Inhomogeneous Strichartz estimates for Schr\"odinger's equation

0.26 MB·

by  Youngwoo Koh

INHOMOGENEOUS STRICHARTZ ESTIMATES FOR
SCHRO¨DINGER’S EQUATION
6
1
0 YOUNGWOOKOHANDIHYEOKSEO
2

r Abstract. Foschi and Vilela in their independent works ([3],[13]) showed
p
that the range of (1/r,1/r) for which the inhomogeneous Strichartz estimate
3 A t(cid:13)(cid:13)hRe0tcelio(ste−ds)p∆eFnt(a·,gso)ndsw(cid:13)(cid:13)iLtqthLevrxer.tickeFskAL,qteB′L,rxeB′′,hPol,dPs′feoxrcseopmtethqe,qepoisinctsonPta,Pin′ed(seine
2 Figure1). WeobtaintheestimateforthecornerpointsP,P′.

]
P
A
1. Introduction
.
h
In this paper we consider the following Cauchy problem for the Schr¨odinger
t
a equation:
m
i∂ u+∆u=F(x,t),
t
[
(u(x,0)=f(x),

2 where (x,t)∈Rn×R, n≥1. By Duhamel’s principle, we have the solution
v
3 t
4 u(x,t)=eit∆f(x)−i ei(t−s)∆F(·,s)ds,
6 Z0
7 where eit∆ is the free Schr¨odinger propagator defined by
0
. eit∆f(x)=(2π)−n eix·ξ−it|ξ|2f(ξ)dξ.
1
0 ZRn
6 The Strichartz estimates for the solution play impobrtant roles in the study of
1 well-posedness for nonlinear Schr¨odinger equations (cf. [1, 12]). They actually
: consist of two parts, homogeneous (F =0) and inhomogeneous (f = 0) part. The
v
i homogeneous Strichartz estimate
X
r keit∆fkLqtLrx .kfkL2
a
holds if and only if (r,q) is admissible pair, that is,
r,q ≥2, (n,r,q)6=(2,∞,2) and n/r+2/q=n/2
(see[11,4,9,6]andreferencestherein). Butdeterminingtheoptimalrangeof(r,q)
and (r,q) for which the inhomogeneous Strichartz estimate
t
ei(t−s)∆F(·,s)ds .kFk (1.1)
e e Lqe′Lre′
(cid:13)Z0 (cid:13)LqtLrx t x
(cid:13) (cid:13)
holds is not complete(cid:13)d yet when n≥3. It(cid:13)was observed that this estimate is valid
(cid:13) (cid:13)
on a wider range than what is given by admissible pairs (r,q), (r,q) (see [2], [5]).
2010 Mathematics Subject Classification. Primary35B45,35Q40.
e e
Key words and phrases. Strichartzestimates,Schro¨dinger equation.
IhyeokSeowassupportedbytheTJParkScienceFellowshipofPOSCOTJParkFoundation.
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2 YOUNGWOOKOHANDIHYEOKSEO
Figure 1. The range of (1/r,1/r) for (1.1) when n ≥ 3. Here
P =( n−2 , (n−2)2 ) and P′ =( (n−2)2 , n−2 ).
2(n−1) 2n(n−1) 2n(n−1) 2(n−1)
e
Foschi and Vilela in their independent works ([3],[13]) showed that the range of
(1/r,1/r) for which (1.1) is valid for some q,q is contained in the closed pentagon
with verticesA,B,B′,P,P′ except the points P,P′ (see Figure 1). The aim ofthis
paper is to obtain (1.1) for the points P,P′. Our result is the following.
e e
Theorem 1.1. Let n≥3. Then (1.1) holds when
1 1 n−2 (n−2)2 n−2 1 1 n
( , )=P =( , ) if ≤ = < ,
r r 2(n−1) 2n(n−1) 2(n−1) q q′ 2(n−1)
and when
e e
1 1 (n−2)2 n−2 n−2 1 1 n
( , )=P′ =( , ) if < = ≤ .
r r 2n(n−1) 2(n−1) 2(n−1) q q′ 2(n−1)
Remark 1.2. Since 1/r+1/r = (n−2)/n, the condition q = q′ follows from the
e e
scaling condition
1 1 n 1 1 n
e e
+ + ( + )= . (1.2)
q q 2 r r 2
The conditions 1/q < n/2(n−1) and (n−2)/2(n−1) < 1/q′ when (1/r,1/r) =
P and P′ correspond to the knowne necessaryeconditions ([3, 13])
1 n 2 1 n 2 e e
< (1− ) and < (1− ),
q 2 r q 2 r
respectively. In Section 3 we also give new necessary conditions for (1.1).
Remark 1.3. Ourproofcanbeeasilymodifiedto coverthe rangeof(r,q)and(r,q)
obtained by Foschi and Vilela. But we have chosen to present the proof only for
the points P,P′ to keep the exposition as simple as possible. The case n ≥ 4 in
e e
Theorem 1.1 was already shown in [7] but the argument there does not suffice to
obtain the same result in dimension n=3.
INHOMOGENEOUS STRICHARTZ ESTIMATES 3
2. Proof of Theorem 1.1
Under the same conditions in Theorem 1.1, we will show
t
ei(t−s)∆F(·,s)ds .kFk (2.1)
Lqe′Lre′
(cid:13)Z−∞ (cid:13)LqtLrx t x
(cid:13) (cid:13)
whichimplies(1.1). I(cid:13)ndeed,toobtain(1.1)f(cid:13)rom(2.1),firstdecomposetheLq norm
(cid:13) (cid:13) t
in the left-hand side of (1.1) into two parts, t ≥ 0 and t < 0. Then the latter can
be reducedto the former by changing the variable t7→−t,and so it is only needed
to consider the first part t≥0. But, since [0,t)=(−∞,t)∩[0,∞), applying (2.1)
with F replaced by χ (s)F, one can bound the first part as desired.
[0,∞)
Let φ be a cut-off function with φ(ξ) = 1 if |ξ| ≤ 1, φ(ξ) = 0 if |ξ| > 2, and
0≤φ(ξ)≤1. Then it is enough to show that
t
eix·ξ−i(t−s)|ξ|2|φ(ξ)|2F\(·,s)(ξ)dξds .kFk . (2.2)
Lqe′Lre′
(cid:13)Z−∞ZRn (cid:13)LqtLrx t x
(cid:13) (cid:13)
Once w(cid:13)e have this estimate, by the usual scaling we(cid:13)see that for all j ≥0
(cid:13) (cid:13)
t
eix·ξ−i(t−s)|ξ|2|φ(ξ/2j)|2F\(·,s)(ξ)dξds
(cid:13)Z−∞ZRn (cid:13)LqtLrx
(cid:13) (cid:13)
(cid:13)(cid:13) .2−2j(1q+q1e−(cid:13)(cid:13)n2(1−r1−r1e))kFkLqe′Lre′
t x
.kFk .
Lqe′Lre′
t x
Here, for the last inequality, we used the scaling condition (1.2). Since we may
assumethatF is containedinthe SchwartzspaceonRn+1,by alimiting argument
(j →∞), we now get (2.1) from the above estimate.
Now, for fixed t, we define
T f(x)= eix·ξ−it|ξ|2φ(ξ)f(ξ)dξ
t
Z
and note that its adjoint operator Tt∗ is given by b
T∗f(x)= eix·ξ+it|ξ|2φ(ξ)f(ξ)dξ.
t
Z
Then the desired estimate (2.2) can be rewritten asb
t
T T∗F ds .kFk
t s s Lqe′Lre′
(cid:13)Z−∞ (cid:13)LqtLrx t x
(cid:13) (cid:13)
where we use the notat(cid:13)ion F to denot(cid:13)e F (·) = F(·,s). By duality we are now
(cid:13) s (cid:13) s
reduced to showing the bilinear form estimate
t
hT∗F ,T∗G i dsdt .kFk kGk (2.3)
(cid:12)ZRZ−∞ s s t t L2x (cid:12) Lqte′Lrxe′ Lqt′Lrx′
(cid:12) (cid:12)
underthe same(cid:12)conditionsinTheorem1.1. T(cid:12)oshow(2.3),wewillusethe following
(cid:12) (cid:12)
lemma.
Lemma 2.1. Let n≥3, and let 2≤r,r ≤∞ and 1≤q,q ≤∞. Define
t−2j
Bj(F,G)=ZRZt−2je+1hTs∗Fs,Tt∗GtiL2xedsdt
4 YOUNGWOOKOHANDIHYEOKSEO
and assume one of the following conditions for (r,r;q,q):
1 n−21 1 1 1
i) ≤ and ≤ ≤ ≤1,
r n r r q eq′ e
n−21 1 1 n 1 1 1 1
ii) ≤ ≤ and − ( − )≤ ≤ ≤1,
n r r re e 2 r er q q′
1 1 n 1 1 1 n 1 1
iii) ≤ ≤ and 0≤ ≤ ≤1− ( − ),
r re n−2re q q′e 2 er r
n 1 1 1 1 1
iv) ≤ and 0≤ ≤ ≤1− .
e n−2r re q qe′ r e
Then we have
|Bj(F,Ge)|.2jβ(r,re,q,qe)kFkLqe′Lre′ekGkLq′Lr′, (2.4)
t x t x
where
1 1 n−1 n
+ + − if i) holds,
q q r 2

1 1 n 1 1
β(r,r,q,q)=q + qe− 2(e1− r − r) if ii) or iii) holds,
1 1 n−1 n
e e + + − if iv) holds.
Remark 2.2. The ranges oqf (1qe/r,1/rr′) in i)2, eii), iii) and iv) correspond to the
triangular regions OB′C′, OABe′, OAB and OBC in Figure 1, respectively.
e
Proof of Lemma 2.1. One can easily get the above lemma by interpolating the es-
timates (2.4) in the following four cases:
(a) r =r =∞ (point O) and 1≤q′ ≤q ≤∞,
(b) r =r =2 (point A) and 1≤q′ ≤q ≤∞,
(c) r =2e, 2n ≤r ≤∞ (segment BCe) and 2≤q′ ≤q ≤∞,
n−2
(d) 2n e≤r ≤∞, r =2 (segment Be′C′) and 1≤q′ ≤q ≤2.
n−2
e e
Thefirstandsecondones,(a)and(b),werealreadyshownin[7](seeLemma2.1
there). So we only needeto show (c) and (d). For (c) weedecompose F and G as
Fk(x,s)=F(x,s)χ (s) and Gk(x,t)=G(x,t)χ (t)
{2jk≤s<2j(k+1)} {2jk≤t<2j(k+1)}
for fixed j. Then we see that
t−2j
|B (F,G)| = hT∗Fk,T∗G i dsdt
j k∈ZZRZt−2j+1 s s t t L2x
X
≤ hT∗Fk,T∗Gk+1i dsdt+ hT∗Fk,T∗Gk+2i dsdt
k∈ZZRZR s s t t L2x k∈ZZRZR s s t t L2x
X X
because t∈(s+2j,s+2j+1). Using Ho¨lder’s inequality in x, we note that
hT∗Fk,T∗Gk+1i dsdt ≤ T∗Fkds T∗Gk+1dt .
(cid:12)(cid:12)Xk∈ZZRZR s s t t L2x (cid:12)(cid:12) Xk∈Z(cid:13)(cid:13)ZR s s (cid:13)(cid:13)L2x(cid:13)(cid:13)ZR t t (cid:13)(cid:13)L2x
We(cid:12)also note that (cid:12) (cid:13) (cid:13) (cid:13) (cid:13)
(cid:12) (cid:12) (cid:13) (cid:13) (cid:13) (cid:13)
kTtfkLqLr .kfkL2 (2.5)
t x
holdsforr,q ≥2andn/r+2/q≤n/2. Indeed,bythestationaryphasemethod(see
p. 344 in [10]), we see kT fk . (1+|t|)−n/2kfk . Then (2.5) follows directly
t L∞x L1
INHOMOGENEOUS STRICHARTZ ESTIMATES 5
fromtheabstractStrichartzestimatesofKeelandTao[6]. Usingthedualestimate
of (2.5),
T∗F ds .kFk ,
t s Lq′Lr′
(cid:13)ZR (cid:13)L2x t x
(cid:13) (cid:13)
we now get (cid:13) (cid:13)
(cid:13) (cid:13)
hT∗Fk,T∗Gk+1i dsdt . kFkk kGk+1k
(cid:12)k∈ZZRZR s s t t L2x (cid:12) k∈Z L2tLrxe′ L1tL2x
(cid:12)X (cid:12) X
where 2n(cid:12)(cid:12) ≤r ≤∞. On the other hand, b(cid:12)(cid:12)y Ho¨lder’s inequality in time, it follows
n−2
that
k∈ZkFekkL2tLrxe′kGk+1kL1tL2x ≤k∈Z2j(q1+q1e−12)kFkkLqte′Lrxe′kGk+1kLqt′L2x
X X
≤2j(1q+1qe−12)kFkLqe′Lre′kGkLq′L2
t x t x
if 2≤q′ ≤q ≤∞. Here, for the last inequality we used that
|A B |≤ |A |p p1 |B |pe p1e if 1 + 1 ≥1. (2.6)
e n n n n
p p
Xn (cid:16)Xn (cid:17) (cid:16)Xn (cid:17)
Hence,
e
(cid:12)k∈ZZRZRhTs∗Fsk,Tt∗Gkt+1iL2xdsdt(cid:12)≤2j(q1+1qe−12)kFkLqte′Lrxe′kGkLqt′L2x.
(cid:12)X (cid:12)
Similarl(cid:12)y, (cid:12)
(cid:12) (cid:12)
(cid:12)k∈ZZRZRhTs∗Fsk,Tt∗Gkt+2iL2xdsdt(cid:12)≤2j(q1+1qe−12)kFkLqte′Lrxe′kGkLqt′L2x.
(cid:12)X (cid:12)
Consequ(cid:12)(cid:12)ently, we get (c). The case (d) ca(cid:12)(cid:12)n be shown in a similar way as (c). (cid:3)
Now we return to (2.3). It suffices to show that
|B (F,G)| .kFk kGk . (2.7)
j Lqe′Lre′ Lq′Lr′
j∈Z t x t x
X
Weonlyconsiderthecase(1/r,1/r)=P sincethecase(1/r,1/r)=P′ followsfrom
the same argument. Now let
1 1 n−2 (n−2)2 e n−2 1 e 1 n
( , )=( , )=P and ≤ = < .
r r 2(n−1) 2n(n−1) 2(n−1) q q′ 2(n−1)
NotethatthepointP liesonthesegmentOB,andsowewilluseLemma2.1under
the coneditions iii) and iv) (see Figure 1 and Remark 2.2). Sincee 1− n(1 − 1) =
2 r re
1− 1 = n > 1, if we choose ǫ > 0 small enough so that ǫ≤ 1 ( n − 1),
r 2(n−1) qe′ 20 2(n−1) qe′
we can use Lemma 2.1 under the conditions iii) and iv) for (a,b;q,q) with all
(a,b)∈B((1,1),10ǫ), where
r re
1 1 1 1 1 1 1 1 1 1 e
B ( , ),10ǫ ={( , )∈[0, ]×[0, ] : | − |,| − |<10ǫ}.
r r a b 2 2 r a r b
Now, usin(cid:0)g Lemma 2(cid:1).1, we see that
e |B (F,G)| .2jβ(a,b,q,qe)kFk kGk e, (2.8)
j Lqe′Lb′ Lq′La′
t x t x
6 YOUNGWOOKOHANDIHYEOKSEO
where
1 1 n 1 1 1 n 1
+ − (1− − ) if ≤ ,
q q 2 a b a n−2b
β(a,b,q,q)=
1 1 n−1 n 1 n 1
 + + − if > .
q qe a 2 a n−2b
e
Next we decompose F andG using the following lemma whose proof can be found
in [6] (see Lemma 5.1 there): e
Lemma 2.3 (AtomicdecompositionofLp). Let 1≤p<∞. Then any f ∈Lp can
x
be written as
∞
f = c χ
k k
k=−∞
X
whereeach χ is a function boundedby O(2−k/p) andsupported on aset of measure
k
O(2k) and the c are non-negative constants with kc k .kfk .
k k lp Lp
By this lemma, we may write
Fs(x)= fek(s)χek,s(x) and Gt(x)= gk(t)χk,t(x),
Xek∈Z Xk∈Z
where χe (x) is bounded by Oe(2−ek/re′) and supported on a set of measure O(2ek),
k,s
and χ (x) is bounded by O(2−k/r′) and supported on a set of measure O(2k).
k,t
Also, feek and gk satisfy
1 1
|fek(s)|re′ re′ .kFskLrxe′ and |gk(t)|r′ r′ .kGtkLrx′. (2.9)
(cid:16)ekX∈Z (cid:17) (cid:16)Xk∈Z (cid:17)
Combining (2.8) and this decomposition, we now get
|Bj(F,G)|. 2jβ(a,b,q,qe)2ek(r1e−1b)2k(r1−a1)kfekkLqe′kgkkLq′. (2.10)
Xj∈Z Xj∈ZeXk∈ZXk∈Z t t
If 1 ≤ n 1, we use (2.10). But if 1 > n 1, we use (2.10) with 2j replaced by
a n−2b a n−2b
2−j. (Since j ∈Z, we may replace 2j by 2−j in (2.10).) Then we conclude that
|Bj(F,G)|. Hj,ek,k(a,b)kfekkLqe′kgkkLq′,
Xj∈Z Xj∈ZeXk∈ZXk∈Z t t
where
2(ek−n2j)(r1e−1b)+(k−n2j)(r1−a1) if 1 ≤ n 1,
a n−2b
Hj,ek,k(a,b)= 2ek(1re−1b)+(k+(n−1)j)(r1−a1) if 1 > n 1.
a n−2b
First we consider the cases where k 6= nj and k6= nj. Let us set
2 2
n n
U = (j,k,k); k− j >0, k− j >0 ,
1 2 e 2
n n
(cid:8) (cid:9)
U2 = (j,ke,k); k− j <0, ek− j >0 ,
2 2
n n
(cid:8) (cid:9)
U3 = (j,ek,k); k− j <0, ek− j <0 ,
2 2
n n
U4 =(cid:8)(j,ek,k); k− j >0, ek− j <0(cid:9).
2 2
(cid:8) (cid:9)
e e
INHOMOGENEOUS STRICHARTZ ESTIMATES 7
Figure 2.
Then we may write
|Bj(F,G)|≤ Hj,ek,k(a,b)kfekkLqe′kgkkLq′
Xj∈Z (j,ek,k)∈XU1∪U2∪U3 t t
+ Hj,ek,k(a,b)kfekkLqe′kgkkLq′. (2.11)
e t t
(j,kX,k)∈U4
For each (j,k,k) ∈ U ∪U ∪U , we choose1 (1,1) ∈ B((1,1),10ǫ)\B((1,1),ǫ)
1 2 3 a b r re r re
with 1 ≤ n 1 so that
a n−2b
e
2(ek−n2j)(r1e−1b)+(k−n2j)(r1−a1)kfekkLqe′kgkkLq′
t t
U1∪XU2∪U3
≤ 2−ǫ(|ek−n2j|+|k−n2j|)kfekkLqe′kgkkLq′.
t t
U1∪XU2∪U3
Bysumminginj,andusing(2.6)andYoung’sinequalitysinceK(·)=(1+|·|)2−ǫ|·|
is absolutely summable, we see that
2−ǫ(|ek−n2j|+|k−n2j|)kfekkLqe′kgkkLq′
t t
U1∪XU2∪U3
e
. (1+|k−k|)2−ǫ|k−k|kfekkLqe′kgkkLq′
eXk∈ZXk∈Z t t
e 1 1
. kfekkqLe′qe′ qe′ kgekkqL′q′ q′.
(cid:16)Xek∈Z t (cid:17) (cid:16)Xek∈Z t (cid:17)
1ThelineOP intersectstheregions(b)and(c)sinceitsslopeis(n−2)/n(seeFigure2). Hence,
if(j,ek,k)∈U1,choose(a1,1b)thatliesabovethelineOP intheregion(b). If(j,ek,k)∈U2,choose
(a1,1b)inthe region(a). If(j,ek,k)∈U3, choose(a1,1b)thatliesabove thelineOP intheregion
(c).
8 YOUNGWOOKOHANDIHYEOKSEO
Since q′ ≥r′ and q′ ≥r′, by Minkowski’s inequality and (2.9),
1 1 1 1
(cid:16)Xek∈ZekfekkeqLe′qte′(cid:17)qe′(cid:16)Xek∈ZkgekkqL′qt′(cid:17)q′ .(cid:16)eXk∈ZkfekkrLe′qte′(cid:17)re′(cid:16)Xek∈ZkgekkrL′qt′(cid:17)r′
1 q′
. |fe(s)|re′ re′ |ge(t)|r′ r′
k Lqe′ k Lq′
(cid:13)(cid:16)eXk∈Z (cid:17) (cid:13) t (cid:13)(cid:16)ekX∈Z (cid:17) (cid:13) t
(cid:13) (cid:13) (cid:13) (cid:13)
.(cid:13)kFkLqe′Lre′kGkLq′Lr(cid:13)′. (cid:13) ((cid:13)2.12)
t x t x
Consequently, we bound the first term in the right-hand side of (2.11) as desired.
To bound the second term, we first write
Hj,ek,k(a,b)= 2ek(r1e−1b)+(k+(n−1)j)(r1−a1)kfekkLqe′kgkkLq′
XU4 Xj∈ZekX<n2jkX>n2j t t
= 2j(n−1)(r1−a1) 2ek(r1e−1b)kfekkLqe′ 2k(r1−a1)kgkkLq′
Xj∈Z ekX<n2j t kX>n2j t
and note that
2ek(1re−1b)kfekkLqe′ 2k(r1−a1)kgkkLq′
ekX<n2j t kX>n2j t
≤ 2ek(r1e−1b)qe 1qe kfekkqLe′qe′ qe1′ 2k(1r−a1)q q1 kgkkqL′q′ q1′
(cid:16)ekX<n2j (cid:17) (cid:16)ekX<n2j t (cid:17) (cid:16)kX>n2j (cid:17) (cid:16)kX>n2j t (cid:17)
.2n2j(r1e−1b)2n2j(1r−a1)kFkLqe′Lre′kGkLq′Lr′.
t x t x
Here we used (2.12) for the last inequality. Hence we get
Hj,ek,k(a,b). 2n2j(r1e−1b)23n2−2j(r1−a1)kFkLqe′Lre′kGkLq′Lr′.
XU4 Xj∈Z t x t x
Now we choose2 (1,1)∈B((1,1),10ǫ)\B((1,1),ǫ) with
a b r re r re
1 n 1 1 1 1 1
> , −10ǫ< − <−9ǫ, 2ǫ> − >ǫ
a n−2b r a r b
when j ≥0, and with
1 n 1 1 1 1e 1
> , −2ǫ< − <−ǫ, 10ǫ> − >9ǫ
a n−2b r a r b
when j <0. Then we get the desired bound
e
Hj,ek,k(a,b).kFkLqe′Lre′kGkLq′Lr′.
t x t x
XU4
Consequently, we get (2.7).
Finally we consider the cases where k = nj or k = nj. When k = nj, we note
2 2 2
that
Hj,ek,k(a,b)kfekkLqe′kgkkLq′ = e2(ek−k)(r1e−1b)kfekkLqe′kgkkLq′,
Xj∈ZeXk∈ZXk∈Z t t eXk∈ZXk∈Z t t
2Choose(1,1)intheregion(d)whenj≥0andintheregion(e)whenj<0(seeFigure2).
a b
INHOMOGENEOUS STRICHARTZ ESTIMATES 9
where 1 ≤ n 1. Hence if we choose(a,b) inthe region(a) or(c) with 1 ≤ n 1,
a n−2b a n−2b
we see
2(ek−k)(r1e−1b)kfekkLqe′kgkkLq′ ≤ 2−ǫ|ek−k|kfekkLqe′kgkkLq′.
eXk∈ZXk∈Z t t eXk∈ZXk∈Z t t
From this, we get the desired bound as before. The other cases follow easily in a
similar way.
3. Necessary conditions
It was shown in [3] that
n−2 2 n n−2 2 n
− ≤ and − ≤ (3.1)
r q r r q r
are the necessary conditions for which the inhomogeneous estimate
e e e
t
ei(t−s)∆F(·,s)ds .kFk (3.2)
Lqe′Lre′
(cid:13)Z0 (cid:13)LqtLrx t x
(cid:13) (cid:13)
holds. (We refer the r(cid:13)eader to [3, 13, 8] for(cid:13)other necessaryconditions.) Compared
(cid:13) (cid:13)
with (3.1), we give here the following new necessary condition:
n−2 2 n n−2 2 n
− ≤ , − ≤ . (3.3)
r q r r q r
The first condition in (3.3) is stronger than the first one in (3.1) when 1/r ≤ 1/r,
and the second conditionein (3.3) is stronger thanethe esecond one in (3.1) when
1/r≥1/r. e
Proof of (3.3). If (3.2) holds with a pair (r,q) on the left and a pair (r,q) on the
e
right,then it must be alsovalid when one switches the roles of(r,q) and(r,q). By
thisdualityrelation,weonlyneedtoshowthesecondcondition(n−2)/r−2/q≤n/r
e e
in (3.3). For this, we first write
e e
I(F):= tei(t−s)∆F(·,s)ds=(4π)−n2 t |t−s|−n2ei|4x|−t−ys|2|F(y,s)dyeds. e
Z0 Z0 ZRn
Let 0<ǫ<1/2 and F(y,s)=χ . Since
{0<s<ǫ2,|y|<ǫ}
|x−y|2 |x|2 t(|x−y|2−|x|2)+s|x|2
= + ,
4(t−s) 4t 4t(t−s)
for 0<s<ǫ2, |y|<ǫ, 10<t<11 and |x|− 1 <ǫ, we see that
ǫ
t(|x−y|2−|x|2)+s|x(cid:12)|2 1(cid:12)1·3+ǫ2· 2 1
(cid:12) < (cid:12) ǫ2 < .
4t|t−s| 4·10·9 2
(cid:12) (cid:12)
Hence, if 10<t<(cid:12) 11 and |x|− 1 <ǫ, (cid:12)
(cid:12) ǫ (cid:12)
|I(F)(x,t)|≥ (4π)−n2(cid:12)(cid:12)ei|x4|t2 (cid:12)(cid:12)t |t−s|−n2eit(|x−y|42t−(t|−x|s2))+s|x|2F(y,s)dyds
(cid:12) Z0 Z (cid:12)
(cid:12)(cid:12) ǫ2 (cid:12)(cid:12)
&(cid:12) dyds (cid:12)
Z0 Z|y|<ǫ
&ǫn+2.
10 YOUNGWOOKOHANDIHYEOKSEO
This implies that
kI(F)kLqtLrx ≥kI(F)kLqt(10<t<11)Lrx(||x|−1ǫ|<ǫ)
&ǫn+2 (1/ǫ+ǫ)n−(1/ǫ−ǫ)n 1/r
&ǫn+2(cid:0)ǫ−n((1+ǫ2)n−(1−ǫ2(cid:1))n) 1/r
&ǫn+2(cid:0)ǫ−nr+2. (cid:1)
2 n
On the other hand, kFkLqse′Lrye′ ∼ ǫqe′ǫre′. Now the estimate (3.2) leads us to
ǫn+2ǫ−nr+2 .ǫqe2′ǫren′. Byletting ǫ→0,we concludethat2/q+n/r≥(n−2)/r. (cid:3)
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SchoolofMathematics,KoreaInstituteforAdvancedStudy,Seoul130-722,Repub-
licof Korea
E-mail address: [email protected]
Departmentof Mathematics,SungkyunkwanUniversity, Suwon440-746,Republicof
Korea
E-mail address: [email protected]