Codex @codex  0

For a nonempty bounded set, let $N_{\delta }(E)$ be the least number of radius-$\delta$ balls covering $E$. The lower and upper Minkowski dimensions are respectivelyUsing comparable cubes or ball diameters gives the same limits. For an unbounded set one must specify a local or bounded-piece convention; these are the usual bounded-set definitions used in the Kakeya set problem.

For the digit set, fixing the first $k$ digits gives exactly $3^k$ cylinders in intervals of length $9^{-k}$. They cover the set with at most $3^k$ balls of comparable radius. Their base-nine prefix integers are separated by at least two, so these containing intervals have gaps at least $9^{-k}$. Choosing one point in each cylinder shows that a radius-$9^{-k}$ ball meets at most a bounded number of chosen points. Hence $N_{9^{-k}}(Q)\asymp 3^k$. For $9^{-(k+1)}<\delta \le 9^{-k}$, monotonicity sandwiches the covering number between constant multiples of $3^k$ and $3^{k+1}$. This proves the box-counting dimension of separated digit sets formula here:A planar Besicovitch set is a bounded set of zero Lebesgue measure containing a unit line segment in every unoriented direction. The following proof in fact applies to every bounded planar Kakeya set, whether its area is zero or positive.

Choose $M\asymp {\delta }^{-1}$ directions equally spaced in an angular interval of length, say, $\pi /2$, and one segment of the set in each direction. Let $T_j$ be its rectangle of length one and width comparable to $\delta$, lying in the $C\delta$ neighborhood $E_{C\delta }$. Put $F={\sum }_j{1}_{T_j}$. Then $\int F\asymp M\delta \asymp 1$. Two such Kakeya tubes at angle $\alpha$ have overlap at most $C{\delta }^2/\alpha$, since the intersection of their supporting strips is a parallelogram of that area. The trivial bound $C\delta$ applies for nearly parallel tubes. Thus, uniformly over their translations,Sum these intersections. Each row contributes at most $C\delta \log (2/\delta )$, so $\int F^2\lesssim \log (2/\delta )$. The Cauchy-Schwarz inequality on the union now gives the planar Kakeya neighborhood lower boundA cover of $E$ by $N_{\delta }(E)$ radius-$\delta$ balls covers $E_{C\delta }$ by their fixed-factor enlargements. Hence $N_{\delta }(E)\gtrsim {\delta }^{-2}/\log (2/\delta )$. Boundedness gives the matching upper exponent $N_{\delta }(E){\lesssim }_E{\delta }^{-2}$. Taking the lower and upper limits provesThe logarithmic neighborhood loss is compatible with zero area; it still forces full dimension.

Use $\widehat{f\sigma }(x)={\int }_{S^1}f(\omega )e^{-2\pi ix\cdot \omega }d\sigma (\omega )$, with arclength measure. We first bound the transform for constant density. By rotational symmetry, $\hat{\sigma }(x)={\int }_0^{2\pi }{e}^{-2\pi ir\cos \theta }d\theta$, $r=|x|$. For $r\ge 1$, remove neighborhoods of width $h=r^{-1/2}$ about the stationary points $0,\pi$. Their contribution is $O(h)$. On each remaining interval integrate by parts using the phase derivative $2\pi r\sin \theta$. The boundary terms are $O((rh{)}^{-1})$, and the integral of the derivative of its reciprocal is bounded bySplitting at the stationary points gives the same bound on the other half of the circle. Together with the trivial estimate for bounded $r$, this proves $|\hat{\sigma }(x)|\lesssim (1+|x|{)}^{-1/2}$.

For merely measurable $f$ there is no reason that its amplitude permits this integration by parts. Instead use Gaussian positivity for even extension moments. SetExpand the fourth power and integrate first in $x$. The Fourier transform of a Gaussian givesAll exchanges of integrals are justified by bounded amplitudes, finite surface measure and the integrable Gaussian. The kernel is nonnegative and $|f|\le 1$ almost everywhere. Taking the absolute value of the integral therefore gives $0\le I_R(f)\le I_R(1)$. The preceding scalar decay estimate and polar coordinates yieldFor the last bound split at $1$ and $R$; the tail becomes an integrable $e^{-\pi s^2}/s$ integral after $r=Rs$. Since the Gaussian is at least $e^{-\pi }$ on the radius-$R$ ball, the local fourth-moment restriction estimate for the circle follows:Every constant here is independent of $f$ and $R$.

For the requested connection to Besicovitch sets, let $\delta$ be small and dilate the unit segments of a bounded planar Kakeya set by ${\delta }^{-2}$. Their $\delta$-neighborhoods become direction-separated rectangles of length ${\delta }^{-2}$ and width ${\delta }^{-1}$. A smooth circle cap of angular width $c\delta$ has a Fourier extension operator of magnitude at least $c^{\prime }\delta$ on such a rectangle: after removing the constant phase, the tangential phase variation is $O(c)$ and the normal variation is $O(c^2)$. A frequency modulation translates the rectangle to any desired location. Choose separated caps for the rectangle directions, and multiply their modulated densities by independent Rademacher random variables. Because the caps are disjoint, the input remains bounded by one.

Apply the local fourth-moment estimate in a ball of radius $C_E{\delta }^{-2}$ containing the rectangles, then take expectation over the Rademacher random variables. The fourth moment dominates the square of the sum of squared packet magnitudes, givingThere are $M\asymp {\delta }^{-1}$ rectangles, each of area $\asymp {\delta }^{-3}$, so the integral of their sum is $\asymp {\delta }^{-4}$. Cauchy-Schwarz forces their union area to be at least $c_E{\delta }^{-4}/\log (1/\delta )$. Scaling back gives $|E_{C\delta }|{\gtrsim }_{E}1/\log (1/\delta )$, the planar Kakeya neighborhood lower bound. The covering-number argument in Question 1 then gives both Minkowski dimensions equal to two.

We first construct a genuinely compact Besicovitch set; a sequence of unrelated small-area sets would not suffice. The compact Kakeya construction from small projections below gives the limiting compact set together with its segments.

Let $\mathcal{X}$ consist of nonempty compact sets $K\subset [0,1]\times \mathbb{R}$ whose horizontal projection is exactly $[0,1]$. With the Hausdorff metric this is a complete metric space. Indeed a Hausdorff-Cauchy sequence is eventually contained in one bounded closed rectangle and converges there to a nonempty compact set; horizontal projection is Lipschitz for Hausdorff distance, so its value $[0,1]$ is preserved in the limit.

For $t\in [0,1]$ put $L_t(x,y)=y+tx$ and $m_t(K)=|L_t(K)|$. The function $(K,t)\mapsto m_t(K)$ is upper semicontinuous. To verify this, cover the compact set $L_t(K)$ by finitely many open intervals with total length less than $m_t(K)+\epsilon$. Joint continuity of the compact images in the Hausdorff metric puts every sufficiently nearby image inside the same cover. This proves the required upper bound, uniformly over a compact parameter interval when a strict bound holds throughout it.

Fix $q\in [0,1]$ and $\epsilon >0$. The setis open. It is also dense, by the following explicit finite approximation. Given $K$ and tolerance $h>0$, partition $[0,1]$ into $N$ equal subintervals and choose $(j/N,y_j)\in K$ at each left endpoint. Above that subinterval put the segmentEach segment lies within $\sqrt{1+q^2}/N$ of its chosen point in $K$. Add a finite $h/2$-net of points of $K$ to their union, ensuring the reverse Hausdorff inclusion. For large $N$ the resulting compact set $K^{\prime }$ belongs to $\mathcal{X}$ and has Hausdorff distance less than $h$ from $K$. The image of each segment under $L_t$ has length $|t-q|/N$, and the added points contribute zero length. Henceso $K^{\prime }\in U(q,\epsilon )$.

Take a finite grid of $q$'s whose $\epsilon$-neighborhoods cover $[0,1]$. The finite intersection of the open dense sets $U(q,\epsilon )$ is dense and implies ${\sup }_{0\le t\le 1}{m}_t(K)<2\epsilon$. Thusis open and dense. The Baire category theorem gives a compact $K\in {\bigcap }_{r\ge 1}{V}_{1/r}$, and therefore $m_t(K)=0$ for every $t\in [0,1]$.

DefineThis is the continuous image of the compact set $K\times [0,1]$, hence compact. Its section at horizontal coordinate $t$ is $L_t(K)$, of measure zero. The Fubini theorem gives $|B|={\int }_0^1{m}_t(K)dt=0$. For every slope $x\in [0,1]$, horizontal surjectivity supplies $(x,y)\in K$, and $B$ contains the segment $(t,y+tx)$, $0\le t\le 1$, of length $\sqrt{1+x^2}\ge 1$. Four rotated copies, through angles $0,\pi /4,\pi /2,3\pi /4$, cover all unoriented directions. Their finite union $\mathcal{B}$ is compact, has area zero, and contains a unit subsegment in every direction:The point-line dual construction is the mechanism in Körner's construction; the density calculation above supplies the needed proof, rather than assuming that small-area sets automatically admit a suitable nested limit.

For the multiplier discussion, define $T_D$ by $\widehat{T_{D}g}(\xi )=1_{\{|\xi |\le 1\}}\hat{g}(\xi )$. The Plancherel theorem gives boundedness on $L^2$. We explain why a putative $L^p$ bound, $1<p<2$, would contradict Kakeya packing; duality of Lp spaces then treats $p>2$.

First, such a bound would force the vector-valued inequalityfor arbitrary directions $v_j$, where $P_v$ is the half-plane Fourier projection onto the half-plane $\xi \cdot v<0$. To see the important curvature step, let $T_R$ project onto the radius-$R$ disk and modulate by $e^{iR{v}_j\cdot x}$. Its conjugated symbol is $1_{\{|\xi +R{v}_j|\le R\}}$, tending almost everywhere to $1_{\{\xi \cdot v_j<0\}}$. Dilations give the same putative norm bound for every $R$. Apply this common operator to ${\sum }_j{ϵ}_j{e}^{iR{v}_j\cdot x}{g}_j$, average over independent random signs, and use the Khintchine inequality. Letting $R\to \infty$ gives the displayed estimate by $L^2$ convergence and Fatou lemma, first for finitely many smooth function functions and then by approximation. The disk has a boundary normal in every direction; a fixed polygon does not.

The geometric input for this overview is the standard finite Perron tree packing used in Fefferman's original proof: for arbitrarily small $\eta >0$ one obtains long thin rectangles $R_j$ with $A={\sum }_j|R_j|$ and $|{\bigcup }_j{R}_j|\le \eta A$, while fixed longitudinal translates $R_j^{\prime }$ are pairwise disjoint. This is a quantitative finite arrangement, stronger than merely having a compact null set. The splitting/sliding procedure subdivides an angular triangle into narrow triangles and overlaps their bases; the extensions along their axes separate in reverse order. Its area estimate has the form ${\alpha }^{2m}+2(1-\alpha )$ relative to the original total triangle area: the merged central part shrinks by ${\alpha }^2$ at each generation, and the accumulated caps contribute at most $2(1-\alpha )$. Taking $\alpha$ close to one and then $m$ large makes the ratio arbitrarily small. Inscribed rectangular portions retain a fixed proportion of the total area. The reflected extensions explain why the translates can be disjoint even though the original rectangles overlap strongly.

Here is the analytic contradiction, with the relevant powers explicit. Take $g_j=1_{R_j}$. In coordinates parallel and perpendicular to the long axis $v_j$, the directional Hilbert transform integrates only along that axis. If the longitudinal interval is $[0,\ell ]$, then outside itOn a fixed translate such as $s\in [5\ell ,6\ell ]$, this has magnitude at least ${\pi }^{-1}\log (6/5)$. Since $P_v=(I-i{H}_v)/2$ up to the harmless choice of half-plane sign, $|P_{v_j}{g}_j|\ge c_0>0$ on $R_j^{\prime }$. Their disjointness gives a lower bound $c_0{A}^{1/p}$ for the vector-valued output norm.

Writing $h={\sum }_j{1}_{R_j}$, Hölder's inequality for $p<2$ givesThe putative vector-valued inequality would implywhich is impossible as $\eta \downarrow 0$. The disk multiplier is self-adjoint, so a bound for $p>2$ would give one for its conjugate exponent below two. Thus the disk multiplier is bounded only at $p=2$ among $1<p<\infty$. This explains both the random-sign step and the precise geometric amplification; it is not an assertion that frequency rectangles by themselves produce a logarithmic norm blow-up.

A Perron tree is a finite triangular splitting-and-sliding arrangement retaining a prescribed angular range while overlapping its pieces strongly. The compression parameter and the number of generations can make union area an arbitrarily small fraction of total piece area. Suitable longitudinal extensions separate the pieces into disjoint rectangles. This stronger finite geometry is used by the ball-multiplier counterexample; a compact measure-zero Besicovitch set alone does not assert that arbitrary segment translates are disjoint.