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Math-2949 · mathematics · Analysis

Three-dimensional Kakeya maximal estimate

Status OpenLLM status Open unverified

Statement

Definitions and mathematical statement

Statement recorded in the supplied source.
For delta in (0,1), define $K_\delta $ f(v) as the supremum of the averages of |f| over radius-delta, length-one tubes directed along v in $S^2$. For every $\epsilon >0$, is $||K_\delta $ $f||_L3(S2)$ $<=$ $C_\epsilon $ $\delta ^{-\epsilon }||f||_L3(R3)$, with $C_\epsilon $ independent of f and delta?

Short English statement

For delta in (0,1), define $K_\delta $ f(v) as the supremum of the averages of |f| over radius-delta, length-one tubes directed along v in $S^2$. For every $\epsilon >0$, is $||K_\delta $ $f||_L3(S2)$ $<=$ $C_\epsilon $ $\delta ^{-\epsilon }||f||_L3(R3)$, with $C_\epsilon $ independent of f and delta?

Sources

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-2949-A1 · Literature report · LLM-generated · unknown ·

OpenAI claim summary for Math-2949

Three-dimensional Kakeya maxima: This is the three-dimensional tube maximal problem at exponent three on both sides. For every positive loss exponent the constant may depend on that exponent, but the bound is uniform over the small tube radius and over all input functions.

With the hypotheses and conventions above, the claimed result addresses the whole mathematical assertion on this page.

Three-dimensional Kakeya maximal estimate

OpenAI's claim

Three-dimensional Kakeya maxima: The claimed tube maximal estimate in three dimensions has $L^3$ input and output and loses only an arbitrarily small power of the tube radius. In particular, the density exponent in its corresponding shaded-tube estimate is three.

Mathematical statement of the claimed result

Mathematical claim 1.
For $0<\delta<1$, let $T_\delta(a,\omega)=\{a+t\omega+u:|t|\le1/2,\ u\cdot\omega=0,\ |u|\le\delta\}$ and $K_\delta f(\omega)=\sup_a(\pi\delta^2)^{-1}\int_{T_\delta(a,\omega)}|f|$. With ordinary surface measure on $S^2$, the asserted estimate is \[\|K_\delta f\|_{L^3(S^2)}\le C_\epsilon\delta^{-\epsilon}\|f\|_{L^3(\mathbb R^3)}\quad(\epsilon>0,\ f\in L^3(\mathbb R^3)),\] where $C_\epsilon$ does not depend on $\delta$ or $f$.

Outline of the argument

Three-dimensional Kakeya maxima: The proof keeps the fraction of each tube that is actually observed throughout a multiscale concentration model. Near-extremal indexed configurations are made stationary at critical exponents, while their position and direction information are retained jointly. The geometric argument excludes a positive critical loss, and the final reduction passes from tube shadings to general functions without worsening the cubic density dependence.

Formal verification

Three-dimensional Kakeya maxima: The pinned catalogue lists no Lean formalization for this result.

This site has not run the Lean development. These links describe OpenAI’s selected formal statements and should be read with their stated scope.

Status caveat

OpenAI reports that an internal model produced the manuscripts. Their arguments have not been independently verified as proofs by this site. Any portion outside the documented formal scope may contain mistakes, and the catalogue label reports the announced mathematical claim.

Sources

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