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For a direction $e$, the Kakeya maximal function is the supremum over translations of normalized averages of $|f|$ over unit-length Kakeya tubes in direction $e$. It measures how large a function can be on at least one thin tube in each direction. Normalizing by tube volume is essential when comparing different thicknesses.

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.

For a unit direction $e$, let $T_{\delta }(a,e)$ be a length-one Kakeya tube with transverse radius $\delta$, centered at $a$, and define the Kakeya maximal function byUsing normalized surface measure on the direction sphere, the Kakeya maximal conjecture is the following family of estimates, in the formulation relevant to this paper:The constant is independent of $\delta$ and $f$. Replacing round Kakeya tubes by comparable rectangular tubes changes only dimensional constants.

A bounded Kakeya set contains a unit line segment in every direction. The Kakeya Minkowski dimension conjecture says that every such set has full Minkowski dimension $n$. The maximal estimate in fact gives full lower as well as upper Minkowski dimension.

To prove that implication, let $E_{\delta }=\{x:\mathrm{dist}(x,E)<\delta \}$. Each unit segment in $E$ has a thinner tube contained in $E_{\delta }$, so ${\mathcal{K}}_{c\delta }{1}_{E_{\delta }}(e)\ge 1$ for every $e$, with a fixed dimensional $c>0$. Apply the maximal estimate to this indicator function:If $N_{\delta }(E)$ is the smallest number of radius-$\delta$ balls covering $E$, that cover, enlarged by a fixed factor, covers $E_{\delta }$. Thus $|E_{\delta }|{\lesssim }_n{\delta }^n{N}_{\delta }(E)$ andTaking the lower limit of $\log N_{\delta }(E)/\log (1/\delta )$ and then letting $\epsilon \downarrow 0$ gives lower Minkowski dimension at least $n$. Bounded subsets of ${\mathbb{R}}^n$ have upper Minkowski dimension at most $n$. Consequently both dimensions equal $n$, which proves the requested implication.

Choose $M\asymp {\delta }^{-1}$ angularly separated unit segments in a bounded Kakeya set, and let $F$ sum the indicators of their Kakeya tubes. Then $\int F\asymp 1$. Two tubes at angular separation $\alpha$ intersect in area at most $C{\delta }^2/(\delta +\alpha )$, so summing each harmonic angular row gives $\int F^2\lesssim \log (2/\delta )$. The Cauchy-Schwarz inequality on their union proves the displayed neighborhood lower bound. Since $|E_{\delta }|\lesssim {\delta }^2{N}_{\delta }(E)$, both planar Minkowski dimensions equal two.