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For $0<\delta\ll 1$ we define a $\delta$-tube to be any rectangular box $T$ in $\mathbb{R}^d$ with $d-1$ sides length $\delta$ and one side of length $1$, observe that such tubes have volume $\delta^{d-1}$. Let $\mathbb{T}$ be an arbitrary collection of such $\delta$-tubes whose orientations form a $\delta-$ separated set of points on $\mathrm{S}^{d-1}$(i.e. for every direction $e_j, e_k, |e_j-e_k|\leq \delta$). If we use $\#\mathbb{T}$ to denote the cardinality of $\mathbb{T}$ , and $\chi_{T}$ to denote the indicator function of $T$, then the linear Kakeya conjecture says that for each $\frac{d}{d-1}<q\leq\infty$, there is a constant, independent of $\delta$ and the collection $\mathbb{T}$, such that $$ \|\sum_{T\in\mathbb{T}}\chi_{T}\|_{L^{q}(\mathbb{R}^d)}\leq C\delta^{(d-1)/q}(\#\mathbb{T})^{1/q} $$ My first question is that which example can show that the range $\frac{d}{d-1}<q\leq\infty$ is necessary?

Another question is the case $q=\infty$, then this estimate suggest that the LHS should be bounded, which is confusing me a little. Consider a collection of tubes passing through the origin and let the origin be the center of all tubes, then since the intersection of all tubes contains a ball of radius $\delta$, and the number of such tubes can be say $\delta^{-(d-1)}$, then we have $\|\sum_{T\in\mathbb{T}}\chi_{T}\|_{L^{\infty}(\mathbb{R}^d)}\approx\delta^{-(d-1)}$, which is not uniformly bounded, so where am I wrong?

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The (full) Kakeya maximal function conjecture can be more succinctly written as $$|| \sum_{t \in \mathbb{T}} \chi_{t} ||_{L^{\frac{d}{d-1}}(\mathbb{R}^d)} \ll_{\epsilon} \delta^{-\epsilon}.$$ Estimates (such as the one you wrote) involving the number of tubes $\#\mathbb{T}$ are only superficially stronger, thanks to factorization theory.

Note that the estimates are more interesting the smaller $q$ is. When $q=\infty$ the correct estimate is, as you observe, $|| \sum_{t \in T} \chi_{T} ||_{L^{\infty}(\mathbb{R}^d)} \ll \delta^{-(d-1)}.$ The quantity is not uniformly bounded at $L^{\infty}$, as your expression above might suggest.

To see that the estimate you wrote fails when $q=\frac{d}{d-1}$ (or, in other words, one needs the $\delta^{-\epsilon}$ in the formulation I gave above) consider the $\delta$ neighborhood of a Besicovitch set, say $B_{\delta}$. Since a Besicovitch set contains a unit line segment in every direction, for every $t \in \mathbb{T}$ there is translate of a tube in this direction in $B_{\delta}$. Let $\sum_{t \in \mathbb{T}} \chi_{t}$ denote the function obtained from using each such tube from $B_{\delta}$. Now $||\sum_{t \in \mathbb{T}} \chi_t||_{L^{1}} = \delta^{1-d} \delta^{d-1} = 1$, but the support of $\sum_{t \in \mathbb{T}} \chi_t$ tends to $0$ with $\delta$. It follows that $|| \sum_{t \in T} \chi_{T} ||_{L^{\frac{d}{d-1}}(\mathbb{R}^d)} \rightarrow \infty$ as $\delta \rightarrow 0$.

There a number of expository treatments of these topics. See: Tao's course notes and Wolff's survey article on Kakeya.

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  • $\begingroup$ Thank you very much. Can you explain a little what's the factorization theory? so when $q=\infty$, the above inequality indeed fails? $\endgroup$
    – Tomas
    Dec 11, 2013 at 13:37
  • $\begingroup$ @Shanlin, the idea behind factorization theory in this setting is very intuitive. If one has a small collection of $\delta$ separated tubes that have a lot of overlap, then one can get a bigger collection also with a lot of overlap by unioning the original collection with a random rotation of itself. One can repeat this process until one has a tube in every ($\delta$ separated) direction. This is explained on Tao's notes linked above, as well as his blog post on Stein's maximal principle: terrytao.wordpress.com/2011/05/12/steins-maximal-principle. $\endgroup$
    – Mark Lewko
    Dec 11, 2013 at 18:55

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