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The maximal estimate used in the classical Kakeya set problem asks, for every $\epsilon >0$, for a bound from $L^n({\mathbb{R}}^n)$ to $L^n(S^{n-1})$ with loss at most ${\delta }^{-\epsilon }$. Constants may depend on dimension and $\epsilon$, but not tube thickness or the function. Applied to an indicator function of a neighborhood of a Kakeya set, it gives the Kakeya Minkowski dimension conjecture.

The dimension assertion asks that every bounded Kakeya set in ${\mathbb{R}}^n$ have full Minkowski dimension. The Kakeya maximal conjecture implies the stronger lower-dimension conclusion: its application to $E_{\delta }$ gives $|E_{\delta }|{\gtrsim }_{n,\epsilon }{\delta }^{n\epsilon }$, hence $N_{\delta }(E){\gtrsim }_{n,\epsilon }{\delta }^{-n+n\epsilon }$ for every $\epsilon >0$.

A bounded subset of ${\mathbb{R}}^n$ is a Kakeya set if it contains a unit line segment in every direction. Segment positions may vary with direction. Such a set need not have positive Lebesgue measure; the dimension problem concerns its Minkowski dimension or other notions of 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.