# The Kakeya problem in harmonic analysis

In 1917, Abram Besicovitch asked the following question:

If $f:\mathbb{R}^2 \to \mathbb{R}$ is Riemann-integrable, does there exist a pair of orthogonal coordinate axes which would render both $x \mapsto \int f(x,y) \, dy $ and $y \mapsto \int f(x,y) \, dx$ Riemann-integrable?

Besicovitch noted that a counterexample could be given if a compact set of measure zero in $\mathbb{R}^2$ that contains a line segment in every direction could be constructed.

To see this, we assume that $F$ is such a set. We fix a pair of axes and translate $F$ so that the line segments in $F$ parallel to the axes are of irrational distance apart from the axes. Let $F_0$ be the set of points in $F$ consisting of at least one rational coordinate with respect to the axes chosen. For each line segment $l$ in $F$, we observe that both $F_0 \cap l$ and $(F \smallsetminus F_0) \cap l$ are dense in $l$.

This results in severe “one-dimensional discontinuity” of the characteristic function $\chi_{F_0}$ in each direction. This, in turn, implies that $x \mapsto \chi_{F_0}(x,y)$ is not Riemann-integrable regardless of the choice of axes or $y$. $F$ is of planar measure zero, however, and so $\chi_{F_0}$ is Riemann-integrable on $\mathbb{R}^2$ by the Lebesgue criterion for Riemann integrability.

Two years later, Besicovitch succeeded in constructing such a set:

Theorem(Besicovitch, 1919). There exists a compact planar set of measure zero that contains a unit line segment in each direction.

We construct such a set by considering first an equilateral triangle. We slice it up into thin pieces and overlap them in different directions. This yields a set with smaller measure that nevertheless contains a wider variety of unit line segments than the original triangle.

Continuing this way, we end up with a set of measure zero that contains a unit line segment in each direction from, say, 0 to 90 degrees. Piecing together several of these sets yields the desired set. See Terry Tao’s interactive applet on the first ten steps of the construction.

The above construct generalizes straightforwardly to higher dimensions. For the sake of efficiency, it is convenient to make the following definition:

Definition.A(or aKakeya set) in $\mathbb{R}^n$ is a compact set in $\mathbb{R}^n$ that contains a unit line segment in each direction, viz.,Besicovitch set$$\forall e \in \mathbb{S}^{n-1} \, \, \exists x \in \mathbb{R}^n \, \, \forall t \in [-1/2,1/2] \, \, \left(x + te \in E\right)$$

By considering the products of the two-dimensional Besicovitch sets, we obtain the following results:

Theorem(Besicovitch, 1919). If $n \geq 2$, then there are Kakeya sets of measure zero in $\mathbb{R}^n$.

Why the name * Kakeya sets*? Unbeknownst to Besicovitch, the Japanese mathematician Soichi Kakeya was investigating a similar problem in 1917:

Question(Kakeya needle problem). What is the minimum area of a compact planar set in which a unit line segment can be rotated 180 degrees?

The civil war and the blockade prevented the rest of the world from finding out about Besicovitch’s work at the time. For a number of years the three-cusped hypercycloid was considered to be the optimal solution.

Besicovitch eventually learned of the problem, shortly after his departure from Russia in 1924. By slightly modifying his construction of the two-dimensional Kakeya set, Besicovitch was able to establish the following:

Theorem(Besicovitch, 1928). For each $\varepsilon > 0$, there exists a compact planar set of measure $\varepsilon$ in which a unit line segment can be rotated 180 degrees.

And so the needle problem can be solved on a *very small* set. But just how small is *very small*? This seemingly innocent question is, in fact, a starting point of a deep theory, which serves as a meeting ground for several major areas of modern mathematics.

In this post, we explore the connections between the Kakeya problem and the **Fourier transform**

$$\hat{f}(\xi) = \int_{\mathbb{R}^n} f(x) e^{-2 \pi i x \cdot \xi} \, dx,$$

defined for functions $f: \mathbb{R}^n \to \mathbb{C}$. Here $\cdot$ denotes the standard dot product.

If $f \in L^1$, we can make sense of the above definition. Moreover, the Fourir transform $\hat{f}:\mathbb{R}^n \to \mathbb{C}$ is continuous, decays at infinity, and satisfies the estimate

$$\lVert \hat{f} \rVert_\infty \leq \lVert f\rVert_1,$$

which implies that the Fourier transform is a bounded linear operator from $L^1$ to $L^\infty$.

How about functions in $L^p < p < \infty$?

Clearly, the integral definition makes no sense if the function is not in $L^1$. We can nevertheless define the Fourier transform operator by considering a dense subset of $L^p$ on which the integral definition is valid.

The prime candidate is the * Schwartz space* $\mathscr{S}$, which consists of infinitely-differentiable functions on $\mathbb{R}^n$ such that all of their partials, including the functions themselves, decrease more rapidly than polynomials. Since $\mathscr{S}$ is dense in $L^p$ for all $1 \leq p < \infty$, the definition of Fourier transform is valid on $\mathscr{S}$.

In fact, the Fourier transform is a homeomorphism of $\mathscr{S}$ onto itself. The inverse map, called the * Fourier inversion formula*, is defined by

$$f^\vee(x) = \int_{\mathbb{R}^n} f(\xi) e^{2\pi i \xi\cdot x} \, d\xi$$

for each $f \in \mathscr{S}$. The inversion formula establishes the identity

$$\lVert \hat{f} \rVert_2 = \lVert f \rVert_2$$

for all $f \in \mathscr{S}$. It follows that we can extend the Fourier transform operator onto all of $L^2$ such that the above norm estimate holds for all $f \in L^2$. This is * Plancherel’s theorem*.

An “interpolation” theorem (specifically, the Riesz–Thorin theorem) now establishes the **Hausdorff–Young inequality**

$$\lVert \hat{f} \rVert_{p'} \leq \lVert f \rVert_p$$

for all $1 \leq p \leq 2$, where $p'$ is the * conjugate exponent* of $p$, defined by the identity $p^{-1} + (p')^{-1} = 1$.

Unfortunately, the Fourier transform is not a bounded operator into another Lebesgue space for $p>2$. How do we, then, make sense of the equality

$$f = (\hat{f})^\vee$$

for general $f \in L^p$? More precisely, if

$$S_Rf(x) = \int_{\vert \xi \vert \leq R} \hat{f}(\xi) e^{2 \pi i \xi \cdot x} \, d\xi,$$

in what sense does $S_R f$ converge to $f$ as $R \to \infty$?

The $p=1$ case is hopeless, for Andrey Kolmogorov exhibited an $L^1$ functino whose Fourier inversion fails to converge at every point. We thus restrict our attention to $1 < p < \infty$.

For $n = 1$, there is the pointwise almost-everywhere result, established for $L^2$ functions by Lennart Carleson in 1966 and extended to $L^p$ functions for all $1 < p < \infty$ by Richard Hunt two years later.

As for $n > 1$, we must first consider the method of summing the integrals. If the integral is summed over rectangles, then the multivariate almost-everywhere convergence follows from the Carleson–Hunt theorem. The pointwise convergence result over balls is still open.

How about $L^p$ convergence? For $n=1$, we have the classical theorem of Marcel Riesz:

Theorem(M. Riesz, 1928). Fix $1 < p < \infty$, let $f \in L^p(\mathbb{R})$, and assume that $\hat{f}$ exists. Then $S_R f \to f$ in $L^p$.

How about $n > 1$? Again, spherical summation is the difficult part of the multivariate case. The $L^2$ convergence is given by the Plancherel theorem, but the convergence result fails to materialize for all other values of $p$.

Theorem(Fefferman ball multiplier theorem, 1971). The spherical summation of the Fourier inversion formula converges in the norm topology of $L^p$ if and only if $p = 2$.

By a result of de Leeuw, settling the ball multipler problem on $n = k$ settles it for $n = k-1$. It thus suffices to consider the ball multipler problem for $n = 2$.

Moreover, the Riesz representation theorem carries over the ball multipler theorem on $L^p$ to its conjugate exponent, and so it suffices to consider the $p > 2$ case.

Finally, the uniform boundedness principle implies that the convergence of spherical summation is equivalent to the norm estimate

$$\lVert S_R f \rVert_p \leq k_p \lVert f \rVert_p$$

for all sufficiently large $R$. A scaling argument shows that the above estimate is, in fact, equivalent to the single norm estimate

$$\lVert S_1 f \rVert_p \leq k_p \lVert f \rVert_p.$$

It thus suffices to exhibit, for each $N > 0$ and every $p > 2$, a non-zero function $f$ such that

$$\lVert S_1 f \rVert_p \geq N \lVert f \rVert_p.$$

By taking $f$ to be a sum of characteristic functions on thin needles in different directions, we can minimize $S_1 f$ while maximizing its $L^p$ norm. It follows that the above inequality fails.

How badly does the convergence fail? To this end, we consider the **Bochner–Riesz mean**

$$S_R^\delta f(x) = \int_{\vert \xi \vert \leq R} \hat{f}(\xi) e^{2 \pi i \xi \cdot x} \left( 1 - \frac{\vert \xi \vert^2}{R^2} \right)^{\delta} \, d\xi$$

for each $R > 0$ and $\delta \geq 0$.

$S_R^0$ is the regular spherical summation, whose lack of convergence we have discussed above. For the $\delta > 0$ case, a similar reduction argument as above implies that the $L^p$ convergence of the Bochner–Riesz mean is equivalent to the norm estimate

$$\lVert S_1^\delta f \rVert_p \leq k_p \lVert f \rVert_p.$$

By a result of Herz, the above estimate does not hold unless

$$\left\vert \frac{1}{p} - \frac{1}{2} \right\vert < \frac{2\delta +1}{2n}.$$

The * Bochner–Riesz conjecture* posits that the $L^p$ norm estimate does hold for all values of $p$ that satisfy the above restriction.

As it turns out, Fefferman’s construction could disprove the Bochner–Riesz conjecture if it can be modified to solve the $n$-dimensional needle problem on a set *too small to be considered $n$-dimensional*.

To make this notion precise, we introduce the concept of Hausdorff measure and dimension.

Fix $m \in \mathbb{N}$, and let $\omega_m$ be the $m$-dimensional Lebesgue measure of the closed unit ball in $\mathbb{R}^m$. The **$m$-dimensional Hausdorff outer measure** $\mathscr{H}^m$ defined for every subset $E \subseteq \mathbb{R}^n$ by

$$\mathscr{H}^m(E) = \lim_{\delta \to 0} \inf_{\substack{E \subseteq \bigcup S_j \ \operatorname{diam}(S_j) \leq \delta}} \sum_{j=1}^\infty \omega_m \left( \frac{\operatorname{diam}(S_j)}{2}\right)^m,$$

where the infimum is taken over a countable cover $(S_j)_{n=1}^\infty$ of $E$ of diameter at most $\delta$.

More generally, we let

$$\omega_s = \frac{\pi^{s/2}}{\Gamma(s/2+1)}$$

for each $s > 0$ and define the * $s$-dimensional Hausdorff outer measure* $\mathscr{H}^s$ of $E \subseteq \mathbb{R}^n$ to be

$$\mathscr{H}^s(E) = \lim_{\delta \to 0} \inf_{\substack{E \subseteq \bigcup S_j \ \operatorname{diam}(S_j) \leq \delta}} \sum_{j=1}^\infty \omega_s \left( \frac{\operatorname{diam}(S_j)}{2}\right)^s.$$

The * Hausdorff dimension* of $E \subseteq \mathbb{R}^n$ is defined to be either the supremum of all $s$ such that $\mathscr{H}^s(E) = \infty$ or the infimum of all $s$ such that $\mathscr{H}^s(E) = 0$, which always agree.

The Hausdorff measure is defined from the Hausdorff outer measure via the standard Carathéodory construction. Here we have chosen the normalization so that the $n$-dimensional Hausdorff measure coincides with the $n$-dimensional Lebesgue measure.

With this normalization, $\mathscr{H}^0$ is the counting measure. $\mathscr{H}^1$ of a rectifiable curve is its length. In general, the $m$-dimensional Hausdorff measure of an $m$-manifold coincides with its canonical surface measure.

The Hausdorff dimension $E \subseteq \mathbb{R}^n$ specifies the *fair* dimension of $E$ in the following sense. While a curve and a surface both have zero $\mathscr{H}^3$-value, it would be unfair to say that they are *equally small*. Indeed, a curve would have zero $\mathscr{H}^2$-value as well.

We are now ready to state the following

Kakeya set conjecture.Every Kakeya set in $\mathbb{R}^n$ is of Hausdorff dimension $n$.

Let us transform the Kakeya set problem into a problem in harmonic analysis. For each $ \delta > 0$, $e \in \mathbb{S}^{n-1}$, and $a \in \mathbb{R}^n$, we let $T_e^\delta(a)$ be the set of all $x \in \mathbb{R}^n$ such that $\vert (x-a) \cdot e \vert \leq 1/2$ and $\vert x - (x \cdot e) e \vert \leq \delta$. This is essentially the $\delta$-neighborhood of the unit line segment in the direction of $e$ centered at $a$.

We define the * Kakeya maximal function* to be

$$f_\delta^*(e) = \sup_{a \in \mathbb{R}^n} \frac{1}{m(T^\delta_e(a))} \int_{T^\delta_e(a)} \vert f(x) \vert \, dx$$

for each $e \in \mathbb{S}^{n-1}$. The following conjecture implies the Kakeya set conjecture:

Kakeya maximal function conjecture.For all $\varepsilon > 0$, there exists a constant $C_\varepsilon > 0$ such that$$\lVert f_\delta^*\rVert_{L^n(\mathbb{S}^{n-1})} \leq C_\varepsilon \delta^{-\varepsilon } \lVert f \rVert_n.$$

Here $\lVert \cdot \rVert_{L^n(\mathbb{S}^{n-1})}$ is the norm of the Lebesgue space on $\mathbb{S}^{n-1}$, defined to be

$$\lVert f\rVert_{L^n(\mathbb{S}^{n-1})} = \int_{\mathbb{S}^{n-1}} \vert f(x) \vert^n \, d \sigma(x).$$

$d\sigma$ is the canonical surface measure on $\mathbb{S}^{n-1}$, which coincides with the $(n-1)$-dimensional Hausdorff measure on $\mathbb{S}^{n-1}$.

The formulation of the Kakeya maximal function conjecture is, in part, inspired by the problem of restricting the Fourier transform onto a lower-dimensional subset of $\mathbb{R}^n$. Specifically, we consider the spherical Fourier transform

$$\widehat{f \, d\sigma}(\xi) = \int_{\mathbb{S}^{n-1}} f(\xi) e^{-2 \pi i x \cdot \xi} \, d\sigma(x)$$

for functions $f:\mathbb{S}^{n-1} \to \mathbb{C}$. The following conjecture implies the Kakeya maximal conjecture.

Stein’s restriction conjecture.If $f \in L^\infty(\mathbb{S}^{n-1})$, then, for each $q > \frac{2n}{n-1}$, there exists a constant $C_q > 0$ such that$$\lVert\widehat{f \, d\sigma}\rVert_q \leq C_q \lVert f\rVert_\infty.$$

It is known that the Bochner–Riesz conjecture implies the restriction conjecture, whence it follows that the Bochner–Riesz conjecture implies the Kakeya set conjecture.

The Kakeya problem is an active area of research, with contributions from some of the finest analysts of our time. See (Lab) or (Tao1) for a quick survey. More systematic treatments can be found in (Wol) and (Tao2).