Sparse bounds for pseudodifferential operators

Abstract

We prove sparse bounds for pseudodifferential operators associated to Hörmander symbol classes. Our sparse bounds are sharp up to the endpoint and rely on a single scale analysis. As a consequence, we deduce a range of weighted estimates for pseudodifferential operators. The results naturally apply to the context of oscillatory Fourier multipliers, with applications to dispersive equations and oscillatory convolution kernels.2010 Mathematics Subject Classification. Primary: 35S05, Secondary: 42B25. Key words and phrases. Pseudodifferential operators, sparse domination, weighted theory. 1 We use the notation A À B to denote that there is a constant C such that A ď CB and the notation A Àǫ B to specify the dependence of the implicit constant in a certain parameter ǫ. We omit the constant factors of π coming from our normalisation of the Fourier transform. p p,Q :" |Q|´1 ş Q |f | p dx. Given 1 ď r, s ă 8, the pr, sq sparse form Λ S,r,s is defined as Λ S,r,s pf, gq :" ÿ QPS |Q| f r,Q g s,Q , and the r-sparse operator A r,S is defined as A r,S f pxq :" ÿ QPS f r,Q χ Q pxq. Theorem 1.2. Let a P S m ρ,δ for some m ă 0 and 0 ă δ ď ρ ă 1. Then for any compactly supported bounded functions f, g on R n , there exist sparse collections S and r S of dyadic cubes such that | T a f, g | ď Cpm, ρ, r, sqΛ S,r,s 1 pf, gqand | T a f, g | ď Cpm, ρ, r, sqΛ r S,s 1 ,r pf, gq for all pairs pr, s 1 q and ps 1 , rq such that m ă´np1´ρqp1{r´1{2q, 1 ď r ď s ď 2 2 Given 1 ď r ď 8, r 1 denotes its conjugate Hölder exponent, that is 1{r`1`r 1 " 1.

Some Remarks on the Pointwise Sparse Domination

“…Therefore, T a satisfies (5.1) for all s > 1. This result was obtained in [4]. Therefore, assuming additionally that T A is of weak type (q, q), q > 1, we obtain that T A satisfies (5.1) for s = q.…”

Sparse domination and the strong maximal function

“…Observe that any T ∈ Z R must have π 1 (T ) ⊆ π 1 (R). Indeed, if π 1 (T ) π 1 (R), by definition z T ∈ π 1 (T ) (1) × π 2 (T ) (k) and one has π 1 (R) ⊂ π 1 (T ) (1) . But this is impossible since π 1 (T ) (1) ⊆ π 1 (T ) \ π 1 (R).…”

“…Observe that any rectangle T ∈ E ,v, * R intersecting R vertically and whose z T is inside R must intersect (Ĩ) (1) ×J vertically (since othwerwise z T would be too close to p 0 ). According to Lemma 3.5 with = 1, the maximal rectangle with shortest height must have height at least 4|J| = |π 2 (R)| and the heights of these maximal rectangles increase exponentially.…”

“…According to Lemma 3.5 with = 1, the maximal rectangle with shortest height must have height at least 4|J| = |π 2 (R)| and the heights of these maximal rectangles increase exponentially. Therefore the number of maximal rectangles going through (Ĩ) (1) is at most…”

“…The operators that have been studied include but are not limited to rough singular integral operators, Bochner-Riesz multipliers, spherical maximal functions, singular integrals along manifolds, pseudodifferential operators, and the bilinear Hilbert transform. See for example [1,6,17,18,2,19,5,9,8]. Sparse domination provides a fine quantification of the mapping properties of operators that carries much more information than L p bounds.…”