Worked solution: The planar case: every Kakeya set in $\mathbb{R}^2$ has dimension 2 (Davies, 1971)

Step 3 of 4: Córdoba's

L^2

trick: counting how tubes overlap

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In plain words

Two $\delta\times 1$ tubes crossing at a shallow angle $\theta$ overlap in a thin parallelogram of area about $\delta^2/\theta$ — the more nearly parallel they are, the longer they run side by side. Adding up the characteristic functions of all $N$ tubes into a single 'multiplicity function' $f$ , the integral $\int f^2$ exactly counts, with multiplicity, all pairwise overlaps $|T_i\cap T_j|$ . Because the $N$ directions are spread $\delta$ -evenly across a range of angles of size $\sim 1$ , summing $\delta^2/|\theta_i-\theta_j|$ over all pairs produces a harmonic-series divergence — a single factor of $\log N$ — rather than anything worse.

|T_i \cap T_j| \lesssim \min\!\Big(\delta,\ \frac{\delta^2}{|\theta_i-\theta_j|}\Big),\qquad f = \sum_{i=1}^N \chi_{T_i}

Detailed analysis

Order the tubes so $\theta_i \approx i\delta$ for $i=1,\dots,N$ . Fix $T_i$ : for each $k=1,\dots,N$ there are $O(1)$ tubes $T_j$ with $|\theta_i-\theta_j|\approx k\delta$ , each contributing overlap $\lesssim \delta^2/(k\delta)=\delta/k$ (capping at $|T_i\cap T_i|=\delta$ for $k=0$ ). So $\sum_j |T_i\cap T_j| \lesssim \delta\sum_{k=1}^{N}\frac{1}{k} \sim \delta\log N$ . Summing over all $N$ choices of $i$ , $\int f^2 = \sum_{i,j}|T_i\cap T_j| \lesssim N\delta\log N \sim \log N$ (using $N\delta\sim 1$ ). Restoring the more standard length- $N$ , radius- $1$ tube normalization used in Guth's lecture notes (equivalent by rescaling) gives the frequently quoted form $\int f^2 \lesssim N^2\log N$ .