# Kakeya conjecture (Tomas Wolff 1995)

There is a main obstacle to improve the kakeya conjecture,remain in dimension 3,and the  result established by Tomas Wolff in 1995 is almost the best result in $R^3$ even until now.the result establish by Katz and Tao can be view as a corollary of Wolff’s X-ray estimate.

For $f\in L_{loc}^1(R^d)$,for $0<\delta<1$:

$f_{\delta}^*:P^{d-1}\longrightarrow R$.$f_{\delta}^*(e)=\sup_{T}\frac{1}{|T|}\int_{T}|f|$.

$T$ is varise in all cylinders with length 1.radius $\delta$.axis in the $e$ direction.

$f_{\delta}^{**}:R^d\longrightarrow R$.$f^{**}_{\delta}(x)=sup_{T}\frac{1}{|T|}\int_{T}|f|$.

$T$ varise in cylinders contains x,length 1,radius $\delta$.

Keeping this two maximal function in mind,we give the statement of the Kakeya maximal function conjecture:

$||M_{\delta}f||_d\leq C_{\epsilon} \delta^{-\epsilon}||f||_d$

Where $M_{\delta}=f_{\delta}^*$ or $M_{\delta}=f_{\delta}^{**}$.

Because we have the obviously $1-\infty$ estimate:

$||f^*_{\delta}||_{\infty}\leq\frac{||f||_1}{|T|}=\delta^{1-d}||f||_1$.

$||f^{**}_{\delta}||_{\infty}\leq\frac{||f||_1}{|T|}=\delta^{1-d}||f||_1$.

So by the Riesz-Thorin interpolation we have:

$||M_{\delta}f||_{q}\leq C_{\epsilon}\delta^{-(\frac{d}{p}-1+\epsilon)}||f||_p$.           (*)

for $1\leq p\leq d,q\leq(d-1)p'$.the task is establish (*) for $(p,q)$ as large as posible in the range.

for the 2 dimension case,the result is well know.the key estimate is:

$\sum_{j}|T_i\cap T_j|\leq log(\frac{1}{\delta})|T_i|$

for $d\geq 3$ case,the main result of Wolff is:

$||M_{\delta}f||_q\leq C_{\epsilon}\delta^{-(\frac{d}{p}-1+\epsilon)}||f||_p$

hold for $p=\frac{d+2}{2}.q=(d-1)p'$. $M_{\delta}=f_{\delta}^*$ or $f_{\delta}^{**}$.

Now we sketch the proof.

prove $f_{\delta}^*,f_{\delta}^{**}$ cases together.

We can make some reduction:

the first one is we can assume the sup of $f$ is in a fix compact set.

the second is instead of consider $f_{\delta}^{**}$,we can consider $f_{\delta}^{***}(x)=\sup_{T}\frac{1}{|T|}\int_T|f|$.

where $T$ varies in all cylinder with radius $\delta$,length 1,axis $\frac{\pi}{100}$ with a fix direction.

the first reduction is obvious(why?)

the second reduction rely on a observe:

$||f_{\delta}^{***}||_q\leq A(\delta)||f||_p$          $\Longrightarrow$    $||f_{\delta}^{**}||_q\leq CA(\delta)||f|_p$

this is just finite cover by rotation of the coordinate and triangle inequality.

now we begin to establish a frame and put the two situations $f_{\delta}^*,f_{\delta}^{***}$ into it.

Let $M(d,1)$ be all line in $R^d$.

then $M(d,1)=R^d\times S^{d-1}/\sim$ is a $2d-2$ dim manifold.

$M(d,1)\longrightarrow P^{d-1}$

$l \longrightarrow e_l$

$e_l$ is the line parallel to $l$.and the middle point is original.

$dist(l_1,l_2)\sim \theta(l_1,l_2)+d_{mis}(l_1,l_2)$.

Wolf axiom:

$(A,d)$ metric space.

$\mu(D(\alpha,\delta)) \sim \delta^m$.$\alpha\in A$.$\delta \leq diam(A)$.

for certain $m\in R^+$.

$\forall \alpha\in A$.$F_{\alpha} \subset M(d,1)$ is given.and $\bar{\cup_{\alpha}F_{\alpha}}$ is compact.

$d(\alpha,\beta)\lesssim inf_{l\in F_{\alpha};m\in F_{\beta}}dist(l,m)$ for all $\alpha,\beta \in A$.

If $f:R^d\longrightarrow R$ then we define $M_{\delta}f:A\longrightarrow R$ by

$M_{\delta}f(\alpha)=\sup_{l\in F(\alpha)}\frac{1}{|T_{l}^{\delta}|}\int_{|T^{\delta}_l|}|f|$.

Property (**):

If $l_0\in \cup_{\alpha F_{\alpha}}$. $\Pi$ is a 2-plane.containing $l_0$and if $\sigma \geq \delta$ and if $\{\alpha_j\}_{j=0}^N$ is a $\delta-seperated$ subset of $A$ and for each j,there is $l_j\in F_{\alpha_j}$ with $dist(l_j,M(\Pi,l))<\delta$ and $dist(l,l_0)<\sigma$.then

$N\leq \frac{C\sigma}{\delta}$

# Kakeya conjecture in R^3

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

### 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:

$\mu(T_{i}\cap (\cup_{j\in I,j\neq i}T_j)).

where $T_i=T_i(x_i,\theta_i)$ satisfied $\cap_{i\in I}T_i$ is a $\delta$-neibeihood of kakeya set $X$.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 $I_{\delta}$ in different scales(this can be view as a multi-scale apporoach).

this inequality is easy to proof.just observed that $\mu(T_i\cap T_j)\sim \frac{1}{\theta_i-\theta_j}\delta^2$.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 $I$,but a low density subset $\hat I\subset I$,where $\frac{|\hat I|}{|I|}\sim \delta^{\lambda}$,and make $\lambda\to 0^+$.

### Kakeya estimates

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

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

$||\sum_{i\in I}\chi_{T_i}||_{d'}\lesssim \delta^{\frac{d-3}{d}}\theta^{\frac{d+1}{d}}m^{1-\beta}$
for some β > 0, where m is the directional multiplicity of $T_{\delta}$.(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 $\frac{5}{2}$,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 $X_{\delta}=\mu(\cup_{i\in I}T_i)$.

### Reduce to additive combination problem

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

# Kakeya Conjecture

Last year I read a nice blog articles Recent progress on the Kakeya conjecture and have several questions with this article.

follows the proof strategy called Multiscale analysis,although we can use the estimate with large $\delta_1$ to get estimate with small $\delta_2$,(may be loss some $\delta^c$ in the inequality in this way),but the main difficult is we should proof the new tubes with scales $\delta_2$ is contains in the the olders.as soon as we proof this ,to obtain a lower bound of minkwoski dimension with kakeya set, suffice to get following estimate :
the new cubes with scale $\delta_2$ contains a positive constants volumes of every old cubes with scale $\delta_1$.
this type of estimate is easy to attain because it is very similar to the “principle of close packing of spheres”.

in general ,we should not expect this claims:
the new tubes with scales $\delta_2$ is contains in the the older.

but if we can proof in some sense most of new tubes comes from this way maybe we can make progress on the original problem.

roughly speaking,we should partition the whole set of $T_{\delta}$ into two part,comes from old ones or not,for the first kind i.e contains in a old one,use the way explained above to treat.the second kind we need to proof the influence is very some or we can sometimes use the cubes from the first kind to instead the cubes from second kind and the measure of $|A|_{\delta}$ change little.