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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.

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.