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

Frequently asked questions

Why does this argument fail to give the full Kakeya conjecture in

\mathbb{R}^3

and higher?

In the plane, two distinct directions determine an intersection angle, and $\delta$ -separated directions automatically make tube overlaps small on average — the harmonic sum over angles is exactly what keeps the loss to one logarithm. In $\mathbb{R}^3$ and higher, tubes can cluster inside lower-dimensional slabs or plates without any two of them being nearly parallel, escaping this simple pairwise-angle bookkeeping entirely; controlling that possibility needed the far more intricate multiscale and structural arguments of Wolff, Katz–Tao, and eventually Wang–Zahl.

Source

Roy O. Davies (theorem); standard proof exposition after A. Córdoba, 1971

Some remarks on the Kakeya problem

Condensed summary

Step 4 of 4: Cauchy–Schwarz forces the union to stay large

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In plain words

The trivial sum $\int f = \sum_i|T_i| \sim N\delta \sim 1$ just says the $N$ tubes have total area comparable to $1$ if they never overlapped at all. Cauchy–Schwarz relates this trivial sum, the overlap integral $\int f^2$ from the previous step, and the area of the union: $f$ is supported exactly on $\bigcup T_i$ , so $\int f = \int_{\cup T_i} f\cdot 1 \le |\cup T_i|^{1/2}(\int f^2)^{1/2}$ . Squaring and rearranging turns the upper bound on overlap directly into a lower bound on the area of the union — with only a single logarithm lost compared to the impossible best case of total disjointness.

\Big(\int f\Big)^{2} \le |{\textstyle\bigcup_i T_i}| \cdot \int f^2 \quad\Longrightarrow\quad \Big|\bigcup_{i=1}^N T_i\Big| \gtrsim \frac{1}{\log(1/\delta)}

Detailed analysis

From Step 3, $\int f^2 \lesssim \log(1/\delta)$ (in the $\delta$ -tube normalization) and $\int f \sim 1$ . Cauchy–Schwarz on the domain $\bigcup T_i$ gives $1 \sim (\int f)^2 \le |\bigcup T_i|\cdot \int f^2 \lesssim |\bigcup T_i|\cdot \log(1/\delta)$ , hence $|\bigcup T_i| \gtrsim 1/\log(1/\delta)$ . Covering $\bigcup T_i$ by $\sim \delta^{-2}$ squares of side $\delta$ shows that at least $\gtrsim \delta^{-2}/\log(1/\delta)$ of them are needed, so the $\delta$ -covering number of (a neighborhood of) $K$ grows like $\delta^{-2+o(1)}$ as $\delta \to 0$ : this is exactly the statement that $\dim_M(K) = 2$ . Córdoba's 1977 paper carries out this argument in full rigor and with sharp constants; Davies' original 1971 argument reached the same numerical conclusion — and simultaneously the Hausdorff dimension, a priori a stronger statement — by a different, purely measure-theoretic route using projective duality between points and lines together with Marstrand-type projection theorems, without discretizing into tubes at all.