nLab 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 $E\subset\mathbb{R}^n$ has Hausdorff dimension $n$.

X-Ray Transform

The X-ray transform is the operator

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

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

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

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

$$\left\Vert \operatorname{X} f\right\Vert_{L^q(\sigma\mapsto L^r_x(\mathbb{R}^{n-1}))} \le C\left\Vert f\right\Vert_{L^p(\mathbb{R}^n)}.$$

For brevity we will write $L^q_\sigma L^r_x := L^q(\sigma\mapsto L^r_x(\mathbb{R}^{n-1}))$.

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

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

where $\delta_{l} := \lim_{\varepsilon\to 0}\varepsilon^{-n+1}1_{\{x\in\mathbb{R}^n\mid \;l\cap B_\varepsilon(x)\neq\emptyset\}}$. If $(\sigma,x)$ is a line $l\in M_n$, and $P_\sigma$ the projection to the plane $\{y\mid \langle y,\sigma\rangle = 0\}$, then $\delta_l$ is the pullback $P_\sigma^*\delta_x = \delta_x\circ P_\sigma$, where $\delta_x$ is the Dirac delta centered at $x\in\mathbb{R}^{n-1}$.

The operator dual to $\operatorname{X}$, i.e. $\int \operatorname{X}h\,f\,d\mu = \int h\,\operatorname{X}^*f\,dx$, acts on functions $f$ in $M_n$ as

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

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

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

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

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

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

Definition

(Sobolev space)

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

$$1 = \sum_{k\ge 0}\varphi_k,$$

where $\varphi_0 := \tilde{\varphi}$ and $\varphi_k(\xi) := \varphi(\xi/2^k)$, for $k\ge 1$. Let $P_k$ be the projection $(P_k f)^\wedge := \varphi_k \hat{f}$.

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

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

and

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

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 $n-sp\gt 0$. If for every function $f$ such that $supp\,f\subset K$, for $K$ compact, it holds that

(1)$$\Vert \operatorname{X} f\Vert_{L^1_\sigma L^\infty_x(M_n)}\le C_K\Vert f\Vert_{W^{s,p}(\mathbb{R}^n)},$$

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

Proof

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

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

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

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

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

$$1_E \le \sum_k 1_{A_k} := \sum_k \Big(\sum_{Q_k\in\mathcal{C}_k}1_{Q_k}\Big).$$

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

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

$$1\le \Vert \operatorname{X}1_E\Vert_{L^1_\sigma L^\infty_x}\le C\sum_{2^k\le\delta}\Vert \operatorname{X}\big(\varphi_k*1_{A_k}\big)\Vert_{L^1_\sigma L^\infty_x}.$$

By (1) we have that

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

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

$$\Vert \varphi_k*f\Vert_{W^{s,p}}\le C2^{-ks}\Vert f\Vert_p;$$

see Lemma below. Hence,

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

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

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

$$1 \le \Big(\sum_{2^k\le\delta} 2^{p'k(\frac{n-d}{p}-s)}\Big)^\frac{1}{p'}\Big(\sum_k 2^{kd}\vert\mathcal{C}_k\vert\Big)^\frac{1}{p} \le C\delta^\alpha\Big(\sum_{Q_k\in\mathcal{C}}l(Q_k)^d\Big)^\frac{1}{p},$$

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

Lemma

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

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

where $C$ depends on $\psi$.

Proof

If $s$ is an integer, then (2) follows from $\partial^\alpha(\psi_l*f) = (\partial^\alpha \psi_l)*f$ and Young Inequality for convolutions.

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

$$\Vert P_k(\psi_l*f)\Vert_p \le \Vert \check{\varphi}_k\Vert_1\Vert \psi_l\Vert_1\Vert f\Vert_p \le C\Vert f\Vert_p.$$

If $k\gt l$ then fix a number $2N\gt s$. We estimate the norm of $P_k f$ as

$$\Vert P_k(\psi_l*f)\Vert_p \le \Vert (\vert \xi\vert^{-2N}\varphi_k)^\vee\Vert_1 \Vert \Delta^N(\psi_l*f)\Vert_p.$$

As before we get $\Vert \Delta^N (\psi_l*f)\Vert_p\le C2^{2Nl}\Vert f\Vert_p$. For the other term, since $(\vert \xi\vert^{-2N}\varphi_k)^\vee(x) = 2^{(n-2N)k}(\vert \xi\vert^{-2N}\varphi)^\vee(2^k x)$ then we get

$$\Vert (\vert \xi\vert^{-2N}\varphi_k)^\vee\Vert_1 = 2^{-k2N}\Vert (\vert \xi\vert^{-2N}\varphi)^\vee\Vert_1;$$

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

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

which concludes the proof.

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

Theorem

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

(3)$$\left\Vert \vert \langle D\rangle^\frac{1}{2}\operatorname{X} f\right\Vert_{L^2(M_n)} \le C_K\left\Vert f\right\Vert_{L^2(\mathbb{R}^n)}.$$

Remark

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

Proof

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

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

The Fourier transform is $(\operatorname{X}g(\cdot,e_n))^\wedge(\eta) = \hat{g}(\eta,0)$. If $\delta_{\perp\sigma} = \lim_{\varepsilon\to 0}\varepsilon^{-1}1_{\{\xi\mid \vert\langle \xi,\sigma\rangle\vert\le\varepsilon\}}$ is the Dirac delta of the plane normal to $\sigma$, then we may write

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

Hence,

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

For high frequencies we have that

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

For low frequencies we have that

$$\int_{\vert\xi\vert\le 1} \langle \xi\rangle\vert\hat{f}\vert^2\frac{d\xi}{\vert \xi\vert} \le C\Vert f\Vert_1^2 \le C_K\Vert f\Vert_2^2.$$

The statement of the Theorem follows.

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

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

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

Definition

(Lorentz Spaces)

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

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

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

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

Theorem

(Drury 1983)

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

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

Remark

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

Proof

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

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

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

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

(6)$$\int 1_F \operatorname{X} f\,\operatorname{X}\delta_0\,d\mu(l) \ge \lambda\int_F \operatorname{X}\delta_0\,d\mu(l)\ge \frac{\lambda}{2}\left\Vert \operatorname{X}^*1_F\right\Vert_\infty.$$

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

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

$$\int 1_F \operatorname{X} f\,\operatorname{X}\delta_0\,d\mu(l) = \int f\operatorname{X}^*(1_F \operatorname{X}\delta_0)\,dx \le \left\Vert f\right\Vert_{L^{n,1}} \left\Vert \operatorname{X}^*(1_F \operatorname{X}\delta_0)\right\Vert_{L^{\frac{n}{n-1},\infty}},$$

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

$$\operatorname{X}^*(1_F \operatorname{X}\delta_0)(x) = \frac{1}{\vert x\vert^{n-1}}1_F(l(\tilde{x}, 0)),$$

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

$$\left\Vert \operatorname{X}^*(1_F \operatorname{X}\delta_0)\right\Vert_{L^{\frac{n}{n-1},\infty}} = c_n\vert \operatorname{X}^*1_F(0)\vert^\frac{n-1}{n}\le c_n\left\Vert \operatorname{X}^*1_F\right\Vert_\infty^\frac{n-1}{n}.$$

Then we get the bound

(7)$$\int 1_F \operatorname{X} f\,\operatorname{X}\delta_0\,d\mu(l) \le c_n \left\Vert f\right\Vert_{L^{n,1}}\left\Vert \operatorname{X}^*1_F\right\Vert_\infty^\frac{n-1}{n}.$$

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

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

This and (5) allow us to conclude that

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

which implies the restricted weak inequality (4).

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

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

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

$$\Vert \operatorname{X} f\Vert_{L^q_\sigma L^\infty_x(M_n)}\le C\Vert f\Vert_{W^{s,p}(\mathbb{R}^n)},$$

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

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 $k$-plane transform, Indiana Univ. Math. J., 33, 1984

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

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