On inhomogeneous Strichartz estimates for fractional Schr\"odinger equations and their applications

0.36 MB·English

by  Chu-Hee Cho

ON INHOMOGENEOUS STRICHARTZ ESTIMATES FOR
FRACTIONAL SCHRO¨DINGER EQUATIONS AND THEIR
APPLICATIONS
5
1
0 CHU-HEECHO,YOUNGWOOKOHANDIHYEOKSEO
2

l
u
J
Abstract. InthispaperweobtainsomenewinhomogeneousStrichartzesti-
8
matesforthefractionalSchro¨dingerequationintheradialcase. Thenweapply

] them to the well-posedness theory for the equation i∂tu+|∇|αu=V(x,t)u,
P 1 < α < 2, with radial H˙γ initial data below L2 and radial potentials
A V ∈LrtLwx underthescaling-criticalrangeα/r+n/w=α.
.
h 1. Introduction
t
a
To begin with, let us consider the following Cauchy problem
m
[ i∂tu+ αu=F(x,t), 1<α<2,
|∇| (1.1)
2 (u(x,0)=f(x),
v associated with the fractional Schr¨odinger equation
9
9 i∂tu+ αu=V(x,t)u (1.2)
|∇|
3 where V : Rn+1 C is a potential. This equation has recently attracted in-
5 →
terest from mathematical physics. This is because fractional quantum mechanics
0
. introduced by Laskin [16] is governedby the equation where it is conjectured that
1
physical realizations may be limited to the cases of 1<α<2. Of course, the case
0
5 α=2 corresponds to the ordinary quantum mechanics.
1 By Duhamel’s principle, the solution of (1.1) is given by
:
v t
u(x,t)=eit|∇|αf(x) i ei(t−s)|∇|αF(,s)ds, (1.3)
i
X − Z0 ·
r wherethe propagatoreit|∇|α isgivenbymeansoftheFouriertransform,asfollows:
a
1
eit|∇|αf(x)= eix·ξ+it|ξ|αf(ξ)dξ.
(2π)n ZRn
Then the standard approach to the problem (1.1) is tobobtain the corresponding
Strichartz estimates which control space-time integrability of (1.3) in view of that
of the initial datum f and the forcing term F.
In the classicalcase α=2,the Strichartzestimates originatedby Strichartz[23]
have been extensively studied by many authors ([9, 14, 2, 12, 8, 24, 15, 17, 18, 5,
6, 20]). Over the past several years, considerable attention has been paid to the
fractional order where 1<α<2 in the radial case (see [21, 11, 13] and references
2010 Mathematics Subject Classification. Primary: 35B45,35A01;Secondary: 35Q40.
Key words and phrases. Strichartzestimates,Well-posedness,Schro¨dingerequations.
ThefirstauthorwassupportedbyPOSTECHMathBK21PlusOrganization.
Thesecondauthor wassupportedbyNRFgrant2012-008373.
1
2 CHU-HEECHO,YOUNGWOOKOHANDIHYEOKSEO
therein). Fromtheseworks,when2n/(2n 1) α<2,thehomogeneousStrichartz
− ≤
estimate
keit|∇|αfkLqtLpx .kfkH˙γ (1.4)
holds for radial functions f H˙γ(Rn) if
∈
α n n 4n 2
+ = γ, 2 q and (q,p)=(2, − ). (1.5)
q p 2 − ≤ ≤∞ 6 2n 3
−
Here the condition (1.5) is optimal if 2n/(2n 1) < α < 2. But when α =
−
2n/(2n 1),(1.4)isunknownforthe endpoint(q,p)=(2,(4n 2)/(2n 3)). Also,
− − −
it is known that the estimate does not hold in general if f does not have radial
symmetry.
Now, by duality and the Christ-Kiselev lemma ([4]), one may use (1.4) with
γ =0 to get some inhomogeneous estimates
t
ei(t−s)|∇|αF(,s)ds . F (1.6)
(cid:13)Z0 · (cid:13)LqtLpx k kLqte′Lpxe′
for (q,p) and (q,p)(cid:13)(cid:13)which satisfy (1.5) with(cid:13)(cid:13)γ = 0 and q > q′. This means that
(q,p)and(q,p)lieonthesegmentAD inFigure1. However,thesetrivialestimates
are not enough to imply the well-posedness for the equation (1.2) with the initial
e e e
data f H˙eγ(eRn) beyond the case γ = 0. When γ = 0 we need to obtain (1.6) on
∈ 6
a wider range of (q,p) and (q,p). See Section 2 for details.
Let us first mention the following necessary conditions for (1.6):
e1e 1 1 1
α( + ) n(1 )=0 (1.7)
q q − − p − p
and
1 ne n 1 n e n
+ < , + < . (1.8)
q p 2 q p 2
The first is just the scaling condition and the second will be shown in Section 4.
Our main result in this paper is the folloewingetheoremwhere we obtain (1.6) on
the open segment BC in Figure 1.
Theorem 1.1. Let n 2 and 2n/(2n 1) α < 2. Assume that F(x,t) is a
≥ − ≤
radial function with respect to the spatial variable x. Then we have
t
ei(t−s)|∇|αF(,s)ds . F (1.9)
(cid:13)Z0 · (cid:13)Lqx,t k kLqxe′,t
(cid:13) (cid:13)
if (cid:13) (cid:13)
1 1 n 1 1 n
, < and + = . (1.10)
q q 2(n+1) q q n+α
Remark 1.2. Itshouldbe notedthat the range(1.10)is sharp. Namely, the second
e e
condition in (1.10) is the scaling condition for (1.9) (see (1.7)), and the first one is
the necessary condition (1.8) when q =p and q =p.
From interpolation between (1.6) and (1.9), we can directly obtain further es-
e e
timates when (q,p) and (q,p) are contained in the open quadrangle with vertices
A,B,D,C. Precisely, we have the following corollary.
e e
INHOMOGENEOUS STRICHARTZ ESTIMATES 3
Figure 1. The range for Corollary 1.3. Here the open segment
(B,C) is the range for (1.9), and [A,D] is the range for (1.6).
Corollary 1.3. Let n 2 and 2n/(2n 1) α < 2. Assume that F(x,t) is a
≥ − ≤
radialfunctionwithrespecttothespatialvariable x, andthat(q,p)and(q,p)satisfy
the necessary conditions (1.7) and (1.8). Then we have
t e e
ei(t−s)|∇|αF(,s)ds . F (1.11)
(cid:13)Z0 · (cid:13)LqtLpx k kLqte′Lpxe′
if the following cond(cid:13)itions hold: (cid:13)
(cid:13) (cid:13)
For (q,p),
•
n(n+2 α) 1 1 1 n 1 n α 1
− − ( )< < − ( − )+ , (1.12)
(2α 1)n+α p − 2 q (α 1)n+α p − 2n 2
− −
and
1 (α 1)n2 α2n α21 α(n+2 α)
> − − − + − . (1.13)
p ((2α 1)n+α)n q 2((2α 1)n+α)
− −
Similarly for (q,p).
•
Letusgivemoredetailsabouttheconditionsintheabovecorollary. ThelineBD
inFigure1iswhentheeeqeualityholdsinthefirstinequalityof (1.12). Similarly,the
lines AC and AB correspond to the second inequality in (1.12) and the inequality
(1.13),respectively. Finally,theline CD issharpbecauseitisdeterminedfromthe
necessary condition (1.8).
Now we apply the above Strichartz estimates to the well-posedness theory for
the fractional Schr¨odinger equation in the radial case:
i∂ u+ αu=V(x,t)u, 1<α<2,
t
|∇| (1.14)
(u(x,0)=f(x),
where we assume that u, f and V are radial functions with respect to the spatial
variablex. Weobtainthefollowingwell-posednessfor(1.14)withH˙γ initialdataf
belowL2 andpotentialsV LrLw underthe scaling-criticalrangeα/r+n/w=α.
∈ t x
The Cauchy problem (1.14) was studied in [7, 19] particularly when α = 2 and
γ =0.
4 CHU-HEECHO,YOUNGWOOKOHANDIHYEOKSEO
Theorem 1.4. Let 2n α < 2 and −(α−1)n < γ 0 for n 2. Assume that
2n−1 ≤ 2(n+1) ≤ ≥
f H˙γ(Rn) and V Lr([0,T];Lw(Rn)) for some T > 0. Then there exists a
un∈ique solution u C∈([0,Tt ];H˙γ(Rnx)) Lq([0,T];Lp(Rn)) of (1.14) if
∈ ∩ t x
α n n α n
+ = γ, + =α, (1.15)
q p 2 − r w
γ 1 γ 1
− < < + , (1.16)
α 1 q (α 1)n 2
− −
and
1 n(n+α)(α 1)+2γ((2α 1)n+α) 1 1 γ(n+2 α)
1 − − < <1 + − . (1.17)
−q− 2n(n+α)(α 1) r −q (n+α)(α 1)
− −
Remark 1.5. The condition α/r+n/w=α on the potential is critical in the sense
ofscaling. Indeed, u (x,t)=u(ǫx,ǫαt)takes(1.14)intoi∂ u + αu =V (x,t)u
ǫ t ǫ ǫ ǫ ǫ
|∇|
with V (x,t)=ǫαV(ǫx,ǫαt). Hence the norm
ǫ
Vǫ LrLw = Vǫ Lr(R;Lw(Rn)) =ǫα−α/r−n/w V LrLw
k k t x k k t x k k t x
is independent of ǫ precisely when α/r+n/w=α.
Remark 1.6. InProposition3.9of[11],theinhomogeneousestimateswereshownin
certainregionthat lies below the segmentED inFigure 1. (Note that (1/q,1/p)
∈
ED if and only if q,p 2 and 2/q +(2n 1)/p = n 1/2. Also, the point E
≥ − −
is the same as A when α = 2n/(2n 1).) By interpolation between these and
−
our estimates, we can also obtain further estimates in the triangle with vertices
A,C,E. We omit the details since it does not affect the range −(α−1)n <γ 0 of
2(n+1) ≤
γ in Theorem 1.4 (see Section 2).
The rest of the paper is organized as follows. In Section 2, we prove Theorem
1.4bymakinguse ofthe Strichartzestimates (1.4)and(1.11). Section3isdevoted
to proving Theorem 1.1, and in Section 4 we show the necessary condition (1.8).
In the final section, Section 5, we show Lemma 3.5 which gives some estimates for
Bessel functions and is used for the proof of Theorem 1.1.
Throughout the paper, we shall use the letter C to denote positive constants
which may be different at each occurrence. We also use the symbol f to denote
the Fourier transform of f, and denote A . B and A B to mean A CB and
∼ ≤
CB A CB, respectively, with unspecified constants C >0. b
≤ ≤
2. Application
In this section we prove Theorem 1.4. The proof is quite standard but we need
to observe that if (q,p) and (q,p) satisfy the inhomogeneous estimate (1.11), then
the midpoint of them lies on the segment AD. Note that (1/q,1/p) AD if and
only if q,p 2 and α/q+n/p = n/2. Hence, if α/q+n/p = n/2 ∈γ for γ R,
e e
≥ − ∈
then (q,p) should satisfy α/q+n/p = n/2+γ to give (1.11). In what follows, it
will be convenient to keep in mind this key observation.
By Duhamel’s principle, the solution of (1.14) is given by
e e e e
t
Φ(u):=eit|∇|αf(x) i ei(t−s)|∇|αV(,s)u(,s)ds. (2.1)
− · ·
Z0
INHOMOGENEOUS STRICHARTZ ESTIMATES 5
Thenthestandardfixed-pointargumentistochoosethesolutionspaceonwhichΦ
is a contractionmapping. The Strichartz estimates play a centralrole in this step.
Indeed, by the estimates (1.4) and (1.11), we see that
kΦ(u)kLqt([0,T];Lpx) ≤CkfkH˙γ +CkVukLqte′([0,T];Lpxe′) (2.2)
if
(α 1)n
− − <γ 0 (2.3)
2(n+1) ≤
α n n α n n
+ = γ, + = +γ, (2.4)
q p 2 − q p 2
γ 1 γ 1
− < < + , (2.5)
α 1 q (αe 1)en 2
− −
and
γ(n+2 α) 1 n(n+α)(α 1)+2γ((2α 1)n+α)
− < < − − . (2.6)
(n+α)(α 1) q 2n(n+α)(α 1)
− −
Here,the conditions(2.3)and(2.5)aregivenfromthatthe line α+n = n γ lies
q p 2 −
intheclosedtrianglewithverteicesA,C,Dexcepttheclosedsegments[A,C],[C,D].
Note that γ = −(α−1)n when this line passes through the point C. Similarly, the
2(n+1)
condition (2.6) is givenfrom that the line α + n = n +γ lies in the closedtriangle
qe pe 2
with vertices A,B,D except the closed segments [A,B],[B,D].
By Ho¨lder’s inequality, we now get
kΦ(u)kLqt([0,T];Lpx) ≤CkfkH˙γ +CkVkLrt([0,T];Lwx)kukLqt([0,T];Lpx)
if α/r+n/w = α and the condition (1.17) holds. Indeed, when applying Ho¨lder’s
inequality to the second term in the right-hand side of (2.2), the conditions α/r+
n/w=α and (1.17) follow from (2.4) and (2.6), respectively.
From the above argument and the linearity, it follows that
kΦ(u)−Φ(v)kLqt([0,T];Lpx) ≤CkVkLrt([0,T];Lwx)ku−vkLqt([0,T];Lpx)
1
≤ 2ku−vkLqt([0,T];Lpx),
whichsaysthatΦisacontractionmapping,ifT issufficientlysmall. Buthere,since
the above process works also on time-translated small intervals if u(,t) H˙γ(Rn)
· ∈
forallt 0,thesmallnessassumptiononT canberemovedbyiteratingtheprocess
≥
a finite number of times. For this we will show that
kukL∞t H˙xγ .kfkH˙γ +kVkLrtLwxkukLqtLpx. (2.7)
From (2.1), we first see that
t
u . eit|∇|αDγf + ei(t−s)|∇|αDγ(V(,s)u(,s))ds .
k kL∞t H˙xγ k kL∞t L2x (cid:13)Z0 · · (cid:13)L∞t L2x
Since eit|∇|α is an isometry in L2, th(cid:13)(cid:13)e first term in the right-hand side(cid:13)(cid:13)is clearly
bounded by f . On the other hand, by the inhomogeneous estimate (1.6) the
k kH˙γ
second term is bounded by Dγ(Vu) , where u,v 2 and α/u+n/v =n/2.
k kLute′Lvxe′ ≥
Here we use the Sobolev embedding
kgkLb .kDβgkLa, e e e e
6 CHU-HEECHO,YOUNGWOOKOHANDIHYEOKSEO
where 1/a 1/b=β/n with 0 β <n/a and 1<a< , and Ho¨lder’s inequality
− ≤ ∞
to get
Dγ(Vu) . Vu
k kLute′Lvxe′ k kLute′Lax
≤kVkLrtLwxkukLqtLpx.
The required conditions here are summarized as follows:
α n n
u,v 2, + = ,
≥ u v 2
1 1 γ n
= −e e, 0 γ < , 1<a< ,
a − v′ n ≤−e ea ∞
1 1 1 1 1 1
= + , = + .
u′ r q a w p
e
But, the inequalities u,v 2 and 1 < a < are satisfied automatically from
≥ ∞
the conditions on q,r,p,w ein Theorem 1.4. On the other hand, the inequality
0 γ < n is redundant because v 2. The remaining four equalities is reduced
≤− a e e ≥
to the following one equality
α n n α n
e
+ +γ = ( + α)
q p − 2 − r w −
which is clearly satisfied from the condition (1.15). Consequently, we get (2.7).
3. Inhomogeneous Strichartz estimates
In this sectionwe proveTheorem 1.1. Let us first consider the multiplier opera-
tors P for k Z defined by
k
∈
P f =φ( /2k)f,
k
|·|
where φ:R [0,1]is a smoothcut-off function which is supportedin (1/2,2)and
satisfies →φ(/2k) = 1. Thedn we will obtainbthe following frequency localized
k∈Z ·
estimates (Proposition 3.1) which imply Theorem 1.1.
P
Proposition 3.1. Let n 2 and 2n/(2n 1) α < 2. Assume that F(x,t) is a
≥ − ≤
radial function with respect to the spatial variable x. Then we have
ei(t−s)|∇|αP F(,s)ds . F (3.1)
(cid:13)ZR k · (cid:13)Lqx,t k kLqxe′,t
uniformly in k Z if(cid:13) (cid:13)
∈ (cid:13) (cid:13)
1 1 n 1 1 n
, < and + = . (3.2)
q q 2(n+1) q q n+α
Indeed, since q > 2 from the first condition in (1.10), by the Littlewood-Paley
theorem and Minkowskieintegral inequality, one caen see that
2 2 1/2 2
ei(t−s)|∇|αF(,s)ds C ei(t−s)|∇|αP F(,s)ds
k
(cid:13)ZR · (cid:13)Lqx,t ≤ (cid:13)(cid:16)Xk∈Z(cid:12)ZR · (cid:12) (cid:17) (cid:13)Lqx,t
(cid:13) (cid:13) (cid:13) (cid:12) (cid:12) (cid:13) 2
(cid:13) (cid:13) C(cid:13) (cid:12) ei(t−s)|∇|αP P(cid:12)F(,s)(cid:13)ds .
k j
≤ Xk∈Z(cid:13)(cid:13)ZR (cid:0)|j−Xk|≤1 · (cid:1) (cid:13)(cid:13)Lqx,t
Now, by (3.1), the right-hand side in t(cid:13)he above is bounded by (cid:13)
2
C PjF Lqe′ .
x,t
Xk∈Z(cid:13)|j−Xk|≤1 (cid:13)
(cid:13) (cid:13)
INHOMOGENEOUS STRICHARTZ ESTIMATES 7
Sinceq′ <2,usingtheMinkowskiintegralinequalityandLittlewood-Paleytheorem,
this is bounded by C F 2 . By this boundedness and q′ < 2 < q, one may now
k kLqe′
x,t
use thee Christ-Kiselev lemma ([4]) to obtain
e
t
ei(t−s)|∇|αF(,s)ds . F
(cid:13)Z0 · (cid:13)Lqx,t k kLqxe′,t
as desired. (cid:13) (cid:13)
(cid:13) (cid:13)
Now it remains to prove the above proposition.
3.1. Proof of Proposition 3.1. Since we are assuming the scaling condition in
(3.2), by rescaling (x,t) (λx,λαt), we may show (3.1) only for k =0.
→
Let us first consider x = rx′, y = λy′ and ξ = ρξ′ for x′,y′,ξ′ Sn−1, where
∈
r= x, λ= y and ρ= ξ . Then by using the fact (see [22], p. 347) that
| | | | | |
e−irρx′·ξ′dx′ =cn(rρ)−n−22Jn−2(rρ),
ZSn−1 2
where J denotes the Bessel function with order m, it is easy to see that
m
ei(t−s)|∇|αP F(,s)ds (3.3)
0
R ·
Z
= eiρ(rx′−λy′)·ξ′+i(t−s)ραφ(ρ)F(λy′,s)(λρ)n−1dy′dξ′dρdλds
ZRZRZRZSn−1ZSn−1
=r−n−22 λ−n−22 ei(t−s)ραJn−2(rρ)Jn−2(λρ)ρφ(ρ)dρ λn−1F(λy′,s) dλds.
R R 2 2
Z (cid:18)Z (cid:19)
Now we define the operators T h, j 0, as (cid:2) (cid:3)
j
≥
∞
T0h(r,t)=χ(0,1)(r) r−n−22 eitραJn−2(rρ)ϕ(ρ)h(ρ)dρ (3.4)
Z0 2
and for j 1
≥
∞
Tjh(r,t)=χ[2j−1,2j)(r) r−n−22 eitραJn−2(rρ)ϕ(ρ)h(ρ)dρ, (3.5)
Z0 2
where χ denotes the characteristic function of a set A and ϕ2(ρ) = ρφ(ρ). Then
A
the adjoint operator T∗ of T is given by
k k
∞
T0∗H(ρ)=ϕ(ρ) e−isρα χ(0,1)(λ) λ−n−22Jn−2(λρ)H(λ,s)dλds
ZR Z0 2
and for k 1
≥
∞
Tk∗H(ρ)=ϕ(ρ) e−isρα χ[2k−1,2k)(λ) λ−n−22Jn−2(λρ)H(λ,s)dλds. (3.6)
ZR Z0 2
Hence, since F(λy′,s) is independent of y′ Sn−1, by setting H(λ,s)=F(λy′,s),
∈
it follows from (3.3) that
ei(t−s)|∇|αP F(,s)ds= T T∗(λn−1H).
R 0 · j k
Z j,k≥0
X
Now we are reduced to showing that
T T∗(λn−1H) . H , (3.7)
(cid:13)jX,k≥0 j k (cid:13)LqtLqr k kLqte′Lqre′
(cid:13) (cid:13)
where we denote by L(cid:13)q the space Lq(rn−1(cid:13)dr).
r
8 CHU-HEECHO,YOUNGWOOKOHANDIHYEOKSEO
Figure 2. The range of q,q for Lemma 3.2. Here the open seg-
ment (A,B) is the range for Proposition 3.1 (see (3.2)).
e
From now on, we will show (3.7) by making use of the following lemma which
will be obtained in Subsection 3.2.
Lemma 3.2. Let n 2 and 2n/(2n 1) α<2. Then we have for j,k 0
≥ − ≤ ≥
kTjTk∗(λn−1H)kLqtLqr
.2j(2n2+q1−2n4−1)2k(2n2+qe1−2n4−1)2−|j2−k|(21−max(q1,q1e))kHkLqte′Lqre′
if 2 q,q 6 (see Figure 2).
≤ ≤
The case 2n/(2n 1) < α < 2. We first decompose the sum over j,k into two
e −
parts, j k and j k:
≤ ≥
kTjTk∗(λn−1H)kLqtLqr
j,k≥0
X
∞ ∞ ∞ ∞
≤ kTjTk∗(λn−1H)kLqtLqr + kTjTk∗(λn−1H)kLqtLqr.
j=0k=j k=0j=k
XX XX
When j k, using Lemma 3.2, we then have
≤
∞ ∞
kTjTk∗(λn−1H)kLqtLqr
j=0k=j
XX
∞ ∞
. 2j(2n2+q1−2n4−1)2k(2n2+qe1−2n4−1)2−|j2−k|(21−max(q1,q1e))kHkLqte′Lqre′
j=0k=j
XX
∞ ∞
= 2j(2n2+q1−2n4−2−21max(q1,q1e)) 2k(2n2+qe1−n2+12max(q1,q1e))kHkLqte′Lqre′.
j=0 k=j
X X
Note here that the first condition in (3.2) implies
2n+1 n 1 1 1 1 1 n
+ max( , ) (n+1)max( , ) <0. (3.8)
2q − 2 2 q q ≤ q q − 2
e e e
INHOMOGENEOUS STRICHARTZ ESTIMATES 9
From this, it follows that
∞ ∞
2j(2n2+q1−2n4−2−21max(q1,q1e)) 2k(2n2+qe1−n2+21max(q1,1qe))
j=0 k=j
X X
∞
. 2j(2n2+1(1q+q1e)−2n2−1).
j=0
X
On the other hand, the second condition in (3.2) implies
2n+1 1 1 2n 1
( + ) − <0 (3.9)
2 q q − 2
since α>2n/(2n 1). Consequently, we get
− e
∞ ∞ ∞
kTjTk∗(λn−1H)kLqtLqr . 2j(2n2+1(q1+q1e)−2n2−1)kHkLqte′Lqre′
j=0k=j j=0
XX X
. H
k kLqte′Lqre′
as desired. The other part where j k follows clearly from the same argument.
≥
The case α = 2n/(2n 1). The previous argument is no longer available in this
−
case,since the left-hand side in (3.9) becomes zero. But here we deduce (3.7) from
bilinear interpolation between bilinear form estimates which follow from Lemma
3.2. This enables us to gain some summability as before.
Let us first define the bilinear operators B by
j,k
B (H,H)= T∗(λn−1H),T∗(λn−1H) ,
j,k k j
L2
r,t
D E
where , denotes the usuaelinner product on the spaceeL2 . Then it is enough to
h i r,t
show the following bilinear form estimate
B (H,H) . H H . (3.10)
j,k k kLqte′Lqre′k kLqt′Lqr′
(cid:12)jX,k≥0 (cid:12)
(cid:12) e (cid:12) e
In fact, from (3.10) w(cid:12)e get (cid:12)
(cid:13)(cid:13)jX,k≥0TjTk∗(λn−1H)(cid:13)(cid:13)LqtLqr =kHeksLuqrp,′t=1ZZ jX,k≥0TjTk∗(λn−1H) rn−q1He(r,t)drdt
(cid:13) (cid:13)
= sup T∗(λn−1H),T∗(rn−q1H)
k j
e L2
kHkLqr,′t=1jX,k≥0D E r,t
e
.kHeksLuqrp,′t=1kHkLqte′Lqre′kr−nq−′1HkLqt′Lqr′ =kHkLqte′Lqre′
e
as desired.
For(3.10), wefirstdecomposethe sumoverj,k into twoparts,j k andj k:
≤ ≥
∞ ∞ ∞ ∞
B (H,H) B (H,H) + B (H,H) .
j,k j,k j,k
≤
(cid:12)jX,k≥0 (cid:12) Xj=0(cid:12)Xk=j (cid:12) Xk=0(cid:12)Xj=k (cid:12)
(cid:12)(cid:12) e (cid:12)(cid:12) (cid:12)(cid:12) e (cid:12)(cid:12) (cid:12)(cid:12) e (cid:12)(cid:12)
10 CHU-HEECHO,YOUNGWOOKOHANDIHYEOKSEO
Thenwe will use the following estimate whichfollowsfromHo¨lder’s inequality and
Lemma 3.2:
|Bj,k(H,H)|= H(r,s)rnq−′1rn−q1TjTk∗(λn−1H)drds
ZZ
e ≤kHkLeqt′Lqr′ TjTk∗(λn−1H) LqtLqr
.2je(2n2+q1−2n(cid:13)(cid:13)4−1)2k(2n2+qe1−2n4−(cid:13)(cid:13)1)2−|j2−k|(12−max(q1,q1e))kHkLqte′Lqpe′kHkLqt′Lqr′
for2 q,q 6. By usingthis and(3.8), the firstpartwherej k is nowbounded
e
≤ ≤ ≤
as follows:
∞ ∞e
B (H,H)
j,k
Xj=0(cid:12)Xk=j (cid:12)
. ∞(cid:12)(cid:12) ∞ 2j(2n2+q1−e2(cid:12)(cid:12)n4−1)2k(2n2+qe1−2n4−1)2−|j2−k|(12−max(q1,q1e))kHkLqte′Lqre′kHkLqt′Lqr′
j=0k=j
XX
∞ ∞ e
= 2j(2n2+q1−2n4−2−21max(q1,q1e)) 2k(2n2+qe1−n2+21max(1q,q1e))kHkLqte′Lqre′kHkLqt′Lqr′
j=0 k=j
X X
∞ e
. 2j(2n2+1(q1+1qe)−2n2−1)kHkLqte′Lqre′kHkLqt′Lqr′ (3.11)
j=0
X
e
for 2(n+1)/n < q,q 6. If one applies this bound directly for q,q satisfying the
≤
conditions in Proposition 3.1 as in the previous case, then one can not sum over j
because 2n2+1(q1+q1e)e−2n2−1 =0 whenα=2n/(2n−1). Butherewee willmakeuse
of the following bilinear interpolation lemma (see [1], Section 3.13, Exercise 5(b))
together with (3.11) to give
∞ ∞
B (H,H) . H H . (3.12)
j,k k kLqte′Lqre′k kLqt′Lqr′
Xj=0(cid:12)Xk=j (cid:12)
Lemma 3.3. For i =(cid:12)(cid:12)0,1, let A ,Be,C(cid:12)(cid:12) be Banach speaces and let T be a bilinear
i i i
operator such that
T :A B C ,
0 0 0
× →
T :A B C ,
0 1 1
× →
T :A B C .
1 0 1
× →
Then one has, for θ =θ +θ and 1/q+1/r 1,
0 1
≥
T :(A ,A ) (B ,B ) (C ,C ) .
0 1 θ0,q× 0 1 θ1,r → 0 1 θ,1
Here, 0<θ <θ <1 and 1 q,r .
i
≤ ≤∞
Indeed, let us first consider the vector-valued bilinear operator T defined by
T(H,H)= T (H,H) ,
j j≥0
where T = ∞ B . Then, (3.12) is(cid:8)equivalent(cid:9)to
j k=j j,k e e
P T :Lqe′Lqe′ Lq′Lq′ ℓ0(C), (3.13)
t r × t r → 1