Kakeya conjecture in R^3

Kakeya conjecture in $latex R^3$ is very subtle.in fact wolff stay the best(but not very difficult to get,just use the structure so-called hairbrush)result $latex \frac{5}{2}$ until the result of Katz and Tao $latex \frac{5}{2}+\epsilon$.Where $latex \epsilon$ is a constant independent with kakeya set.and in the article of Tao,they proved $latex \epsilon>\frac{1}{10^{10}}$.

Two-dimensional case

first we overview the case of dimension 2,these is the only case that is proved.and the key point is the estimate:

$latex \mu(T_{i}\cap (\cup_{j\in I,j\neq i}T_j))<log(\frac{1}{\delta})\mu(T_i)$.

where $latex T_i=T_i(x_i,\theta_i)$ satisfied $latex \cap_{i\in I}T_i$ is a $latex \delta$-neibeihood of kakeya set $latex X$.to remember one thing:this is equivalent to the maximal function version of kakeya conjecture,but for the minkoski version,there is a extra structure for the group $latex I_{\delta}$ in different scales(this can be view as a multi-scale apporoach).

this inequality is easy to proof.just observed that $latex \mu(T_i\cap T_j)\sim \frac{1}{\theta_i-\theta_j}\delta^2$.and to remember one thing:the inequality can be view as a uniformly estimate of overlap of the kakeya set,that is just mean the overlap would not concentrate to much at a lonely stick.this is enough to get a proof of the 2 dimension case just by a density decrement trick:we just not consider about the whole set $latex I$,but a low density subset $latex \hat I\subset I$,where $latex \frac{|\hat I|}{|I|}\sim \delta^{\lambda}$,and make $latex \lambda\to 0^+$.

Kakeya estimates

Let $latex \sigma\leq \delta\leq \theta<<1$,and let $latex T_{\delta}$ be a collection of $latex \delta$-tubes.whose set of directions all lie in a cap of radius $latex \theta$. Let $latex 2<d<3$ be fixed.
• If we have a Kakeya estimate at some dimension d, and if the collection $latex     T_{\delta}$ is direction-separated, then


$latex ||\sum_{i\in I}\chi_{T_i}||_{d’}\lesssim \delta^{\frac{d-3}{d}}\theta^{\frac{d+1}{d}}$(1)
• If we have an X-ray estimate at some dimension d, and if $latex T_{\delta}$ consists ofessentially distinct tubes, then


$latex  ||\sum_{i\in I}\chi_{T_i}||_{d’}\lesssim \delta^{\frac{d-3}{d}}\theta^{\frac{d+1}{d}}m^{1-\beta}$
for some β > 0, where m is the directional multiplicity of $latex T_{\delta}$.(2)

So obviously the X-ray estimate is stronger than the kakeya estimate.it is just give the information of the overlap of the sticks with the same direction.

in fact wolff have establish the X-ray estimate at dimension $latex \frac{5}{2}$,so (2) just come from a rescaling argument.

The sticky reduction

renormalization process,just consider the process to make the thin sticks to be fat.and to proof this structure nearly has Markov property.but with a very small error term when change the scale.this is proved by the X-ray estimate.

Triple intersection estimate

Use Hardy-Litterwood-Soblev inequality,we can get a so called triple intersection estimate in general,said the triple intersection is smaller than the situation the 3 lines move together.and we just accosiate this to the cap-cup principle to get some information of the volume of $latex X_{\delta}=\mu(\cup_{i\in I}T_i)$. img_0012

Reduce to additive combination problem

The right problem is just you have a $latex n\times n$ cubes,and there is some sticks according them,if the distance of sticks is