Carleson's theorem
Statement
If , then for almost every . (Hunt, 1968, extended this to every with .)
Why is it true?
Given Kolmogorov's 1926 example of an function that diverges everywhere, it seemed entirely plausible that — barely a stronger condition — would fail the same way, or at least almost everywhere. Carleson's theorem is a genuine surprise: the tiny extra assumption of square-integrability rules out divergence everywhere except a set of measure zero. It closed a problem that had been open since Luzin conjectured it in 1913, and is regarded as one of the deepest theorems of twentieth-century analysis.
Proof sketch
A full proof is one of the hardest arguments in twentieth-century analysis; Fefferman gave a celebrated simplification in 1973 whose strategy can be sketched in three moves.
Step 1 (control by a maximal operator). It suffices to bound the Carleson maximal operator on , since a weak-type bound on plus density of nice functions (for which convergence is easy) implies the full a.e. convergence statement.
Step 2 (time–frequency decomposition into tiles). Decompose using wave packets adapted to dyadic tiles in the time–frequency plane — rectangles of area , exactly the atoms discussed in the 'Time and frequency' section below. Each tile carries a piece of localized to both a time interval and a frequency band.
Step 3 (organize tiles into trees and sum). Because the cutoff frequency can vary with , tiles relevant to form 'trees' ordered by time-frequency containment. Fefferman's key combinatorial estimate bounds the total energy carried by all trees, controlling in and completing the proof.
Topics that use this theorem
Step-by-step proofs
No step-by-step proof yet for this theorem.
References
- Terence Tao (2003). Recent progress on the restriction conjecture · arXiv:math/0311181
- Elias M. Stein, Guido Weiss (1971). Introduction to Fourier Analysis on Euclidean Spaces
- Loukas Grafakos (2014). Classical Fourier Analysis · DOI:10.1007/978-1-4939-1194-3