Convolution with measures on some complex curves

Transcription

1 Convolution with measures on some complex curves Seheon Ham School of Mathematics, Korea Institute for Advanced Study joint work with Hyunuk Chung August 6, 2014 Chosun University, Gwangju

2 Complex curves A model example : h(z) = (z, z 2,..., z d 1, z d ) h : Ω C d, Ω is a domain in C. This is regarded as a 2-dimensional surface in R 2d defined by a real mapping z = (u, v) h(u, v) = (u, v, u 2 v 2, 2uv,..., Re(z d ), Im(z d )).

3 Convolution operator Let h(z) = (z, z 2,..., z d 1, φ(z)) in C d, d 2 where φ(z) is an analytic function defined on Ω C. We consider Af(x) = f(x h(z)) D φ(d) (z) d(d+1) dµ(z) =: f σ(x). 4 D is the unit ball in C. dµ(z) is the surface measure dµ(z) = dµ(h(z)) = dudv for z = u + iv. 4 4 d(d+1) dµ(z), φ (d) (z) d(d+1) dµ(z) det(h (z), h (z),..., h (d) (z)) an analogue of the affine arclength measure for real curves.

4 Convolution operator Let h(z) = (z, z 2,..., z d 1, φ(z)) in C d, d 2 where φ(z) is an analytic function defined on Ω C. We consider Af(x) = f(x h(z)) D φ(d) (z) d(d+1) dµ(z) =: f σ(x). 4 trivial estimates : Af L (R 2d ) f L (R 2d ), Af L 1 (R 2d ) f L 1 (R 2d ) By interpolation, Af L p (R 2d ) f L p (R 2d ) for all 1 p. How much the integrability can be improved? (We expect the same results as those for real curves in R d.)

5 Complex curves Motivation : surfaces of half the ambient dimension (u, v) (u, v, u 2 v 2, 2uv), (u, v) R 2 x (x, φ(x)), x R d, φ : R d R d Restriction estimates : Prestini, Christ, Oberlin Convolution estimates : Drury-Guo The same results as those in 2-dimensional case (curves in R 2 ).

6 Some earlier results for real curves Space curves when d 3 : model examples Moment curve : γ(t) = (t, t 2,..., t d ). Monomial curves : For distinct real numbers a i s, γ(t) = (t a 1, t a 2,..., t a d). Polynomial curves of simple type : For a polynomial P (t) of degree at most N, γ(t) = (t, t 2,..., t d 1, P (t)). General polynomial curves : γ(t) = (P 1 (t),..., P d (t)).

7 Convolution estimates for real curves Moment curve γ(t) = (t, t 2,..., t d ) Af(x) = I f(x γ(t))dt = f µ(x). I = [0, 1], µ(ϕ) = I ϕ(γ(t))dt Littman (d = 2), Oberlin (d = 3) : p d = d+1 2, q d = d(d+1) 2(d 1), Af L q d C f L p d Christ : For d 2, Af L q d, C f L p d,1 Stovall : For d 2, Af L q d,v C f L p d,u whenever u < q d, v > p d, and u < v.

8 Convolution estimates for real curves Monomial curve γ(t) = (t a 1, t a 2,..., t a d) Gressman : For positive integers a 1 < < a d, Af L qa, C f L pa,1 where p a = 1 d (a a d ) and q a = 1 d 1 (a a d ). If a i = i, then p a = p d and q a = q d. Generally, p a > p d and q a > q d. To obtain the boundedness at (p d, q d ), we need the affine arclength measure of γ(t).

9 The affine arclength measure The affine arclength measure w(t)dt = [ det ( γ (t), γ (t),..., γ (d) ) ] 2 (t) d(d+1) dt invariant under diffeomorphic reparametrization (e.g. γ(t) γ(φ(s))) compensating for degeneracy of γ uniform bounds which are the same as those for nondegenerate situation

10 Convolution estimates for real curves Polynomial curve γ(t) = (P 1 (t),..., P d (t)) Af(x) = I f(x γ(t))w(t)dt = f µ(x). Affine arclength measure : µ(ϕ) = = I ϕ(γ(t))w(t)dt I ϕ(γ(t))[det ( γ (t), γ (t),..., γ (d) (t) ) ] 2 d(d+1) dt Oberlin (d = 2) : Af L q 2 C f L p 2. Dendrinos-Laghi-Wright (d = 2, 3): Af L q d,p d +ε C f L p d Stovall (d 4): Af L q d,v C f L p d,u where u < q d, v > p d, and u < v.

11 Main ingredient from restriction estimate Let γ(t) = (P 1 (t),..., P d (t)), L(t) = det ( γ (t), γ (t),..., γ (d) ) d(d+1) (t) = (w(t)) J Φ (t 1,..., t d ) = det ( γ (t 1 ), γ (t 2 ),..., γ (t d ) ). 2, Theorem [Dendrinos-Wight, 2010] There exists a decomposition R = Ī j such that I j are pairwise disjoint open intervals, 1. L(t) t b j K j for every t I j 2. Whenever (t 1,..., t d ) I d j, J(t 1,..., t d ) d k=1 L(t k ) 1 d l<k t k t l 3. Φ = γ(t k ) has bounded generic multiplicities.

12 Restriction estimates for complex curves Restriction estimates for complex curves Let h(z) = (z, z 2,..., φ(z)) where φ(z) is an analytic function. ( 4 f(h(z)) q φ (d) q (z) d(d+1) dµ(z))1 C f L R 2 p (R 2d ), whenever q = 2 d(d+1) p, 1 p < d2 +d+2 d 2 +d. dµ(z) is the surface measure dµ(z) = dµ(h(z)) = dudv for z = u + iv 4 φ (d) (z) d(d+1) dµ(z) det(h (z), h (z),..., h (d) (z)) 4 d(d+1) dµ(z).

13 Restriction estimates for complex curves Oberlin, d = 2 φ(z) is an analytic function, φ (z) and the map (z 1, z 2 ) (z 1 z 2, φ(z 1 ) φ(z 2 )) both have generic multiplicities at most N Bak-H, d 3 d 3 and φ(z) = z N, N 0. d = 3 and φ(z) is an arbitrary polynomial of degree at most N. By Bézout s theorem, the transformation (z 1,..., z d ) h(z 1 ) + h(z 2 ) + + h(z d ) has bounded generic multiplicity N (d 1)!.

14 Restriction estimates for complex curves Lemma [Bak-H] Let h(z) = (z, z 2,..., z d 1, z N ) for an integer N d with d 2. Set J d (z 1,..., z d ) = J C (z 1,..., z d ) = det(h (z 1 ),..., h (z d )) where z j C, 1 j d. Then C may be written as the union (ignoring a null-set) of C(d, N) sectors l with vertex at the origin such that for each 1 l C(d, N), we have J d (z 1,..., z d ) c(d, N) max 1 j d z j N d 1 i<j d z j z i where z j l. Here, C(d, N) and c(d, N) are positive constants depending only on d and N.

15 Restriction estimates for complex curves Lemma [Bak-H] Let h(z) = (z, z 2, φ(z)). There exists a positive integer M = M(N) and a collection of convex open sets B 1,..., B M, which are pairwise disjoint, such that C = M l=1 B l ignoring null sets. Moreover, there exists a constant c(n) > 0 such that for 1 l M, J C (z 1, z 2, z 3 ) c(n) max 1 j 3 φ (z j ) 1 i<j 3 z j z i, whenever z 1, z 2, z 3 B l.

16 Restriction estimates for complex curves By the proof of Lemma above, we have that φ (z) N 3 n=k+1 u n z k =: H k z k whenever z B l S E k with E k = G k or D k. Jacobian of the mapping (z 1, z 2, z 3 ) Φ h (z 1, z 2, z 3 ), where h(z) = (z, z 2, z k+3 ) on B l. More precisely, it is known that J C (z 1, z 2, z 3 ) c(n) max 1 j 3 z j k 1 i<j 3 z j z i, if z 1, z 2, z 3 B l for some B l S E k with E k = G k or D k.

17 Convolution with measure on complex curves Let us define Af(x) = D f(x h(z)) φ(d) (z) where h(z) = (z, z 2,..., z d 1, φ(z)). 4 d(d+1) dµ(z), Theorem Let d 2 and φ(z) = z N for a nonnegative integer N. Then there exists a constant C = C(d, N) such that where u < q d, p d < v, and u < v. Af L q d,v (R 2d ) C f L p d,u (R 2d ) (1) Theorem Let h(z) = (z, z 2, φ(z)), where φ(z) is a polynomial of degree at most N. Then (1) holds for d = 3.

18 Convolution with measure on complex curves Comment: Our argument here relies on the combinatorial technique so called band structure argument due to Christ. Since we need to treat weighted measure dσ, we also adapt the method by DLW or Stovall. Dealing with complex variables, the construction of the band structure should be slightly modified. It is not enough to arrange the absolute value of z 1,..., z d because two variables which have the same size can be separated.

19 Sketch of proof Let h(z) = (z, z 2, φ(z)). It is enough to consider, for some k depending on B, T f(x) = B f(x h(z)) dσ(z) = f(x h(z)) B z k 3dµ(z). Restricted weak type estimate T : L 2,1 L 3, T χ E, χ F C E 1 2 F 2 3 and T χ E, χ F C E 2 3 F 1 2, Let us define where T f(x) = B α = T χ E, χ F F and β = χ E, T χ F, E f(x + h(z))dσ(z). Note that α F = β E. Goal : E α 4 β 2 or F α 3 β 3.

20 Sketch of proof : trilinear variant Lorentz space estimate Lemma Let E 1, E 2, G R 6 be measurable sets with finite measure. Suppose that for all x G and α 1 α 2. Then where β = α 1 G E. T χ E1 (x) α 1 and T χ E2 (x) α 2 E 2 α 1 β 2 α 3 2, Lemma Let F 1, F 2, G R 6 be measurable sets with finite measure. Suppose that T χ F1 (y) β 1 and T χ F2 (x) β 2 for all y G and β 1 β 2. Then F 2 α 3 1 β 1 β 2 2, where α 1 = β 1 G F 1.

21 Sketch of proof Since T χ G, χ F1 = χ G, T χ F1 β 1 G, we can define a set F1 1 = {x F 1 : T χ G (x) 1 2 α 1}. Similarly we can define G 1 = {y G : T χ F 1(y) β 1}. For y 0 G 1, we define P = {z 1 B : y 0 + h(z 1 ) F1 1 }, σ(p ) 1 4 β 1, Q z1 = {z 2 B : y 0 + h(z 1 ) h(z 2 ) G}, σ(q z1 ) 1 2 α 1, R z1,z 2 = {z 3 B : y 0 + h(z 1 ) h(z 2 ) + h(z 3 ) F 2 }, σ(r z1,z 2 ) β 2. Recall that σ(p ) = P z k 3dµ(z).

22 Sketch of proof Set S = {(z 1, z 2, z 3 ) : z 1 P, z 2 Q z1, z 3 R z1,z 2 } Φ h (z 1, z 2, z 3 ) = h(z 1 ) h(z 2 ) + h(z 3 ). Then y 0 + Φ h (S) F 2. Since Φ h (z 1, z 2, z 3 ) has bounded generic multiplicity 2N, it follows that F 2 = S J RΦ h (z 1, z 2, z 3 ) dµ(z 1 )dµ(z 2 )dµ(z 3 ) S J CΦ h (z 1, z 2, z 3 ) 2 dµ(z 1 )dµ(z 2 )dµ(z 3 ) S max{ z 1, z 2, z 3 } 2k z 2 z 1 2 z 3 z 1 2 z 3 z 2 2 dµ(z 1 )dµ(z 2 )dµ(z 3 )

23 Sketch of proof Let ν = 3 k+6 and B α = {z B : z (16πν) ν α ν }. 1. z 1 (16πν) ν β ν 1, z 2 (16πν) ν α ν 1, and z 3 (16πν) ν β ν z 2 z 1 cα 2 1 z 2 k 6 ε 1 z 1 k 6 (1 ε 1) where ε 1 = 0 if z 1 B α1 /2, or ε 1 = 1 if z 1 / B α1 / z 3 z 1 cβ 2 2 z 3 k 6 ε 2 z 1 k 6 (1 ε 2) where ε 2 = 0 if z 1 B β2 /2, and ε 2 = 1 if z 1 / B β2 / z 3 z 2 cβ 2 2 z 3 k 6 ε 3 z 2 k 6 (1 ε 3) where ε 3 = 0 if z 2 B β2 /2, and ε 3 = 1 if z 2 / B β2 /2.

24 Sketch of proof Also it is obvious that max{ z 1, z 2, z 3 } 2k z 1 a z 2 b z 3 c for positive constants a,b,c satisfying a + b + c = 2k. Then F 2 S max{ z 1, z 2, z 3 } 2k z 2 z 1 2 z 3 z 1 2 z 3 z 2 2 dµ(z 1 )dµ(z 2 )dµ(z 3 ) α 1 β 2 2 α 1β 1 β 2 = α 2 1 β 1β 3 2. If β 2 α 1, then F 2 α 3 1 β 1β 2 2. If β 2 α 1, we need further observations on α 1 β 2 2.

25 Sketch of proof 1 We assume that z 1 B α1 /2. Then z 2 z 1 cα2 1 z 1 k 6. We consider two balls such as B(z 1 ) = {z : z z 1 < 1 3 cα 1 21 z 1 k 6} B(z 2 ) = {z : z z 2 < 1 3 cα 1 21 z 1 k 6}. z 1 z 3 z 3 z 3 z 2

26 Sketch of proof If z 3 B(z 1 ), then can be replaced by Then we obtain 1 z 3 z 2 cβ 2 2 z 3 k 6 ε 3 z 2 k 6 (1 ε 3) z 3 z cα 1 21 z 1 k 6. F 2 α 2 1 β 2 α 1 β 1 β 2 = α 3 1 β 1β 2 2. If z 3 B(z 2 ), then we get z 3 z cα 1 21 z 1 k 6. If z 3 (B(z 1 ) B(z 2 )) c, we have z 3 z cα 1 21 z 1 k 6 and z 3 z cα 1 21 z 1 k 6. Thus we see that F 2 α 3 1 α 1β 1 β 2 α 3 1 β 1β 2 2 as desired.