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On the unit circle, the arclength-measure transform satisfies $|\hat{\sigma }(x)|\lesssim (1+|x|{)}^{-1/2}$. This follows by removing angular neighborhoods of width $|x{|}^{-1/2}$ around the two stationary points and integrating by parts on the remainder. Gaussian positivity for even extension moments bounds the weighted fourth moment for any bounded $f$ by that for $f=1$. Radial integration of $(1+r{)}^{-2}{e}^{-\pi r^2/R^2}$ with area element $rdr$ costs only $\log R$, proving the displayed local estimate. Modulated disjoint caps and random signs convert it into the planar Kakeya neighborhood lower bound.

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.