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Kakeya Conjecture

A Kakeya Set is a set that contains a unit line segment for every direction. For example, a ball of radius one half is a Kakeya set.

The Kakeya Set Conjecture asserts that every compact Kakeya set E nE\subset\mathbb{R}^n has Hausdorff dimension nn.

X-Ray Transform

The X-ray transform is the operator

Xf(l):= lfdλ(x),\operatorname{X} f(l) := \int_l f\,d\lambda(x),

where l nl\subset\mathbb{R}^n is a line, and λ\lambda the Lebesgue measure on the line. The manifold M nM_n can be represented as

M n={(σ,x) n1× nx,σ=0}; M_n = \{(\sigma, x)\in \mathbb{P}^{n-1}\times\mathbb{R}^n \mid \langle x,\sigma\rangle = 0\};

here n1:=S n1/{±}\mathbb{P}^{n-1} := S^{n-1}/\{\pm\} is the space of lines in n\mathbb{R}^n that pass through the origin, and (σ,x)(\sigma,x) represents a line that passes through xx with direction σ\sigma. We endow M nM_n with a measure μ\mu invariant under rigid motions. It is an open problem to determine the exponents 1p,q,r1\le p,q,r\le\infty for which the following inequality holds:

Xf L q(σL x r( n1))Cf L p( n). \left\Vert \operatorname{X} f\right\Vert_{L^q(\sigma\mapsto L^r_x(\mathbb{R}^{n-1}))} \le C\left\Vert f\right\Vert_{L^p(\mathbb{R}^n)}.

For brevity we will write L σ qL x r:=L q(σL x r( n1))L^q_\sigma L^r_x := L^q(\sigma\mapsto L^r_x(\mathbb{R}^{n-1})).

We can write the X-ray transform of a function fC c ( n)f\in C_c^\infty(\mathbb{R}^n) as

Xf(l)=fδ l(x)dx \operatorname{X} f(l) = \int f\delta_{l}(x)\,dx

where δ l:=lim ε0ε n+11 {x nlB ε(x)}\delta_{l} := \lim_{\varepsilon\to 0}\varepsilon^{-n+1}1_{\{x\in\mathbb{R}^n\mid \;l\cap B_\varepsilon(x)\neq\emptyset\}}. If (σ,x)(\sigma,x) is a line lM nl\in M_n, and P σP_\sigma the projection to the plane {yy,σ=0}\{y\mid \langle y,\sigma\rangle = 0\}, then δ l\delta_l is the pullback P σ *δ x=δ xP σP_\sigma^*\delta_x = \delta_x\circ P_\sigma, where δ x\delta_x is the Dirac delta centered at x n1x\in\mathbb{R}^{n-1}.

The operator dual to X\operatorname{X}, i.e. Xhfdμ=hX *fdx\int \operatorname{X}h\,f\,d\mu = \int h\,\operatorname{X}^*f\,dx, acts on functions ff in M nM_n as

X *f(x)=fδ {x}(l)dμ(l); \operatorname{X}^*f(x) = \int f\delta_{\{x\}}(l)\,d\mu(l);

here δ {x}:=lim ε0ε n+11 {lM nlB ε(x)}\delta_{\{x\}} := \lim_{\varepsilon\to 0}\varepsilon^{-n+1}1_{\{l\in M_n\mid \;l\cap B_\varepsilon(x)\neq\emptyset\}} —this is a distribution in M nM_n, do not confuse with the usual Dirac delta δ x\delta_x centered at xx. The distribution δ {x}\delta_{\{x\}} is the restriction of the measure μ\mu in M nM_n to the set of lines passing through x nx\in\mathbb{R}^n.

The first bound we can get for the X-ray transform is

|Xf|dμ(l)X|f|1dμ(l)=|f|X *1dx=|f|dx, \int \vert \operatorname{X} f\vert\,d\mu(l) \le \int \operatorname{X}\vert f\vert\,1\,d\mu(l) = \int \vert f\vert\,\operatorname{X}^*1\,dx = \int \vert f\vert\,dx,

since X *1\operatorname{X}^*1 is constant. Alternatively, we can use the fact that the integral of X|f|\operatorname{X}\vert f\vert along all the lines with the same direction equals f 1\left\Vert f\right\Vert_1, which also implies that X:L 1( n)L 1(M n)\operatorname{X}:L^1(\mathbb{R}^n)\to L^1(M_n). Any other bound of the X-ray transform can be interpolated with this bound to get further inequalities.

Besides L pL^p-spaces, we need to consider spaces of regular functions.

Definition

(Sobolev space)

Let φ\varphi be a smooth function with Fourier transform supported in |ξ|1\vert\xi\vert\sim 1, and let φ˜\tilde{\varphi} be a smooth function with Fourier transform supported in |ξ|1\vert\xi\vert\lesssim 1. Suppose that

1= k0φ k, 1 = \sum_{k\ge 0}\varphi_k,

where φ 0:=φ˜\varphi_0 := \tilde{\varphi} and φ k(ξ):=φ(ξ/2 k)\varphi_k(\xi) := \varphi(\xi/2^k), for k1k\ge 1. Let P kP_k be the projection (P kf) :=φ kf^(P_k f)^\wedge := \varphi_k \hat{f}.

The Sobolev-Slobodeckij spaces W s,p( n)W^{s,p}(\mathbb{R}^n) are, for 1<p<1\lt p\lt \infty and <s<-\infty\lt s\lt \infty, defined as

f W s,p( n):=[ k0(2 skP kf L p( n)) p] 1pifsinteger, \Vert f\Vert_{W^{s,p}(\mathbb{R}^n)} := \Big[\sum_{k\ge 0}(2^{sk}\Vert P_k f\Vert_{L^p(\mathbb{R}^n)})^p\Big]^\frac{1}{p}\quad if s \neq integer,

and

f W s,p( n):=[ |α|s αf L p( n) p] 1pifs=integer. \Vert f\Vert_{W^{s,p}(\mathbb{R}^n)} := \Big[\sum_{\vert\alpha\vert\le s}\Vert \partial^\alpha f\Vert_{L^p(\mathbb{R}^n)}^p\Big]^\frac{1}{p} \quad if s = integer.

See (Triebel 1978, Sec. 2.3.1).

The relation between that X-ray transform and the Kakeya set conjecture is the content of the following Theorem

Theorem

(Bourgain 1991, Lemma 2.15)

Let nsp>0n-sp\gt 0. If for every function ff such that suppfKsupp\,f\subset K, for KK compact, it holds that

(1)Xf L σ 1L x (M n)C Kf W s,p( n), \Vert \operatorname{X} f\Vert_{L^1_\sigma L^\infty_x(M_n)}\le C_K\Vert f\Vert_{W^{s,p}(\mathbb{R}^n)},

then the Hausdorff dimension of a Kakeya set is at least nspn-sp.

Proof

To compute the Hausdorff dimension we can use either balls B r(x)B_r(x) or dyadic cubes Q k:=[l2 k,(l+1)2 k] nQ_k := [l2^k,(l+1)2^k]^n, for kk integer. In fact, we can cover a cube Q kQ_k with a ball of radius n2 k1\sqrt{n}2^{k-1}, and conversely we can cover a ball of radius rr with a few cubes Q kQ_k, for 2 k1<r2 k2^{k-1}\lt r\le 2^k.

Let E nE\subset\mathbb{R}^n be a Kakeya set. Give any covering 𝒞={Q k}\mathcal{C} = \{Q_k\} of EE at scale 0<δ10\lt\delta\le 1, i.e. a covering such that for every cube Q k𝒞Q_k\in\mathcal{C} its side-length is 2 kδ2^k\le\delta, the goal is to show that if 0d<nsp0\le d\lt n-sp then

Q k𝒞l(Q k) dasδ0, \sum_{Q_k\in\mathcal{C}}l(Q_k)^d \to \infty \quad as\;\delta\to 0,

where l(Q k)=2 kl(Q_k) = 2^k is the side-length of the cube.

Let 𝒞\mathcal{C} be a covering of EE at scale δ\delta. We denote by 𝒞 k\mathcal{C}_k the collection of all the cubes in 𝒞\mathcal{C} with side-length 2 k2^k, and we denote by A kA_k the union of all the cubes in 𝒞 k\mathcal{C}_k; therefore,

1 E k1 A k:= k( Q k𝒞 k1 Q k). 1_E \le \sum_k 1_{A_k} := \sum_k \Big(\sum_{Q_k\in\mathcal{C}_k}1_{Q_k}\Big).

Since our hypotheses involve the spaces W s,pW^{s,p}, we should mollify 1 A k1_{A_k}. We take a suitable smooth function φ\varphi with compact support, define the dilations φ k(x):=2 nkφ(x/2 k)\varphi_k(x) := 2^{-nk}\varphi(x/2^k), and replace 1 A k1_{A_k} by φ k*1 A k\varphi_k*1_{A_k}.

Since EE contains a unit line segment in every direction σ\sigma, then for every σ n1\sigma\in \mathbb{P}^{n-1} it holds that X1 E(,σ) L 1\Vert \operatorname{X}1_E(\cdot,\sigma)\Vert_{L^\infty} \ge 1, and then that

1X1 E L σ 1L x C 2 kδX(φ k*1 A k) L σ 1L x . 1\le \Vert \operatorname{X}1_E\Vert_{L^1_\sigma L^\infty_x}\le C\sum_{2^k\le\delta}\Vert \operatorname{X}\big(\varphi_k*1_{A_k}\big)\Vert_{L^1_\sigma L^\infty_x}.

By (1) we have that

1C 2 kδφ k*1 A k W s,p. 1\le C\sum_{2^k\le\delta} \Vert \varphi_k*1_{A_k}\Vert_{W^{s,p}}.

In general, the W s,pW^{s,p}-norm of φ k*f\varphi_k*f is

φ k*f W s,pC2 ksf p; \Vert \varphi_k*f\Vert_{W^{s,p}}\le C2^{-ks}\Vert f\Vert_p;

see Lemma below. Hence,

1C 2 kδφ k*1 A k W s,pC 2 kδ2 ks+nkp|𝒞 k| 1p 1\le C\sum_{2^k\le\delta} \Vert \varphi_k*1_{A_k}\Vert_{W^{s,p}} \le C\sum_{2^k\le\delta}2^{-ks+\frac{nk}{p}}\vert \mathcal{C}_k\vert^\frac{1}{p}

where |𝒞 k|\vert \mathcal{C}_k\vert denotes the number of cubes in 𝒞 k\mathcal{C}_k.

By hypothesis nsp>0n-sp\gt 0, then for every 0d<nsp0\le d \lt n-sp we can apply Hölder inequality to get

1( 2 kδ2 pk(ndps)) 1p( k2 kd|𝒞 k|) 1pCδ α( Q k𝒞l(Q k) d) 1p, 1 \le \Big(\sum_{2^k\le\delta} 2^{p'k(\frac{n-d}{p}-s)}\Big)^\frac{1}{p'}\Big(\sum_k 2^{kd}\vert\mathcal{C}_k\vert\Big)^\frac{1}{p} \le C\delta^\alpha\Big(\sum_{Q_k\in\mathcal{C}}l(Q_k)^d\Big)^\frac{1}{p},

where α=ndps>0\alpha = \frac{n-d}{p}-s\gt 0. The statement of the Theorem follows.

Lemma

If ψ\psi is a smooth function and ψ l(x):=2 nlψ(2 lx)\psi_l(x) := 2^{nl}\psi(2^l x), for l0l\ge 0, then

(2)ψ l*f W s,p( n)C2 lsf p, \Vert \psi_l*f\Vert_{W^{s,p}(\mathbb{R}^n)}\le C2^{ls}\Vert f\Vert_p,

where CC depends on ψ\psi.

Proof

If ss is an integer, then (2) follows from α(ψ l*f)=( αψ l)*f\partial^\alpha(\psi_l*f) = (\partial^\alpha \psi_l)*f and Young Inequality for convolutions.

If ss\neq integer, then we have to estimate the norm of P kfP_k f; recall that (P kf) :=φ kf^(P_k f)^\wedge := \varphi_k \hat{f}, where φ k(ξ):=φ(ξ/2 k)\varphi_k(\xi) := \varphi(\xi/2^k). If klk\le l then by Young Inequality for convolutions

P k(ψ l*f) pφˇ k 1ψ l 1f pCf p. \Vert P_k(\psi_l*f)\Vert_p \le \Vert \check{\varphi}_k\Vert_1\Vert \psi_l\Vert_1\Vert f\Vert_p \le C\Vert f\Vert_p.

If k>lk\gt l then fix a number 2N>s2N\gt s. We estimate the norm of P kfP_k f as

P k(ψ l*f) p(|ξ| 2Nφ k) 1Δ N(ψ l*f) p. \Vert P_k(\psi_l*f)\Vert_p \le \Vert (\vert \xi\vert^{-2N}\varphi_k)^\vee\Vert_1 \Vert \Delta^N(\psi_l*f)\Vert_p.

As before we get Δ N(ψ l*f) pC2 2Nlf p\Vert \Delta^N (\psi_l*f)\Vert_p\le C2^{2Nl}\Vert f\Vert_p. For the other term, since (|ξ| 2Nφ k) (x)=2 (n2N)k(|ξ| 2Nφ) (2 kx)(\vert \xi\vert^{-2N}\varphi_k)^\vee(x) = 2^{(n-2N)k}(\vert \xi\vert^{-2N}\varphi)^\vee(2^k x) then we get

(|ξ| 2Nφ k) 1=2 k2N(|ξ| 2Nφ) 1; \Vert (\vert \xi\vert^{-2N}\varphi_k)^\vee\Vert_1 = 2^{-k2N}\Vert (\vert \xi\vert^{-2N}\varphi)^\vee\Vert_1;

we conclude so that P k(ψ l*f) pC2 2N(lk)f p\Vert P_k(\psi_l*f)\Vert_p \le C2^{2N(l-k)}\Vert f\Vert_p. Therefore, the W s,pW^{s,p}-norm of ψ l*f\psi_l*f is

ψ l*f W s,p( n) p 0kl2 pksP k(ψ l*f) p p+ k>l2 pksP k(ψ l*f) p p C( 0kl2 pks+2 p2Nl k>l2 p(s2N)k)f p p =C2 pslf p p, \begin{aligned} \Vert \psi_l*f\Vert_{W^{s,p}(\mathbb{R}^n)}^p &\le \sum_{0\le k\le l}2^{pks}\Vert P_k(\psi_l*f)\Vert_p^p + \sum_{k\gt l}2^{pks}\Vert P_k(\psi_l*f)\Vert_p^p \\ &\le C\Big(\sum_{0\le k \le l}2^{pks} + 2^{p2Nl}\sum_{k\gt l}2^{p(s-2N)k}\Big)\Vert f\Vert_p^p \\ &= C2^{psl}\Vert f\Vert_p^p, \end{aligned}

which concludes the proof.

For a fixed direction σ\sigma, the X-ray transform maps ff to a function xXf(x,σ)x\mapsto \operatorname{X} f(x,\sigma) in n1\mathbb{R}^{n-1}. The function Xf(,σ)\operatorname{X} f(\cdot,\sigma) turns out to be more regular than ff.

Theorem

If fL 2( n)f\in L^2(\mathbb{R}^n) has support suppfKsupp f\subset K, for KK compact, then

(3)|D 12Xf L 2(M n)C Kf L 2( n). \left\Vert \vert \langle D\rangle^\frac{1}{2}\operatorname{X} f\right\Vert_{L^2(M_n)} \le C_K\left\Vert f\right\Vert_{L^2(\mathbb{R}^n)}.
Remark

The Japanese bracket \langle\cdot\rangle is defined as ξ:=(1+|ξ| 2) 12\langle \xi\rangle := (1+\vert \xi\vert^2)^\frac{1}{2}, and then D 2=IΔ\langle D\rangle^2 = I-\Delta.

Proof

We set g:=D 12fg := \langle D\rangle^\frac{1}{2} f, and for a fixed direction σ n1\sigma\in \mathbb{P}^{n-1} we compute the Fourier transform of the function x n1Xg(x,σ)x\in\mathbb{R}^{n-1}\mapsto \operatorname{X}g(x,\sigma). By rotational symmetry we can assume that σ=e n\sigma = e_n, so

Xg(x,e n)=g(x,t)dt. \operatorname{X}g(x,e_n) = \int g(x,t)\,dt.

The Fourier transform is (Xg(,e n)) (η)=g^(η,0)(\operatorname{X}g(\cdot,e_n))^\wedge(\eta) = \hat{g}(\eta,0). If δ σ=lim ε0ε 11 {ξ|ξ,σ|ε}\delta_{\perp\sigma} = \lim_{\varepsilon\to 0}\varepsilon^{-1}1_{\{\xi\mid \vert\langle \xi,\sigma\rangle\vert\le\varepsilon\}} is the Dirac delta of the plane normal to σ\sigma, then we may write

Xg(,σ) L 2( n1) 2=|g^| 2δ σdξ. \Vert \operatorname{X}g(\cdot,\sigma)\Vert_{L^2(\mathbb{R}^{n-1})}^2 = \int \vert \hat{g}\vert^2\delta_{\perp\sigma}\,d\xi.

Hence,

Xg(,σ) L 2(M n) 2=|g^| 2( S n1δ σ(ξ)dS(σ))dξ=ξ|f^| 2dξ|ξ|. \Vert \operatorname{X}g(\cdot,\sigma)\Vert_{L^2(M_n)}^2 = \int\vert \hat{g}\vert^2\Big(\int_{S^{n-1}}\delta_{\perp\sigma}(\xi)\,dS(\sigma)\Big)\,d\xi = \int \langle \xi\rangle\vert\hat{f}\vert^2\frac{d\xi}{\vert \xi\vert}.

For high frequencies we have that

|ξ|1ξ|f^| 2dξ|ξ|Cf 2 2. \int_{\vert\xi\vert\ge 1} \langle \xi\rangle\vert\hat{f}\vert^2\frac{d\xi}{\vert \xi\vert} \le C\Vert f\Vert_2^2.

For low frequencies we have that

|ξ|1ξ|f^| 2dξ|ξ|Cf 1 2C Kf 2 2. \int_{\vert\xi\vert\le 1} \langle \xi\rangle\vert\hat{f}\vert^2\frac{d\xi}{\vert \xi\vert} \le C\Vert f\Vert_1^2 \le C_K\Vert f\Vert_2^2.

The statement of the Theorem follows.

Sobolev embedding Theorem h L ( n1)Ch W s,2( n1)\Vert h\Vert_{L^\infty(\mathbb{R}^{n-1})}\le C\Vert h\Vert_{W^{s,2}(\mathbb{R}^{n-1})}, for s>n12s\gt \frac{n-1}{2}, and the inequality (3) allow us to conclude that for every ff such that suppfKsupp\,f\subset K, for KK compact, it holds that

Xf L σ 2L x C Kf W s,2( n), \Vert \operatorname{X} f\Vert_{L^2_\sigma L^\infty_x}\le C_K \Vert f\Vert_{W^{s,2}(\mathbb{R}^n)},

for s>n21s\gt \frac{n}{2}-1. Hence, by Theorem the Hausdorff dimension of a Kakeya set E nE\subset\mathbb{R}^n is at least 2. This result is the best possible in 2\mathbb{R}^2 (Córdoba 1977), but far away from the expected dimension for n3n\ge 3.

Definition

(Lorentz Spaces)

Let (X,ν)(X,\nu) be a measure space. The quasinorm of the space L p,q(ν)L^{p,q}(\nu), for 1p<1\le p\lt\infty and 1q<1\le q \lt\infty, is

f p,q:=(p 0 s qν{|f|s} qpdss) 1q, \left\Vert f\right\Vert_{p,q} := \Big(p\int_0^\infty s^q\nu\{\vert f\vert\ge s\}^\frac{q}{p}\frac{ds}{s}\Big)^\frac{1}{q},

and, for 1p<1\le p\lt\infty and q=q=\infty, is

f p,:=sup s>0(sν{|f|s} 1p) \left\Vert f\right\Vert_{p,\infty} := \sup_{s\gt 0} (s\nu\{\vert f\vert\ge s\}^\frac{1}{p})
Theorem

(Drury 1983)

If E nE\subset \mathbb{R}^n is a measurable set, and 1 E1_E is the characteristic function of EE, then

(4)X(1 E) L n+1,(M n)C|E| 2n+1. \left\Vert \operatorname{X}(1_E)\right\Vert_{L^{n+1,\infty}(M_n)} \le C\left\vert E\right\vert^\frac{2}{n+1}.
Remark

This inequality corresponds to the restricted weak version of the point (p,q,r)=(n+12,n+1,n+1)(p,q,r) = (\frac{n+1}{2}, n+1, n+1).

Proof

Let f=1 Ef = 1_E and define the set F:={lM nXf(l)λ}F := \{l\in M_n\mid \operatorname{X} f(l)\ge\lambda\}, then

(5)λ|F| FXfdμ=fX *1 FdxX *1 F f 1. \lambda\vert F\vert \le \int_F \operatorname{X} f\,d\mu = \int f\operatorname{X}^*1_F \,dx \le \left\Vert \operatorname{X}^*1_F\right\Vert_\infty\left\Vert f\right\Vert_1.

This establishes a lower bound for X *1 F \left\Vert \operatorname{X}^*1_F\right\Vert_\infty, and we will get now an upper bound.

Choose a point x 0x_0 such that X *1 F(x 0)12X *1 F \operatorname{X}^*1_F(x_0)\ge \frac{1}{2}\left\Vert \operatorname{X}^*1_F\right\Vert_\inftyx 0x_0 is the center of the Bourgain’s bush. By translation symmetry we can assume that x 0=0x_0 = 0. If δ 0\delta_0 is the Dirac delta centered at the origin, then

(6)1 FXfXδ 0dμ(l)λ FXδ 0dμ(l)λ2X *1 F . \int 1_F \operatorname{X} f\,\operatorname{X}\delta_0\,d\mu(l) \ge \lambda\int_F \operatorname{X}\delta_0\,d\mu(l)\ge \frac{\lambda}{2}\left\Vert \operatorname{X}^*1_F\right\Vert_\infty.

Here we can think of Xδ 0\operatorname{X}\delta_0 as the distribution we defined earlier δ {0}:=lim ε0ε n+11 {lM nlB ε(0)}\delta_{\{0\}} := \lim_{\varepsilon\to 0}\varepsilon^{-n+1}1_{\{l\in M_n\mid \;l\cap B_\varepsilon(0)\neq\emptyset\}} supported over the lines passing through the point 00. We can also justify the notation Xδ 0\operatorname{X}\delta_0 by a limiting process since δ 0:=lim ε0ε n1 B ε(0)\delta_0 := \lim_{\varepsilon\to 0}\varepsilon^{-n}1_{B_\varepsilon(0)}.

On the other hand, we can also estimate the left side of (6) as

1 FXfXδ 0dμ(l)=fX *(1 FXδ 0)dxf L n,1X *(1 FXδ 0) L nn1,, \int 1_F \operatorname{X} f\,\operatorname{X}\delta_0\,d\mu(l) = \int f\operatorname{X}^*(1_F \operatorname{X}\delta_0)\,dx \le \left\Vert f\right\Vert_{L^{n,1}} \left\Vert \operatorname{X}^*(1_F \operatorname{X}\delta_0)\right\Vert_{L^{\frac{n}{n-1},\infty}},

The function X *(1 FXδ 0)\operatorname{X}^*(1_F \operatorname{X}\delta_0) can be computed explicitly

X *(1 FXδ 0)(x)=1|x| n11 F(l(x˜,0)), \operatorname{X}^*(1_F \operatorname{X}\delta_0)(x) = \frac{1}{\vert x\vert^{n-1}}1_F(l(\tilde{x}, 0)),

where x˜:=x/|x|\tilde{x} := x/\vert x\vert and l(x˜,0)l(\tilde{x}, 0) is the line that passes through 00 and xx. Then, we can estimate the norm of X *(1 FXδ 0)\operatorname{X}^*(1_F \operatorname{X}\delta_0) as

X *(1 FXδ 0) L nn1,=c n|X *1 F(0)| n1nc nX *1 F n1n. \left\Vert \operatorname{X}^*(1_F \operatorname{X}\delta_0)\right\Vert_{L^{\frac{n}{n-1},\infty}} = c_n\vert \operatorname{X}^*1_F(0)\vert^\frac{n-1}{n}\le c_n\left\Vert \operatorname{X}^*1_F\right\Vert_\infty^\frac{n-1}{n}.

Then we get the bound

(7)1 FXfXδ 0dμ(l)c nf L n,1X *1 F n1n. \int 1_F \operatorname{X} f\,\operatorname{X}\delta_0\,d\mu(l) \le c_n \left\Vert f\right\Vert_{L^{n,1}}\left\Vert \operatorname{X}^*1_F\right\Vert_\infty^\frac{n-1}{n}.

By (6) and (7) we get the desired upper bound for X *1 F \left\Vert \operatorname{X}^*1_F\right\Vert_\infty:

c nλX *1 F 1nf L n,1. c_n\lambda \left\Vert \operatorname{X}^*1_F\right\Vert_\infty^\frac{1}{n} \lesssim \left\Vert f\right\Vert_{L^{n,1}}.

This and (5) allow us to conclude that

|F|1λ n+1f 1f L n,1 n=1λ n+1|E| 2, \vert F\vert \lesssim \frac{1}{\lambda^{n+1}}\left\Vert f\right\Vert_1\left\Vert f\right\Vert_{L^{n,1}}^n = \frac{1}{\lambda^{n+1}}\vert E\vert^2,

which implies the restricted weak inequality (4).

Drury’s Theorem and the bound L 1L 1L^1\to L^1 imply that

Xf L q(M n)Cf L p( n), \Vert \operatorname{X} f\Vert_{L^q(M_n)}\le C\Vert f\Vert_{L^p(\mathbb{R}^n)},

for 1p<n+121\le p\lt \frac{n+1}{2} and n1q+1=np\frac{n-1}{q}+1 = \frac{n}{p}. By Sobolev embedding Theorem h L ( n1)h W s,q( n1)\Vert h\Vert_{L^\infty(\mathbb{R}^{n-1})} \le \Vert h\Vert_{W^{s,q}(\mathbb{R}^{n-1})}, for s>n1qs\gt \frac{n-1}{q}, we get further

Xf L σ qL x (M n)Cf W s,p( n), \Vert \operatorname{X} f\Vert_{L^q_\sigma L^\infty_x(M_n)}\le C\Vert f\Vert_{W^{s,p}(\mathbb{R}^n)},

for 1p<n+121\le p\lt \frac{n+1}{2}, n1q+1=np\frac{n-1}{q}+1 = \frac{n}{p} and s>np1s\gt \frac{n}{p}-1. Hence, by Theorem the Hausdorff dimension of a Kakeya set E nE\subset\mathbb{R}^n is at least n+12\frac{n+1}{2}.

References

  • Bourgain, J.?, Besicovitch type maximal function operators and applications to Fourier Analysis, Geom. Funct. Anal., 1, 1991

  • Córdoba, A.?, The Kakeya Maximal Function and the Spherical Summation Multipliers, Amer. J. Math., 99, 1977

  • Christ, M.?, Estimates for the kk-plane transform, Indiana Univ. Math. J., 33, 1984

  • Drury, S. W.?, L pL^p estimates for the X-ray transform, Illinois J. Math., 27, 1983

  • Triebel, H?, Interpolation Theory, Function Spaces, Differential Operators, North-Holland publishing company, 1978

Last revised on May 12, 2020 at 03:40:19. See the history of this page for a list of all contributions to it.