On the Schrodinger maximal function in higher dimension

0.19 MB·English

by  Jean Bourgain

ON THE SCHRO¨DINGER MAXIMAL
FUNCTION IN HIGHER DIMENSION
2
1
0 J. Bourgain
2

n (dedicated to B. Kasin for his 60th)
a
J

6 Abstract. NewestimatesonthemaximalfunctionassociatedtothelinearSchro¨dinger
1
equation are established.

]
P
A
.
h
1. Introduction
t
a
m Recall that the solution of the linear Schr¨odinger equation
[ iu ∆u = 0
t − (1.0)
1 u(x,0) = f(x)
v (cid:26)
2 with (x,t) Rn R is given by
4 ∈ ×
3
3 eit∆f(x) = (2π)−n ei(x.ξ+t|ξ|2)f(ξ)dξ. (1.1)
.
1 Z
0 Assuming f Hs(Rn) for suitable s, when does the albmost convergence property
∈
2
1 limeit∆f = f a.e. (1.2)
: t 0
v →
hold? This problem originates from Carleson’s paper [C] who proved convergence
i
X for s 1 when n = 1. Dahlberg and Kenig [D-K] showed that this result is sharp.
r ≥ 4
a In dimensions n 2, Sh¨olin [S] and Vega V established independently convergence
≥
for s > 1, while similar examples as considered in [D-K] show failure of convergence
2
for s < 1.
4
The problem for n = 2 has been studied by various authors. Proofof convergence
for some s < 1 appears first in the author’s papers [B1], [B2]. Subsequently,
2
the required threshold was lowered by Moyua, Vargas, Vega [M-V-V], Tao, Vargas
[T-V1,2] and S. Lee [L]. The strongest result to date appears in [L] and asserts
a.e. convergence for f Hs(R2),s > 3. On the other hand, no improvements of
∈ 8
Sj¨olin-Vega result for n 3 seem to have been obtained so far.
≥
We will prove the following
Typeset by AMS-TEX
1
Theorem 1. For all n, there is ε > 0 such that the a.e. convergence property
n
holds for f Hs(Rn),s > 1 ε .
∈ 2 − n
In fact, one may take ε = 1 .
n 4n
Note that for n = 2, the exponent corresponds to the one obtained in [L]. The
reason for formulating Theorem 1 this way is that the first (qualitative) statement
allows for a less involved argument that will be presented first.
Most of the progress on this problem relates to the advances around the re-
striction theory for the paraboloid. In particular, the method used in [L] depends
on Tao’s bilinear restriction estimate [T]. The proof of Theorem 1 will be based
on the results and techniques from [B-G] and hence ultimately on the multi-linear
restriction theory developed in [B-C-T].
As pointed out in [L], it should also be observed that all results obtained in
the convergence problem for (1.0) hold equally well for generalized Schr¨odinger
equations
iu Φ(D)u = 0 (1.3)
t
−
with Φ(D) a Fourier multiplier operator satisfying Dαφ(ξ) C ξ 2 α,
−
| | ≤ | |
φ(ξ) c ξ . This is also the case for Theorem 1.
|∇ | ≥ | |
Perhaps the most interesting point in this Note is a disproof of what one seemed
to believe, namely that f Hs(Rn),s > 1, should be the correct condition in
∈ 4
arbitrary dimension n.
Theorem 2. For n > 4, the a.e. convergence property for the Schr¨odinger group
requires f Hs(Rn) with s n 2.
−
∈ ≥ 2n
Hencethisexponent 1 forn . Theexamplesareofaquitedifferentnature
→ 2 → ∞
from previous constructions and rely on an arithmetical input, namely lattice prints
on spheres.
Returning to the generalized setting, we exhibit in the last section of the paper
some further examples setting further restrictions to what may be proven in the
generality of our approach to Theorem 1.
2. Braking the 1-barrier
2
Before getting to more quantitative analysis, we start by establishing the first
part of Theorem 1. The main input in that argument is the multi-linear bound
from [B-C-T] and it is technically rather easy.
A few preliminary reductions. Assume R > 1 large and
suppf [ ξ R]. (2.0)
⊂ | | ∼
2
b
Our aim is to establish a maximal inequality of the form
sup eit∆f CRα f (2.1)
0<t<1| | L2(B1) ≤ k k2
(cid:13) (cid:13)
(cid:13) (cid:13)
where B = [x Rn; x K]. From the stationary phase, (2.1) may be deduced
K
∈ | | ≤
from the restricted inequality
sup eit∆f CRα′ f (2.2)
0<t<1 | | L2(B1) ≤ k k2
R
(cid:13) (cid:13)
(cid:13) (cid:13)
for any α < α (see [L], Lemma 2.3). Hence we restrict 0 < t < 1 in (2.1).
′ R
Next, performing a rescaling, we need to show that
sup eit∆f CRα f (2.3)
0<t<R| | L2(BR) ≤ k k2
(cid:13) (cid:13)
(cid:13) (cid:13)
assuming
suppf [ ξ 1]. (2.4)
⊂ | | ∼
This format corresponds to [B-G]. What follows is closely related to arguments in
b
2, 3 of that paper.
§ §
For notational convenience, set x = t and
n+1
ψ(x,ξ) = x ξ + +x ξ +x ξ 2. (2.5)
1 1 n n n+1
··· | |
Hence
(eit∆f)(x ,... ,x ) = f(ξ)eiψ(x,ξ)dξ Tf (2.6)
1 n
≡
ZΩ
with b b
Ω = [ ξ 1].
| | ∼
Fix 1 K = K(R) R to be specified later. Partition Ω in boxes Ω of size 1
≪ ≪ α K
centered at ξ . Then
α
Tf = eiψ(α,ξα)T f (2.7)
α
α
X
with
T f(x) = f(ξ)ei[ψ(x,ξ) ψ(x,ξα)]dξ. (2.8)
α −
ZΩα
Note that [ψ(x,ξ) ψ(x,ξ )] . 1 and hence T f may be viewed as ‘approxi-
|∇x − α K α
mately constant’ on balls of size O(K) in B Rn+1. We refer the reader to [B-G],
R
⊂
2, for the technical details needed to formalize this last statement.
§
3
Fix a ball B B . The main idea is to exploit a dichotomy similar to the one
K R
⊂
used in 2, 3 of [B-G]. More precisely, we distinguish the following two alternatives.
§ §
Case I: ‘(n 1)-transversality’
−
Viewing the T f as essentially constant on Ω , this means that there are indices
α α
α ,... ,α such that
1 n+1
(2.9) ν(ξ ) ... ν(ξ ) > K C whenever ξ Ω .
| 1′ ∧ ∧ n′+1 | − j′ ∈ αj
n
Here ν(ξ)// 2 ξ e +e denotes the normal of the hypersurface
− i=1 i i n+1
(ξ, ξ 2)
| | P
and
(2.10) T f ,... , T f K nmax T f .
| α1 | | αn+1 | ≥ − α| α |
Hence the following majoration holds on B
K
n+1 n+1 1
Tf . Knmax T f . K2n T f n+11 K2n T f n+1.
| | α | α | | αj | ≤ | αj |
(cid:2)jY=1 (cid:3) α1,.X..,αn+1 hjY=1 i
transversal
Recallthemulti-linearrestrictionboundfrom[B-C-T],implyingthatforα ,... ,α
1 n+1
‘transversal’ as defined in (2.9)
n+1 1
T f n+1 KCRε f (2.12)
(cid:13)hjY=1| αj |i (cid:13)Lq(BR) ≪ k k2
(cid:13) (cid:13)
(cid:13) (cid:13)
2(n+1)
holds, with q = .
n
Hence, denoting x = (x ,x ) and using H¨older’s inequality, the corresponding
′ n+1
contribution to the maximal function may be bounded by
sup (2.11) .
L2
k|xn+1|<R k [|x′|<R]
(2.11) Lq .
k kL2[|x′|<R] [|xn+1|<R]
(2.12)
Rn(21−q1) (2.11) Lq
k k [x<R] ≪
| |
Rn(21−q1)+εKC f 2 = R2(nn+1)+εKC f 2. (2.13)
k k k k
Case II: Failure of (n+1)-transversality
In this situation, there is an (n 1)-dimensional affine hyperplane in Rn such
− L
that dist(Ω , ) . 1 if T f K nmax T f on B .
α L K | α | ≥ − α| α | K
4
Denoting ˜ and 1 -neighborhood of , it follows that on B
L K L K
Tf(x) eiψ(x,ξα)T f(x) +max T f(x) (2.14)
α α
| | ≤ α | |
(cid:12)ΩXα ˜ (cid:12)
(cid:12) ⊂L (cid:12)
(cid:12) (cid:12)
and we write on B
K
1/2
eiψ(x,ξα)T f(x) = φ (x). T f 2 . (2.15)
α BK | α |
ΩXα ˜ (cid:16)ΩXα ˜ (cid:17)
⊂L ⊂L
Next, define a function φ by
φ = φ
|BK BK
for B satisfying alternative II. It follows from (2.14) that the corresponding con-
K
tribution to the maximal function is bounded by
1
max φ T f 2 2 . (2.16)
α
(cid:13)|xn+1|<Rn (cid:16)Xα | | (cid:17) o(cid:13)L2[|x′|<R]
(cid:13) (cid:13)
Let B ,B = B (cid:13) I Rn R be a partition(cid:13)of B(0,R) in K-cubes. Clearly
γ,ℓ γ,ℓ γ ℓ
{ } × ⊂ ×
1/2
(2.16) T f 2 max φ 2dx . (2.17)
α γ,ℓ ′
≤ nXγ,ℓ (cid:16)Xα | | (cid:17)(cid:12)Bγ,ℓhZBγ xn+1∈Iℓ| | io
(cid:12)
(cid:12)
Assume we dispose over a bound
1/2
max φ 2dx < A (2.18)
γ,ℓ ′
hZ6 Bγ xn+1∈Iℓ| | i
where stands for 1 .
6 Bγ mesBγ Bγ
TheRn (2.17) is bounded byR
1 1/2
A T f 2 dx
α
K | |
n Xγ,ℓ ZBγ,ℓ (cid:16)Xα (cid:17) o
1 1 1/2
= A 2 T f 2
α
K | | L2(x<R)
(cid:16) (cid:17) (cid:13)(cid:16)Xα (cid:17) (cid:13) | |
R (cid:13)1/2 (cid:13)1/2
CA (cid:13) f 2 (cid:13)
≤ K k |Ωαk2
(cid:16) (cid:17) (cid:16)Xα (cid:17)
R 1/2
= CA f . (2.19)
2
K k k
(cid:16) (cid:17) 5
It remains to establish a bound on A in (2.18).
Again, since in (2.15) the T f are viewed as constant on B , we need a bound
α K
of the form
1/2
max aαeiψ(x,ξα) AKn2 aα 2 (2.20)
(cid:13)|xn+1|<K(cid:12)Xα (cid:12)(cid:13)L2[|x′|<K] ≤ (cid:16)X| | (cid:17)
(cid:13) (cid:12) (cid:12)(cid:13)
where th(cid:13)e points (cid:12)ξ , ξ 1,(cid:12)a(cid:13)re 1 -separated and in an (n 1)-dim affine
hyperplane Rn{. α} | α| ∼ K −
L ⊂
Performing a rotation around the e -axis, we may assume ξ = c = constant
n+1 α
for ξ = (ξ ,... ,ξ ) . Hence the left side of (2.20) becomes
1 n
∈ L
K12 max aαei x1ξα,1+···+xn−1ξα,n−1+xn+1(ξα2,1+···+ξα2,n−1) .
(cid:13)(cid:13)|xn+1|<K(cid:12)(cid:12)Xα (cid:0) (cid:1)(cid:12)(cid:12)(cid:13)(cid:13)L2[|x1|,...,|xn(−21.|2<1K)]
(cid:13) (cid:12) (cid:12)(cid:13)
At this point, the dimension is reduced from n to n 1.
−
Let Ω bedisjoint 1 -neighborhoodsofthepoints (ξ ,... ,ξ ) inRn 1.
{ ′α} K { α,1 α,n−1 } −
Let g on [ ξ 1] Rn 1 be defined by
′ −
| | ∼ ⊂
a Ω 1 if ξ Ω
g(ξ ) = α | ′α|− ′ ∈ ′α
′
0 otherwise.
(
Hence
1/2
g 2 Kn−21 aα 2 (2.22)
k k ∼ | |
(cid:16)X (cid:17)
and
(2.21) K12 max g(ξ′)ei(y.ξ′+t|ξ′|2) . (2.23)
∼ t<K L2[y <K]
(cid:13)| | (cid:12)Z (cid:12)(cid:13) | |
Denote θ an exponent for(cid:13)which(cid:12)(2.3) holds in dimen(cid:12)s(cid:13)ion n 1.
n 1 (cid:13) (cid:12) (cid:12)(cid:13)
− −
From (2.22), it follows that
1/2
(2.23) . K21+θn−1Kn−21 aα 2
| |
(cid:16)Xα (cid:17)
which shows that we may take A Kθn 1. Hence in (2.19), we obtain the estimate
∼ −
CR21Kθn−1−21 f 2. (2.24)
k k
Together with the Case I contribution (2.13), we obtain the bound
sup |eit∆f| L2 ≪ [R2(nn+1)+εKC +R21Kθn−1−21]kfk2 (2.25)
0<t<R (BR)
(cid:13) (cid:13) 6
(cid:13) (cid:13)
where K is a parameter. Recall that θ < 1. Hence an appropriate choice of
n 1 2
−
K = Rδ permits to obtain (2.3) for some θ = θ < 1.
n 2
3. A quantitative estimate
The proof of the second part of Theorem 1 depends on the more refined analysis
from 4 in [B-G] and its higher dimensional version.
§
Before stating the relevant result from [B-G], we fix some terminology. Let
S = (ξ, ξ 2), ξ 1 Rn+1.
{ | | | | ∼ } ⊂
For fixed δ > 0, consider a partition of ξ : ξ 1 Rn in cells Q of size δ and
{ | | ∼ } ⊂
denote
τ = (ξ, ξ 2);ξ Q (3.1)
{ | | ∈ }
the corresponding partition of S in δ-caps. Thus the convex hull conv(τ) is a
(δ δ δ2)-box tangent toS andwedenote τ◦ Rn+1 thepolarbox ofconv(τ)
×···× × ⊂
n
shifted with 0 as center. Thus τ◦ is a 1 1 1 )-box.
| {z } δ ×···× δ × δ2
(cid:0)
...............................................................................................................................................................................................................................................................................................................................................................................................................................S...................•....................................................................................................................................................................................................................... τ
Denote also
f = f
τ Q
|
with Q and τ related by (3.1).
Let the operator T be as in previous section.
7
Next recall (3.4) - (3.8) from 4 in [B-G], providing an estimate for Tf on B in
R
§
3D (i.e. n = 2). Thus
Tf
| | ≪
1/2
Rε max max (φ Tf 1/3 Tf 1/3 Tf 1/3)2
√1T<δ<1 Eδ hτX∈Eδ τ| τ1| | τ2| | τ3| i
1/2
Rεmax (φ Tf )2 (3.2)
τ τ
|
1
E√R hτX∈E i
where
(3.3) consists of at most 1 disjoint δ-caps.
Eδ δ
(2.4) τ ,τ ,τ τ are 3-transversal δ -caps (K is a large constant).
(3.5) F1or 2eac3h⊂τ,φ 0 is a functionKon Rn satisfying
τ
≥
φ4 Rε
6 τ ≪
ZB
for all B taken in a tiling of Rn+1 with translates of τ◦.
Of course (3.2) certainly implies
1/2
Tf Rε (φ Tf 1/3 Tf 1/3 Tf 1/3)2 (3.6)
| | ≪ τ| τ1| | τ2| | τ3|
δ dXyadic hτδX−cap i
1 <δ<1
√R
1/2
+Rε (φ Tf )2 (3.7)
τ τ
| |
hτ√1XR−cap i
where refers to summation over a partition of S in δ-caps τ.
τ δ cap
−
P
Formula (3.2) has a higher dimensional version, deduced in a similar way from
3 in [B-G]. In particular, we get for general n 2
§ ≥
Tf
| | ≪
n+1 2 1
Rε φτ |Tfτj|n+11 2 (3.8)
Xδ hτδX−cap(cid:16) jY=1 (cid:17) i
1
+Rε (φ Tf )2 2 (3.9)
τ τ
| |
hτ√1XR−cap i
8
where now τ ,... ,τ τ are (n + 1)-transversal δ -caps and φ 0 on Rn+1
1 n+1 ⊂ K τ ≥
satisfies
2n
φq Rε with q = (3.10)
6 τ ≪ n 1
ZB −
for all B in a tiling of Rn+1 from τ◦-translates.
It is now clear from (3.8), (3.9) and convexity that in order to bound
Tf
k kL2[|x′|<R]L∞[|xn+1|<R]
it suffices to estimate
n+1
1
φτ |Tfτj|n+1 L2 L (3.11)
(cid:13) (cid:16)jY=1 (cid:17)(cid:13) [|x′|<R] ∞[|xn+1|<R]
(cid:13) (cid:13)
(cid:13) (cid:13)
with τ ,... ,τ τ as above, and also
1 n+1
⊂
φ Tf (3.12)
τ τ L2 L
k | |k [|x′|<R] ∞[|xn+1|<R]
with τ or 1 -cap.
√R
n+1 1
Denote φ = φτ and F = j=1 |Tfτj|n+1.
Let B be a tiling of BQ(0,R) Rn+1 with π (B ) = I Rn a partition of
jk x jk j
{ } ⊂ ′ ⊂
B(0,R) Rn in 1 1 1 )-boxes and B translate of τ◦
⊂ δ ×···× δ ×δ2 jk ∼
(cid:0) n 1
−
| {z R }
..............................
...............................................................................................................................................B..............................................................j...........................,...................k...........................................................................................................................................................................................
.............................. Rn
I
j
9
Let
2(n+1) 2n
p = and q = .
n n 1
−
By H¨older’s inequality
1
(3.11) = φF 2 2
hXj k kL2IjL∞[|xn+1|<R]i
1
≤ δ−12 kφFk2LpIjL∞xn+1 2. (3.13)
hXj i
Writing for fixed x I
′ j
∈
1
sup φF (x ,x ) φF p p
xn+1| | ′ n+1 ≤ (cid:16)Xk k kL∞(Bj,k(x′))(cid:17)
it follows that
2 1
1 p p 2
(3.13) ≤ δ−2hXj (cid:16)Xk kφFkLpx′L∞xn+1(Bj,k)(cid:17) i . (3.14)
[
Since suppTf τ, we may view F as essentially constant on each B . From
τj ⊂ jk
(3.10)
kφkLpx′L∞xn+1(Bjk) ≤ kφkLpx′Lqxn+1(Bjk)
1 1
≤ |Ij|p−qkφkLq(Bjk)
Rε Ij p1−q1 Bjk q1
≪ | | | |
Rε 1 n2+n2−n1
≪ δ
(cid:16) (cid:17)
and hence
kφFkLpx′L∞xn+1(Bjk) ≪ Rε 1δ n2+n2−n1|Ij|−12δp1kFkL2x′Lpxn+1(Bjk)
(cid:16) (cid:17)
≪ Rεδ21+21n−2(n1+1)kFkL2x′Lpxn+1(Bjk). (3.15)
Substitution of (3.15) in (3.14) gives
2 1
Rεδ21n−2(n1+1)hXj (cid:16)Xk kFkpL2x′Lpxn+1(Bjk)(cid:17)pi2
Rεδ21n−2(n1+1) F L2 Lp . (3.16)
≤ k k [|x′|<R] [|xn+1|<R]
10