1} R)^\alpha \end{aligned}$$
(18)

for all weights H on \(\mathbb R^n\) of fractal dimension \(\alpha \).

Proof

Suppose \(R_1, \ldots , R_n > 0\) and \(\Psi \) is a non-negative Schwartz function. For \(l= 0, 1, 2, \ldots \), we let \(\chi _l\) be the characteristic function of the box in \(\mathbb R^n\) of center 0 and dimensions \(2^{l+1} R_1 \times \ldots \times 2^{l+1} R_n\), and \(B_l=B(0,2^l)\). Then

$$\begin{aligned}{} & {} {\Psi \Big ( \frac{x_1-\nu _1}{R_1}, \ldots , \frac{x_n-\nu _n}{R_n} \Big )} \\{} & {} \quad \le \Big ( \sup _{B_0} \Psi \Big ) \chi _{B_0} \Big ( \frac{x_1-\nu _1}{R_1}, \ldots , \frac{x_n-\nu _n}{R_n} \Big ) \\{} & {} \qquad + \sum _{l=1}^\infty \Big ( \sup _{B_l \setminus B_{l-1}} \Psi \Big ) \chi _{B_l \setminus B_{l-1}} \Big ( \frac{x_1-\nu _1}{R_1}, \ldots , \frac{x_n-\nu _n}{R_n} \Big ) \\{} & {} \quad {\mathop {\sim }\limits ^{\textstyle <}}\sum _{l=0}^\infty 2^{-Nl} \chi _l(x-\nu ) \end{aligned}$$

for all \(x, \nu \in \mathbb R^n\) and \(N \in \mathbb N\), so that

$$\begin{aligned} \int \Psi \Big ( \frac{x_1-\nu _1}{R_1}, \ldots , \frac{x_n-\nu _n}{R_n} \Big ) H(x) dx {\mathop {\sim }\limits ^{\textstyle <}}\sum _{l=0}^\infty 2^{-Nl} \int _{P_l} H(x) dx, \end{aligned}$$

where \(P_l\) is the box in \(\mathbb R^n\) of center \(\nu \) and dimensions \(2^{l+1} R_1 \times \ldots \times 2^{l+1} R_n\).

In the special case \(R_1= \ldots = R_{n-1} = R/K\) and \(R_n=R\) with \(R \ge K^2 \ge 1\) (as in (17)), this gives

$$\begin{aligned} \int \Psi \Big ( \frac{x_1-\nu _1}{R K^{-1}}, \ldots , \frac{x_{n-1}-\nu _{n-1}}{R K^{-1}} \frac{x_n-\nu _n}{R} \Big ) H(x) dx {\mathop {\sim }\limits ^{\textstyle <}}K A_\alpha (H) (K^{-1} R)^\alpha \end{aligned}$$
(19)

for all weights H on \(\mathbb R^n\) of fractal dimension \(\alpha \).

6.2 A Property of \(R/K \times \cdots \times R/K \times R\) Boxes

Suppose \(R \ge K^2 \ge 1\), Q is an \(R/K \times \cdots \times R/K \times R\) box in \(\mathbb R^n\). A box \(Q^* \subset \mathbb R^n\) of dimensions \((R/K)^{-1} \times \cdots \times (R/K)^{-1} \times R^{-1}\) and with the same axes as Q is called a dual box of Q. This subsection is about the following observation.

Lemma 6.2

Suppose \(Q^*\) is a dual box of Q whose \((R/K)^{-1} \times \cdots \times (R/K)^{-1}\)-face is tangent to the unit sphere \(\mathbb S^{n-1} \subset \mathbb R^n\) at some point e. Then \(Q^*\) lies in the \(R^{-1}\)-neighborhood of \(\mathbb S^{n-1}\).

Proof

Let \(\delta = K^{-1}\). Then \(Q^*\) has dimensions \((R \delta )^{-1} \times \ldots \times (R \delta )^{-1} \times R^{-1}\) and its \((R \delta )^{-1} \times \ldots \times (R \delta )^{-1}\)-face is tangent to \(\mathbb S^{n-1}\) at e.

Without any loss of generality, we may assume that \(e=(0, \ldots , 0, 1)\).

Suppose \(y \in Q^*\). Then

$$\begin{aligned} |y|^2= & {} y_1^2 + \ldots + y_{n-1}^2 + (y_n-1+1)^2 = y_1^2 + \ldots + y_{n-1}^2 + (y_n-1)^2\\ {}{} & {} \quad + 2(y_n-1) +1 \end{aligned}$$

so that

$$\begin{aligned} \big | |y|^2-1 \big | \le y_1^2 + \ldots + y_{n-1}^2 + |y_n-1|^2 + 2 |y_n-1| \end{aligned}$$

so that

$$\begin{aligned} \big | |y|-1 \big | \, \big | |y|+1 \big | \le y_1^2 + \ldots + y_{n-1}^2 + 3 |y_n-1| \end{aligned}$$

so that

$$\begin{aligned} \big | |y|-1 \big | \le y_1^2 + \ldots + y_{n-1}^2 + 3 |y_n-1| \le \frac{n-1}{(R \delta )^2} + \frac{3}{R} {\mathop {\sim }\limits ^{\textstyle <}}\frac{1}{R}, \end{aligned}$$

where we have used the fact that

$$\begin{aligned} \frac{1}{(R \delta )^2} = \frac{1}{R} \frac{K^2}{R} \le \frac{1}{R}. \end{aligned}$$

6.3 The Kakeya Information Underlying the Bilinear Estimate

Suppose \(\delta > 0\), \(R \ge \delta ^{-1}\), and \(J_1\) and \(J_2\) are subsets of the circular arc \(\{ e^{i\theta } : \pi /4 \le \theta \le 3\pi /4 \}\) such that \(\text{ Dist }(J_1,J_2) \ge 3\delta \).

Let \(N_1\) and \(N_2\) be the \(R^{-1}\)-neighborhoods of \(J_1\) and \(J_2\), respectively. In this subsection, we derive the following well-known bound on the Lebesgue measure of the set \((x+N_1) \cap N_2\) for \(x \in \mathbb R^2\).

Lemma 6.3

We have

$$\begin{aligned} |(x+N_1) \cap N_2| \le \frac{\pi }{2 R^2 \delta } \end{aligned}$$
(20)

for a.e. \(x \in \mathbb R^2\).

Proof

Since we are interested in the \(L^\infty \)-norm of the function

$$\begin{aligned} x \longmapsto \int \chi _{x+N_1}(y) \chi _{N_2}(y) dy, \end{aligned}$$

we let \(h \in L^1(\mathbb R^2)\) be a non-negative function and consider the integral

$$\begin{aligned} I= \int \int \chi _{x+N_1}(y) \chi _{N_2}(y) dy h(x) dx. \end{aligned}$$

Writing

$$\begin{aligned} I = \int \int \chi _{N_1}(y-x) \chi _{N_2}(y) h(x) dy dx = \int \chi _{N_2}(y) \int \chi _{N_1}(y-x) h(x) dx dy, \end{aligned}$$

and applying the change of variables \(u=y-x\) in the inner integral, we see that

$$\begin{aligned} I= \int \chi _{N_2}(y) \int \chi _{N_1}(u) h(y-u) du dy = \int _{N_2} \int _{N_1} h(y-u) du dy. \end{aligned}$$

Changing into polar coordinates, this becomes

$$\begin{aligned} I= \int _{1-R^{-1}}^{1+R^{-1}} \int _{1-R^{-1}}^{1+R^{-1}} \int _{\tilde{J_1}} \int _{\tilde{J_2}} h(re^{i\theta }-se^{i\varphi }) r s d\theta d\varphi dr ds, \end{aligned}$$

where \(\tilde{J_1}=N_1 \cap \mathbb S^1\) and \(\tilde{J_2}=N_2 \cap \mathbb S^1\).

We define

$$\begin{aligned} T(\theta ,\varphi ) = re^{i\theta }-se^{i\varphi } = (r \cos \theta - s \cos \varphi , r \sin \theta - s \sin \varphi ). \end{aligned}$$

The Jacobian of this transformation is

$$\begin{aligned} J_T(\theta ,\varphi ) = \left| \begin{array}{cc} -r \sin \theta &{} s \sin \varphi \\ r \cos \theta &{} -s \cos \varphi \end{array} \right| = r s \sin (\theta -\varphi ). \end{aligned}$$

So

$$\begin{aligned} \int _{\tilde{J_1}} \int _{\tilde{J_2}} r s h(re^{i\theta }-se^{i\varphi }) d\theta d\varphi = \int _{\tilde{J_1} \times \tilde{J_2}} \frac{h(T(\theta ,\varphi )) |J_T|}{|\sin (\theta -\varphi )|} d(\theta ,\varphi ). \end{aligned}$$

But \(|\theta -\varphi | \le \pi /2\), so

$$\begin{aligned} |\sin (\theta -\varphi )| \ge \frac{2}{\pi } |\theta -\varphi | \ge \frac{2}{\pi } \text{ Dist }(\tilde{J_1},\tilde{J_2}) \ge \frac{2\delta }{\pi }, \end{aligned}$$

and so

$$\begin{aligned} \int _{\tilde{J_1}} \int _{\tilde{J_2}} r s h(re^{i\theta }-se^{i\varphi }) d\theta d\varphi\le & {} \frac{\pi }{2\delta } \int _{\tilde{J_1} \times \tilde{J_2}} h \circ T(\theta ,\varphi ) |J_T(\theta ,\varphi )| d(\theta ,\varphi ) \\= & {} \frac{\pi }{2\delta } \int _X h(x,y) d(x,y) \; \le \; \frac{\pi }{2\delta } \Vert h \Vert _{L^1}. \end{aligned}$$

Thus

$$\begin{aligned} I \le \int _{1-R^{-1}}^{1+R^{-1}} \int _{1-R^{-1}}^{1+R^{-1}} \frac{\pi }{2\delta } \Vert h \Vert _{L^1} dr ds = \frac{\pi }{2\delta R^2} \Vert h \Vert _{L^1}, \end{aligned}$$

and (20) follows by duality.

7 Proof of Theorem 5.1

As the paragraph following the statement of Theorem 5.1 says, our proof of this theorem relies on ideas from [1, 5, 16, 12], and [4].

7.1 The Bilinear Estimate

Following [1, pp. 1281–1283], we write the ball B(0, R) as a disjoint union of two sets, one broad, the other narrow (see Subsection 7.3 below for the definition of these two sets). To estimate the \(L^p(H dx)\)-norm of Ef on the broad set, we consider a bilinear estimate.

For the rest of the paper, we will use the following notation. If \(\phi \) is a function on \(\mathbb R^2\) and \(\rho > 0\), then \(\phi _\rho \) is the function given by \(\phi _\rho (\cdot ) = \rho ^{-2} \phi (\rho ^{-1} \cdot )\).

Lemma 7.1

Suppose f is supported in an arc I and g is supported in an arc J with \(\sigma (I) \sim \sigma (J) \sim \delta \) and \(\delta \le \text{ Dist }(I,J) \le R^\epsilon \delta \). Also, suppose that

$$\begin{aligned} (10) R^\epsilon \le \frac{1}{\delta } \le \frac{R \delta }{10}. \end{aligned}$$
(21)

Then

$$\begin{aligned} \int _{B(0,R)} |Ef(x) Eg(x)|^{p/2} H(x) dx \le R^\epsilon C_B A_\alpha (H)^\beta R^{\alpha /2} \Vert f \Vert _{L^p(\sigma )}^{p/2} \Vert g \Vert _{L^p(\sigma )}^{p/2}. \end{aligned}$$
(22)

Proof

Let \(\eta \) be a \(C_0^\infty \) function on \(\mathbb R^2\) satisfying \(|\widehat{\eta }| \ge 1\) on B(0, 1). Then

$$\begin{aligned}{} & {} {\int _{B(0,R)} |Ef(x) Eg(x)| H(x) dx \; = \; \int _{B(0,R)} |\widehat{fd\sigma }(x) \widehat{gd\sigma }(x)| H(x) dx} \\{} & {} \quad \le \int _{B(0,R)} |\widehat{fd\sigma }(x) \widehat{gd\sigma }(x)| \, |\widehat{\eta }(x/R)|^2 H(x) dx \\{} & {} \quad = \int _{B(0,R)} |( \eta _{R^{-1}} *fd\sigma )\;\widehat{}\;(x) (\eta _{R^{-1}} *gd\sigma )\;\widehat{}\;(x)| H(x) dx \\{} & {} \quad = \int _{B(0,R)} |\widehat{F}(x) \widehat{G}(x)| H(x) dx, \end{aligned}$$

where \(F=\eta _{R^{-1}} *fd\sigma \) and \(G=\eta _{R^{-1}} *gd\sigma \).

Applying the Cauchy-Schwarz inequality in the convolution integral with respect to the measure \(|\eta _{R^{-1}}(\xi -\cdot )|d\sigma \), we see that

$$\begin{aligned} \Vert F \Vert _{L^2}^2&\le \!\! \int \Big ( \int |f(\theta )|^2 \, |\eta _{R^{-1}}(\xi -\theta )| d\sigma (\theta ) \Big ) \Big ( \int |\eta _{R^{-1}}(\xi -\theta )| d\sigma (\theta ) \Big ) d\xi \\&{\mathop {\sim }\limits ^{\textstyle <}} \!\! R \int \int |f(\theta )|^2 \, |\eta _{R^{-1}}(\xi -\theta )| d\sigma (\theta ) d\xi \\&= \!\! R \int |f(\theta )|^2 \int |\eta _{R^{-1}}(\xi -\theta )| d\xi d\sigma (\theta ) \; = \; R \, \Vert \eta \Vert _{L^1} \, \Vert f \Vert _{L^2(\sigma )}^2, \end{aligned}$$

where in the second inequality we used the fact that

$$\begin{aligned} \int |\eta _{R^{-1}}(\xi -\theta )|d\sigma (\theta ) {\mathop {\sim }\limits ^{\textstyle<}}R^2 \sigma (B(\xi ,R^{-1}) {\mathop {\sim }\limits ^{\textstyle <}}R. \end{aligned}$$

Therefore,

$$\begin{aligned} \Vert F \Vert _{L^2} {\mathop {\sim }\limits ^{\textstyle<}}R^{1/2} \Vert f \Vert _{L^2(\sigma )} \hspace{0.25in} \text{ and } \hspace{0.25in} \Vert G \Vert _{L^2} {\mathop {\sim }\limits ^{\textstyle <}}R^{1/2} \Vert g \Vert _{L^2(\sigma )}. \end{aligned}$$
(23)

Since F is supported in the \(R^{-1}\)-neighborhood of I and G is supported in the \(R^{-1}\)-neighborhood of J, we see (via (21)) that F is supported in a ball of radius \((\delta /2) + (\delta /10) = (3\delta /5)\) and similarly for G. So \(F *G\) is supported in a ball of radius \((6\delta /5)\), say \(B(\xi _0,(6\delta /5))\). Via the locally constant property of the Fourier transform, this fact tells us that the Fourier transform of \(F *G\) is essentially constant at scale \(K=\delta ^{-1}\), and hence allows us to implement the localization of the weight argument that we described in Section 5 at the intuitive level, and which we now carry out rigorously.

Let \(\phi \) be a Schwartz function which is equal to 1 on B(0, 6/5). Then \(\phi _\delta (\xi -\xi _0)=\delta ^{-2}\) on \(B(\xi _0,\frac{6\delta }{5})\), so that

$$\begin{aligned} F *G = \delta ^2 \phi _\delta (\cdot - \xi _0) \big ( F *G \big ) \end{aligned}$$

and

$$\begin{aligned} \widehat{F}(x) \widehat{G}(x) = \delta ^2 \Big ( \phi _\delta (\cdot - \xi _0) \big ( F *G \big ) \widehat{\Big )}(x) = \delta ^2 \big ( \phi _\delta (\cdot - \xi _0) \widehat{\big )} *\widehat{F *G}(x). \end{aligned}$$

Since \(\big ( \phi _\delta (\cdot - \xi _0) \widehat{\big )}(x)= e^{-2\pi i x \cdot \xi _0} \widehat{\phi }(\delta x)\), it follows that

$$\begin{aligned} \widehat{F}(x) \widehat{G}(x)= & {} \delta ^2 \int \big ( \phi _\delta (\cdot - \xi _0) \widehat{\big )}(x-y) \widehat{F *G}(y) dy \\= & {} \delta ^2 \int e^{-2\pi i (x-y) \cdot \xi _0} \widehat{\phi }(\delta (x-y)) \widehat{F *G}(y) dy, \end{aligned}$$

so that

$$\begin{aligned} |\widehat{F}(x) \widehat{G}(x)| \le \delta ^2 \int |\widehat{\phi }(\delta (x-y))| \, |\widehat{F *G}(y)| dy. \end{aligned}$$

Therefore,

$$\begin{aligned} \int _{B(0,R)} \!\! |Ef(x) Eg(x)| H(x) dx \le \delta ^2 \int |\widehat{F *G}(y)| \!\! \int |\widehat{\phi }(\delta (x-y))| H(x) dx dy.\nonumber \\ \end{aligned}$$
(24)

For \(l = 0, 1, 2, \ldots \), we let \(B_l=B(y,2^l \delta ^{-1})\) and write

$$\begin{aligned}{} & {} {\int |\widehat{\phi }(\delta (x-y))| H(x) dx} \\{} & {} \quad = \int _{B_0} |\widehat{\phi }(\delta (x-y))| H(x) dx + \sum _{l=1}^\infty \int _{B_l \setminus B_{l-1}} |\widehat{\phi }(\delta (x-y))| H(x) dx \\{} & {} \quad \le \int _{B_0} \frac{C_N H(x)}{(1+\delta |x-y|)^N} dx + \sum _{l=1}^\infty \int _{B_l \setminus B_{l-1}} \frac{C_N H(x)}{(1+\delta |x-y|)^N} dx \\{} & {} \quad \le C_N \int _{B_0} H(x) dx + \sum _{l=1}^\infty \frac{C_N}{\big ( 1+\delta \frac{2^{l-1}}{\delta } \big )^N} \int _{B_l} H(x) dx. \end{aligned}$$

We now let \(0 \le \theta \le 1\) be a parameter that will be determined later and write

$$\begin{aligned} \int _{B_l} H(x) dx= & {} \Big ( \int _{B_l} H(x) dx \Big )^{1-\theta } \Big ( \int _{B_l} H(x) dx \Big )^\theta \\\le & {} |B_l|^{1-\theta } \Big ( A_\alpha (H) \Big ( \frac{2^l}{\delta } \Big )^\alpha \Big )^\theta \\\le & {} C_\theta \Big ( \frac{2^l}{\delta } \Big )^{2(1-\theta )+\alpha \theta } A_\alpha (H)^\theta , \end{aligned}$$

where we have used the fact that \(1/\delta \ge 1\), and we obtain

$$\begin{aligned}{} & {} {\int |\widehat{\phi }(\delta (x-y))| H(x) dx} \\{} & {} \quad \le C_{N,\theta } \Big ( \frac{1}{\delta } \Big )^{n(1-\theta )+\alpha \theta } A_\alpha (H)^\theta + \sum _{l=1}^\infty \frac{C_N}{(1+2^{l-1})^N} \Big ( \frac{2^l}{\delta } \Big )^{2(1-\theta )+\alpha \theta } A_\alpha (H)^\theta \\{} & {} \quad \le C_{N,\theta } A_\alpha (H)^\theta \Big ( \frac{1}{\delta } \Big )^{2(1-\theta )+\alpha \theta }. \end{aligned}$$

Also,

$$\begin{aligned} \int _{B(x_0,r)} \int |\widehat{\phi }(\delta (x-y))| H(x) dx dy = \int \int \chi _{B(x_0,r)}(y) |\widehat{\phi }(\delta (x-y))| dy H(x) dx. \end{aligned}$$

Applying the change of variables \(z=\delta (x-y)\) in the inner integral, we get

$$\begin{aligned} \int _{B(x_0,r)} \int |\widehat{\phi }(\delta (x-y))| H(x) dx dy= & {} \frac{1}{\delta ^2} \int \int \chi _{B(x_0,r)} \big ( x-\frac{z}{\delta } \big ) |\widehat{\phi }(z)| dz H(x) dx \\= & {} \frac{1}{\delta ^2} \int |\widehat{\phi }(z)| \int \chi _{B(x_0,r)} \big ( x-\frac{z}{\delta } \big ) H(x) dx dz. \end{aligned}$$

But

$$\begin{aligned} \int \chi _{B(x_0,r)} \big ( x-\frac{z}{\delta } \big ) H(x) dx = \int _{B(x_0+\frac{z}{\delta },r)} H(x) dx \le A_\alpha (H) r^\alpha \end{aligned}$$

for all \(x_0 \in \mathbb R^n\) and \(r \ge 1\), so

$$\begin{aligned} \int _{B(x_0,r)} \int |\widehat{\phi }(\delta (x-y))| H(x) dx dy \le \frac{1}{\delta ^2} \, \Vert \widehat{\phi } \Vert _{L^1} \, A_\alpha (H) \, r^\alpha \end{aligned}$$

for all \(x_0 \in \mathbb R^2\) and \(r \ge 1\).

For \(y \in \mathbb R^2\), define

$$\begin{aligned} {\mathcal H}(y)= \frac{\delta ^{2(1-\theta )+\alpha \theta }}{C_{N,\theta } A_\alpha (H)^\theta } \int |\widehat{\phi }(\delta (x-y))| H(x) dx. \end{aligned}$$

In view of the above discussion, we have

$$\begin{aligned} \Vert {\mathcal H} \Vert _{L^\infty } \le 1 \hspace{0.5in} \text{ and } \hspace{0.5in} \int _{B(x_0,r)} {\mathcal H}(y) dy \le C A_\alpha ^{1-\theta } \delta ^{(\alpha -2)\theta } r^\alpha \end{aligned}$$

for all \(x_0 \in \mathbb R^2\) and \(r \ge 1\). Thus \({\mathcal H}\) is a weight on \(\mathbb R^2\) of fractal dimension \(\alpha \) with

$$\begin{aligned} A_\alpha ({\mathcal H}) \le C A_\alpha (H)^{1-\theta } \delta ^{(\alpha -2)\theta }. \end{aligned}$$

Going back to (24), we now have

$$\begin{aligned} \int _{B(0,R)} |Ef(x) Eg(x)| H(x) dx\le & {} \delta ^2 \frac{C_{N,\theta } A_\alpha (H)^\theta }{\delta ^{2(1-\theta )+\alpha \theta }} \int |\widehat{F *G}(y)| {\mathcal H}(y) dy \nonumber \\= & {} C_{N,\theta } \, \delta ^{(2-\alpha )\theta } A_\alpha (H)^\theta \int |\widehat{F *G}(y)| {\mathcal H}(y) dy.\nonumber \\ \end{aligned}$$
(25)

Next, we let \(Q^*\) be the box in frequency space (where the circle is located) of dimensions \((R\delta )^{-1} \times R^{-1}\), centered at the origin, and with the \((R\delta )^{-1}\)-side (i.e. the long side) parallel to the line segment that connects the midpoint of I to that of J. We also let \(\{ Q_l \}\) be a tiling of \(\mathbb R^2\) by boxes dual to \(Q^*\) (i.e. each \(Q_l\) is an \(R\delta \times R\) box whose \(R\delta \)-side is parallel to the \((R\delta )^{-1}\)-side of \(Q^*\)) with centers \(\{ \nu _l \}\), \(\psi \) be a \(C_0^\infty \) function on \(\mathbb R^2\), and we define

$$\begin{aligned} \psi _l(\xi )= (R\delta ) R \; \psi (R\delta \xi _1, R\xi _2) \; e^{2\pi i \nu _l \cdot \xi }. \end{aligned}$$

In the definition of \(\psi _l\), we are assuming that the line joining the midpoint of I to that of J is horizontal (i.e. parallel to the \(\xi _1\)-axis). This assumption makes the presentation a little smoother and, of course, does not cost us any loss of generality.

We assume further that the Fourier transform of \(\psi \) is non-negative and satisfies \(\widehat{\psi } \ge 1/2\) on \([-1/2,1/2] \times [-1/2,1/2]\). Then

$$\begin{aligned} \widehat{\psi _l}(x) = \widehat{\psi } \Big ( \frac{x_1-\nu _{l,1}}{R\delta }, \frac{x_2-\nu _{l,2}}{R} \Big ) \ge \frac{1}{2} \hspace{0.25in} \text{ if } \hspace{0.25in} x \in Q_l. \end{aligned}$$

By the Schwartz decay of \(\widehat{\psi }\), we have \(\sum _{m \in \mathbb Z^2} \widehat{\psi }(\cdot - m)^k {\mathop {\sim }\limits ^{\textstyle <}}1\) for any \(k \in \mathbb N\). Also, \(\{ \nu _l \}\) is basically \(R\delta \mathbb Z \times R \mathbb Z\), so

$$\begin{aligned} \sum _{l=1}^\infty \widehat{\psi _l}(R\delta x_1, R x_2)^k = \sum _{l=1}^\infty \widehat{\psi } \Big ( \frac{R\delta x_1-\nu _{l,1}}{R\delta }, \frac{Rx_2-\nu _{l,2}}{R} \Big )^k = \sum _{m \in \mathbb Z^2} \widehat{\psi }(x - m)^k {\mathop {\sim }\limits ^{\textstyle <}}1, \end{aligned}$$

and so

$$\begin{aligned} \sum _{l=1}^\infty \widehat{\psi _l}(x)^k {\mathop {\sim }\limits ^{\textstyle <}}1 \end{aligned}$$
(26)

for all \(x \in \mathbb R^2\).

Going back to (25), we can now write

$$\begin{aligned} \int _{B(0,R)} |Ef(x) Eg(x)| H(x) dx {\mathop {\sim }\limits ^{\textstyle <}}\delta ^{(2-\alpha )\theta } A_\alpha (H)^\theta \sum _{l=1}^\infty \int |\widehat{F}(x) \widehat{G}(x)| \widehat{\psi _l}(x)^3 {\mathcal H}(x) dx. \end{aligned}$$

Letting \(F_l=\psi _l *F\) and \(G_l=\psi _l *G\), this becomes

$$\begin{aligned} \int _{B(0,R)} |Ef(x) Eg(x)| H(x) dx {\mathop {\sim }\limits ^{\textstyle <}}\delta ^{(2-\alpha )\theta } A_\alpha (H)^\theta \sum _{l=1}^\infty \int |\widehat{F_l}(x) \widehat{G_l}(x)| \widehat{\psi _l}(x) {\mathcal H}(x) dx. \end{aligned}$$

By Cauchy–Schwarz,

$$\begin{aligned} \int |\widehat{F_l}(x) \widehat{G_l}(x)| \widehat{\psi _l}(x) {\mathcal H}(x) dx \le \Vert \widehat{F_l} \widehat{G_l} \Vert _{L^2} \Vert \widehat{\psi _l}(x) {\mathcal H} \Vert _{L^2}. \end{aligned}$$

Applying (19) from Subsection 6.1 with \(n=2\) and \(K = \delta ^{-1}\), we have

$$\begin{aligned}{} & {} {\int \widehat{\psi _l}(x)^2 {\mathcal H}(x)^2 dx \; {\mathop {\sim }\limits ^{\textstyle<}}\; \int \widehat{\psi _l}(x) {\mathcal H}(x) dx \; {\mathop {\sim }\limits ^{\textstyle<}}\; A_\alpha ({\mathcal H}) \frac{R}{R\delta } (R\delta )^\alpha } \\{} & {} {\mathop {\sim }\limits ^{\textstyle <}}\; A_\alpha (H)^{1-\theta } \delta ^{(\alpha -2)\theta } R^\alpha \delta ^{\alpha -1} \; = \; A_\alpha (H)^{1-\theta } \delta ^{(\alpha -2)\theta +\alpha -1} R^\alpha , \end{aligned}$$

so that

$$\begin{aligned} \Vert \widehat{\psi _l}(x) {\mathcal H} \Vert _{L^2} {\mathop {\sim }\limits ^{\textstyle <}}A_\alpha (H)^{(1-\theta )/2} \delta ^{((\alpha -2)\theta +\alpha -1)/2} R^{\alpha /2}. \end{aligned}$$

Therefore,

$$\begin{aligned} \int _{B(0,R)} |Ef(x) Eg(x)| H(x) dx {\mathop {\sim }\limits ^{\textstyle <}}A_\alpha (H)^{(1+\theta )/2} \delta ^{((2-\alpha )\theta +\alpha -1)/2} R^{\alpha /2} \sum _{l=1}^\infty \Vert \widehat{F_l} \widehat{G_l} \Vert _{L^2}. \end{aligned}$$

Letting \(\beta = (1+\theta )/2\) (since \(0 \le \theta \le 1\), we have \(1/2 \le \beta \le 1\)), this becomes

$$\begin{aligned} \int _{B(0,R)} |Ef(x) Eg(x)| H(x) dx {\mathop {\sim }\limits ^{\textstyle <}}A_\alpha (H)^\beta \delta ^{(2-\alpha )\beta +\alpha -(3/2)} R^{\alpha /2} \sum _{l=1}^\infty \Vert \widehat{F_l} \widehat{G_l} \Vert _{L^2}. \end{aligned}$$

We now let \(A_l\) be the support of \(F_l\), \(B_l\) be the support of \(G_l\), and define the function \(\lambda _l : \mathbb R^2 \rightarrow [0,\infty )\) by \(\lambda _l(\xi )= |(\xi -A_l) \cap B_l|\). Applying Plancherel’s theorem followed by Cauchy–Schwarz, we see that

$$\begin{aligned} \Vert \widehat{F_l} \widehat{G_l} \Vert _{L^2}^2 = \int |F_l *G_l(\xi )|^2 d\xi \le \Vert \lambda _l \Vert _{L^\infty } \int |F_l|^2 *|G_l|^2(\xi ) d\xi . \end{aligned}$$

By Young’s inequality,

$$\begin{aligned} \int |F_l|^2 *|G_l|^2(\xi ) d\xi \le \Vert |F_l|^2 \Vert _{L^1} \Vert |G_l|^2 \Vert _{L^1} = \Vert F_l \Vert _{L^2}^2 \Vert G_l \Vert _{L^2}^2, \end{aligned}$$

so the only problem is to estimate \(\Vert \lambda _l \Vert _{L^\infty }\). We will do this by using the Kakeya bound (20) of Subsection 6.3.

Our assumptions on the arcs I and J imply that the angle between any two points in \(I \cup J\) is \({\mathop {\sim }\limits ^{\textstyle <}}R^\epsilon \delta \). Also, for each l, the function \(\psi _l\) is supported in the \((R\delta )^{-1} \times R^{-1}\) box \(Q^*\) of center (0, 0) and with the long side parallel to the line joining the midpoints of I and J. So, if \(e \in I \cup J\), then the translate \(Q^*+e\) of \(Q^*\) is contained in an \((R \delta )^{-1} \times R^{\epsilon -1}\) box with the \((R \delta )^{-1}\)-side tangent to \(\mathbb S^1\) at e. Therefore, the property of boxes of this form that was presented in Subsection 6.2 tells us that \(Q^*+e\) is contained in the \(R^{\epsilon -1}\)-neighborhood of \(\mathbb S^1\). Therefore, the sets \(A_l\) and \(B_l\) satisfy the requirements needed for us to apply (20) and conclude

$$\begin{aligned} \Vert \lambda _l \Vert _{L^\infty } {\mathop {\sim }\limits ^{\textstyle <}}\frac{R^\epsilon }{R^2\delta }. \end{aligned}$$

Putting together what we have proved in the previous two paragraphs, we obtain

$$\begin{aligned} \Vert \widehat{F_l} \widehat{G_l} \Vert _{L^2}^2 {\mathop {\sim }\limits ^{\textstyle <}}\frac{R^\epsilon }{R^2\delta } \, \Vert F_l \Vert _{L^2}^2 \Vert G_l \Vert _{L^2}^2, \end{aligned}$$

and hence

$$\begin{aligned}{} & {} {\int _{B(0,R)} |Ef(x) Eg(x)| H(x) dx} \\{} & {} \quad {\mathop {\sim }\limits ^{\textstyle <}}R^\epsilon A_\alpha (H)^\beta \delta ^{(2-\alpha )\beta +\alpha -(3/2)} \frac{R^{\alpha /2}}{(R^2 \delta )^{1/2}} \sum _{l=1}^\infty \Vert F_l \Vert _{L^2} \Vert G_l \Vert _{L^2} \\{} & {} \quad = R^\epsilon A_\alpha (H)^\beta \delta ^{(2-\alpha )(\beta -1)} \frac{R^{\alpha /2}}{R} \sum _{l=1}^\infty \Vert F_l \Vert _{L^2} \Vert G_l \Vert _{L^2}. \end{aligned}$$

By Cauchy–Schwarz and Plancherel,

$$\begin{aligned} \sum _{l=1}^\infty \Vert F_l \Vert _{L^2} \Vert G_l \Vert _{L^2} \le \Big ( \sum _{l=1}^\infty \Vert \widehat{F_l} \Vert _{L^2}^2 \Big )^{1/2} \Big ( \sum _{l=1}^\infty \Vert \widehat{G_l} \Vert _{L^2}^2 \Big )^{1/2}. \end{aligned}$$

Also, by (26),

$$\begin{aligned} \sum _{l=1}^\infty \Vert \widehat{F_l} \Vert _{L^2}^2 = \int |\widehat{F}(x)|^2 \sum _{l=1}^\infty \widehat{\psi _l}(x)^2 dx {\mathop {\sim }\limits ^{\textstyle <}}\Vert \widehat{F} \Vert _{L^2}^2 = \Vert F \Vert _{L^2}^2 \end{aligned}$$

and similarly for \(\sum _{l=1}^\infty \Vert \widehat{G_l} \Vert _{L^2}^2\), so

$$\begin{aligned} \int _{B(0,R)} |Ef(x) Eg(x)| H(x) dx {\mathop {\sim }\limits ^{\textstyle <}}R^\epsilon A_\alpha (H)^\beta \delta ^{(2-\alpha )(\beta -1)} \frac{R^{\alpha /2}}{R} \Vert F \Vert _{L^2} \Vert G \Vert _{L^2}. \end{aligned}$$

Recalling (23), our bilinear estimate becomes

$$\begin{aligned} \int _{B(0,R)} |Ef(x) Eg(x)| H(x) dx {\mathop {\sim }\limits ^{\textstyle <}}R^\epsilon A_\alpha (H)^\beta \delta ^{(2-\alpha )(\beta -1)} R^{\alpha /2} \Vert f \Vert _{L^2(\sigma )} \Vert g \Vert _{L^2(\sigma )}. \end{aligned}$$

Writing

$$\begin{aligned}{} & {} {\int _{B(0,R)} |Ef(x) Eg(x)|^{p/2} H(x) dx} \\{} & {} \quad = \int _{B(0,R)} |Ef(x) Eg(x)|^{(p/2)-1} |Ef(x) Eg(x)| H(x) dx \\{} & {} \quad \le \Vert f \Vert _{L^1(S)}^{(p/2)-1} \Vert g \Vert _{L^1(S)}^{(p/2)-1} \int _{B(0,R)} |Ef(x) Eg(x)| H(x) dx \\{} & {} \quad \le C_B R^\epsilon A_\alpha (H)^\beta R^{\alpha /2} \delta ^{(2-\alpha )(\beta -1)} \Vert f \Vert _{L^1(\sigma )}^{(p/2)-1} \Vert f \Vert _{L^2(\sigma )} \Vert g \Vert _{L^1(\sigma )}^{(p/2)-1} \Vert g \Vert _{L^2(\sigma )} \end{aligned}$$

and applying (33) (see the appendix), we arrive at our desired bilinear estimate

$$\begin{aligned} \int _{B(0,R)} |Ef(x) Eg(x)|^{p/2} H(x) dx \le R^\epsilon C_B A_\alpha (H)^\beta R^{\alpha /2} \Vert f \Vert _{L^p(\sigma )}^{p/2} \Vert g \Vert _{L^p(\sigma )}^{p/2}. \end{aligned}$$

7.2 The Linear Estimate

In this subsection, we work in \(\mathbb R^n\) with \(n \ge 2\).

Lemma 7.2

Suppose f is supported in a cap of radius \(\delta /2\). Also, suppose that

$$\begin{aligned} (10) R^\epsilon \le \frac{1}{\delta } \le \frac{R}{10}. \end{aligned}$$
(27)

Then

$$\begin{aligned} \int _{B(0,R)} |Ef(x)|^p H(x) dx \le C_L A_\alpha (H)^\beta \delta ^{-2\alpha /n} (\delta ^2 R) \Vert f \Vert _{L^p(\sigma )}^p. \end{aligned}$$
(28)

Proof

Let \(\eta \) be a \(C_0^\infty \) function on \(\mathbb R^n\) satisfying \(|\widehat{\eta }| \ge 1\) on B(0, 1), and \(F=\eta _{R^{-1}} *f d\sigma \). Then

$$\begin{aligned} \int _{B(0,R)} |Ef(x)|^2 H(x)dx \le \int _{B(0,R)} |\widehat{F}(x)|^2 H(x)dx. \end{aligned}$$

Also, let \(\psi \) be a \(C_0^\infty \) function on \(\mathbb R^n\), and \(\{ B_l \}\) be a finitely overlapping cover of \(\mathbb R^n\) by balls dual to \(B(0,\delta )\) (i.e. \(\delta ^{-1}\)-balls) with centers \(\{ \nu _l \}\), and set

$$\begin{aligned} \psi _l(\xi )=\delta ^{-n} \psi (\delta ^{-1} \xi ) \, e^{2\pi i \nu _l \cdot \xi }. \end{aligned}$$

We assume further that \(\widehat{\psi }\) is non-negative and \(\ge 1/2\) on the unit ball. Then

$$\begin{aligned} \widehat{\psi _l}(x)= \widehat{\psi }(\delta (x-\tau _l)) \ge \frac{1}{2} \end{aligned}$$

if \(|\delta (x-\tau _l)| \le 1\), i.e. if \(x \in B_l\). Thus

$$\begin{aligned} \int _{B(0,R)} |Ef(x)|^2 H(x)dx {\mathop {\sim }\limits ^{\textstyle <}}\sum _{l=1}^\infty \int |\widehat{F}(x) \widehat{\psi _l}(x)|^2 \widehat{\psi _l}(x) H(x) dx. \end{aligned}$$

Since \(1/n \le \beta \le 2/n\), we can apply Hölder’s inequality with the dual exponents \(1/(1-\beta )\) and \(1/\beta \) to get

$$\begin{aligned} \int _{B(0,R)} |Ef(x)|^2 H(x)dx {\mathop {\sim }\limits ^{\textstyle <}}\sum _{l=1}^\infty \Vert \widehat{F *\psi _l} \Vert _{L^{2/(1-\beta )}}^2 \Vert \widehat{\psi _l} H \Vert _{L^{1/\beta }}. \end{aligned}$$

Since \(\Vert H \Vert _{L^\infty } \le 1\), we have

$$\begin{aligned} \Vert \widehat{\psi _l} H \Vert _{L^{1/\beta }}^{1/\beta } \le \int \widehat{\psi _l}(x)^{1/\beta } H(x) dx, \end{aligned}$$

and hence (by the proof of (19))

$$\begin{aligned} \Vert \widehat{\psi _l} H \Vert _{L^{1/\beta }}^{1/\beta } {\mathop {\sim }\limits ^{\textstyle <}}A_\alpha (H) \Big ( \frac{1}{\delta } \Big )^\alpha . \end{aligned}$$

Also, by Hausdorff–Young,

$$\begin{aligned} \Vert \widehat{F *\psi _l} \Vert _{L^{2/(1-\beta )}} \le \Vert F *\psi _l \Vert _{L^{2/(1+\beta )}}. \end{aligned}$$

Therefore,

$$\begin{aligned} \int _{B(0,R)} |Ef(x)|^2 H(x) dx {\mathop {\sim }\limits ^{\textstyle <}}A_\alpha (H)^\beta \delta ^{-\alpha \beta } \sum _{l=1}^\infty \Vert F *\psi _l \Vert _{L^{2/(1+\beta )}}^2. \end{aligned}$$

Since (27) tells us \(1/R \le \delta /10\), it follows that F is supported in a ball of radius \((\delta /2)+(\delta /10)= (3/5) \delta \), say \(B(\xi _0,3\delta /5)\). Moreover, since \(\psi _l\) is supported in \(B(0,\delta )\), it follows by Hölder’s inequality and Plancherel’s theorem that

$$\begin{aligned} \Vert F *\psi _l \Vert _{L^{2/(1+\beta )}}^2 {\mathop {\sim }\limits ^{\textstyle <}}\delta ^{n \beta } \Vert F *\psi _l \Vert _{L^2}^2 = \delta ^{n \beta } \Vert \widehat{F} \widehat{\psi _l} \Vert _{L^2}^2. \end{aligned}$$

Thus

$$\begin{aligned} \int _{B(0,R)} |Ef(x)|^2 H(x) dx&{\mathop {\sim }\limits ^{\textstyle<}} A_\alpha (H)^\beta \delta ^{-\alpha \beta } \delta ^{n \beta } \sum _{l=1}^\infty \int |\widehat{F}(\xi ) \widehat{\psi _l}(\xi )|^2 d\xi \\&= A_\alpha (H)^\beta \delta ^{(n-\alpha )\beta } \int |\widehat{F}(\xi )|^2 \sum _{l=1}^\infty |\widehat{\psi _l}(\xi )|^2 d\xi \\&{\mathop {\sim }\limits ^{\textstyle <}} A_\alpha (H)^\beta \delta ^{(n-\alpha )(\beta -(2/n))} \delta ^{2-(2\alpha /n)} \Vert F \Vert _{L^2}^2. \end{aligned}$$

But we know from (23) (whose proof shows that it is true in \(\mathbb R^n\) for all \(n \ge 2\)) that \(\Vert F \Vert _{L^2} {\mathop {\sim }\limits ^{\textstyle <}}\sqrt{R} \, \Vert f \Vert _{L^2(\sigma )}\), so

$$\begin{aligned} \int _{B(0,R)} |Ef(x)|^2 H(x) dx {\mathop {\sim }\limits ^{\textstyle <}}A_\alpha (H)^\beta \delta ^{-2 \alpha /n} (\delta ^2 R) \delta ^{(n-\alpha )(\beta -(2/n))} \Vert f \Vert _{L^2(\sigma )}^2. \end{aligned}$$

Writing

$$\begin{aligned} |Ef(x)|^p= |Ef(x)|^{p-2} |Ef(x)|^2 \le \Vert f \Vert _{L^1(\sigma )}^{p-2} |Ef(x)|^2 \end{aligned}$$

and using (32) (see the appendix), we now see that

$$\begin{aligned}{} & {} {\int _{B(0,R)} |Ef(x)|^p H(x) dx} \\{} & {} \quad {\mathop {\sim }\limits ^{\textstyle<}}A_\alpha (H)^\beta \delta ^{-2 \alpha /n} (\delta ^2 R) \delta ^{(n-\alpha )(\beta -(2/n))} \Vert f \Vert _{L^1(\sigma )}^{p-2} \Vert f \Vert _{L^2(\sigma )}^2 \\{} & {} \quad {\mathop {\sim }\limits ^{\textstyle <}}A_\alpha (H)^\beta \delta ^{-2 \alpha /n} (\delta ^2 R) \Vert f \Vert _{L^p(\sigma )}^p, \end{aligned}$$

which proves (28).

7.3 The Recursive Process

We let \(0< \epsilon < 10^{-2}\) and \(R \ge 1\) be two numbers satisfying \(R \ge (1000)^{1/(1-4\epsilon )}\). We also let \(\delta \) be as in Lemma 7.1 (so that \(\delta \) obeys (21)). We’re going to prove our estimate by implementing a recursive process over \(\delta \).

Base of the recursion: Here \(\delta = R^{-1/2}\). Plugging this value of \(\delta \) into (28) in dimension \(n=2\), we get

$$\begin{aligned} \int _{B(0,R)} |Ef(x)|^p H(x) dx \le C_L A_\alpha (H)^\beta R^{\alpha /2} \Vert f \Vert _{L^p(\sigma )}^p. \end{aligned}$$

The recursive step: We state this in the following lemma.

Lemma 7.3

Suppose that the estimate

$$\begin{aligned} \int _{B(0,R)} |Ef(x)|^p H(x) dx \le C R^\epsilon A_\alpha (H)^\beta R^{\alpha /2} \Vert f \Vert _{L^p(\sigma )}^p \end{aligned}$$
(29)

holds for every function \(f \in L^1(\sigma )\) that is supported in an arc of \(\sigma \)-measure \(\le \delta \), and \(\delta \) obeys (21). Then the estimate

$$\begin{aligned} \int _{B(0,R)} |Eg(x)|^p H(x) dx \le C' R^\epsilon A_\alpha (H)^\beta R^{\alpha /2} \Vert g \Vert _{L^p(\sigma )}^p \end{aligned}$$
(30)

holds for every function \(g \in L^1(\sigma )\) that is supported in an arc of \(\sigma \)-measure \(\le R^\epsilon \delta \), where

$$\begin{aligned} C'= 3^p C + (10)^p R^{(p+2)\epsilon } C_B. \end{aligned}$$

Proof

Suppose \(\delta \) satisfies the condition (21):

$$\begin{aligned} (10) R^\epsilon \le \frac{1}{\delta } \le \frac{R \delta }{10}, \end{aligned}$$

and (29) is true whenever \(f \in L^1(\sigma )\), f is supported on an arc \(I_\delta \subset \mathbb S^1\), and \(\sigma (I_\delta ) \le \delta \). We need to show that (30) is true whenever \(g \in L^1(\sigma )\), g is supported on an arc \(I_{R^\epsilon \delta } \subset \mathbb S^1\), and \(\sigma (I_{R^\epsilon \delta }) \le R^\epsilon \delta \), where

$$\begin{aligned} C'= 3^p C + (10)^p R^{(p+2)\epsilon } C_B. \end{aligned}$$

We let \(K = R^\epsilon \) and cover the support of g by K arcs \(\tau \) each of measure \(\delta \). We then write \(g= \sum _{\tau } f_\tau \) with each function \(f_\tau \) supported in the arc \(\tau \).

Following [1] and [7], for \(x \in \mathbb R^2\), we define the significant set of x by

$$\begin{aligned} S(x)= \{ \tau : |Ef_\tau (x)| \ge \frac{1}{10 K} |Eg(x)| \}. \end{aligned}$$

Then

$$\begin{aligned} |Eg(x)| \le \Big | \sum _{\tau \in S(x)} Ef_\tau (x) \Big | + \frac{1}{10} |Eg(x)|, \end{aligned}$$

so that

$$\begin{aligned} |Eg(x)| \le \frac{10}{9} \Big | \sum _{\tau \in S(x)} Ef_\tau (x) \Big |. \end{aligned}$$
(31)

The narrow set \({\mathcal N}\) and the broad set \({\mathcal B}\) are now defined as

$$\begin{aligned} {\mathcal N} = B(0,R) \cap \{ x \in \mathbb R^2 : \# S(x) \le 2 \} \hspace{0.25in} \text{ and } \hspace{0.25in} {\mathcal B} = B(0,R) \setminus {\mathcal N}. \end{aligned}$$

We will estimate \(\int _{\mathcal N} |Eg(x)|^p H(x) dx\) by induction and \(\int _{\mathcal B} |Eg(x)|^p H(x) dx\) by using the bilinear estimate.

By (29) and (31),

$$\begin{aligned} \int _{\mathcal N} |Eg(x)|^p H(x) dx\le & {} 2^{p-1} \Big ( \frac{10}{9} \Big )^p \int _N \sum _{\tau \in S(x)} |Ef_\tau (x)|^p H(x) dx \\\le & {} \Big ( \frac{20}{9} \Big )^p \int _N \sum _\tau |Ef_\tau (x)|^p H(x) dx \\\le & {} 3^p \sum _\tau C R^\epsilon A_\alpha (H)^\beta R^{\alpha /2} \Vert f_\tau \Vert _{L^p(\sigma )}^p \\= & {} 3^p C R^\epsilon A_\alpha (H)^\beta R^{\alpha /2} \Vert g \Vert _{L^p(\sigma )}^p. \end{aligned}$$

To every \(x \in {\mathcal B}\) there are two caps \(\tau _x, \tau _x' \in S(x)\) so that \(\text{ Dist }(\tau _x,\tau _x') \ge \delta \). Writing

$$\begin{aligned} |Eg(x)|^p= |Eg(x)|^{p/2} |Eg(x)|^{p/2} \le (10 K |Ef_{\tau _x}(x)|)^{p/2} (10 K |Ef_{\tau _x'}(x)|)^{p/2}, \end{aligned}$$

we see that

$$\begin{aligned} |Eg(x)|^p \le (10 K)^p \sum _{\tau , \tau ': \, \text{ Dist }(\tau ,\tau ') \ge \delta } |Ef_\tau (x)|^{p/2} |Ef_{\tau '}(x)|^{p/2}. \end{aligned}$$

Using the bilinear estimate (22), it follows that

$$\begin{aligned}{} & {} {\int _{\mathcal B} |Eg(x)|^p H(x) dx} \\{} & {} \quad \le (10 K)^p \sum _{\tau , \tau ': \, \text{ Dist }(\tau ,\tau ') \ge \delta } \int _{\mathcal B} |Ef_\tau (x)|^{p/2} |Ef_{\tau '}(x)|^{p/2} H(x) dx \\{} & {} \quad \le (10 K)^p C_B R^\epsilon A_\alpha (H)^\beta R^{\alpha /2} \sum _{\tau , \tau ': \, \text{ Dist }(\tau ,\tau ') \ge \delta } \Vert f_\tau \Vert _{L^p(\sigma )}^{p/2} \Vert f_{\tau '} \Vert _{L^p(\sigma )}^{p/2} \\{} & {} \quad \le (10)^p K^p C_B R^\epsilon A_\alpha (H)^\beta R^{\alpha /2} \sum _{\tau , \tau ': \, \text{ Dist }(\tau ,\tau ') \ge \delta } \Vert g \Vert _{L^p(\sigma )}^{p/2} \Vert g \Vert _{L^p(\sigma )}^{p/2}. \end{aligned}$$

Therefore,

$$\begin{aligned} \int _{\mathcal B} |Eg(x)|^p H(x) dx \le (10)^p K^{p+2} C_B A_\alpha (H)^\beta R^{\alpha /2} \Vert g \Vert _{L^p(\sigma )}^p. \end{aligned}$$

Combining the narrow and broad estimates, we arrive at (30).

The recursion: Starting with the base of the induction, where \(\delta = R^{-1/2}\) and \(C=C_L\), and applying Lemma 7.3k times, we arrive at an estimate that holds for every function \(f \in L^1(\sigma )\) that is supported on an arc of \(\sigma \)-measure \(\le \delta _k = R^{k \epsilon }\delta = R^{k \epsilon }/\sqrt{R}\), with constant

$$\begin{aligned} C_k= 3^{kp} C_L + (10)^p R^{(p+2)\epsilon } C_B \sum _{l=0}^{k-1} 3^{lp} = 3^{kp} C_L + (10)^p R^{(p+2)\epsilon } C_B \frac{1-3^{kp}}{1-3^p}. \end{aligned}$$

At the step before the last, \(k=(1/(2\epsilon )) -2\) and \(\delta _k= R^{[(1/(2\epsilon )) -2] \epsilon }/\sqrt{R}=R^{-2\epsilon }\), which is a valid value of \(\delta \) (i.e. \(\delta _k=R^{-2\epsilon }\) obeys (28), because \(10 R^\epsilon \le 1/R^{-2\epsilon } \le R^{1-2\epsilon }/10\)). Applying Lemma 7.3 one last time, we get the estimate

$$\begin{aligned} \int _{B(0,R)} |Ef(x)|^p H(x) dx \le C R^\epsilon A_\alpha (H)^\beta R^{\alpha /2} \Vert f \Vert _{L^p(\sigma )}^p \end{aligned}$$

for every function \(f \in L^1(\sigma )\) that is supported on an arc of \(\sigma \)-measure \(\le R^{-\epsilon }\), where the constant C satisfies

$$\begin{aligned} C \le 3^{p/(2\epsilon )} \Big ( C_L + \frac{(10)^p R^{(p+2)\epsilon }}{3^p-1} C_B \Big ). \end{aligned}$$

Since the circle \(\mathbb S^1\) can be covered by \(\sim R^\epsilon \) such arcs, (15) follows and Theorem 5.1 is proved.