Global exact controllability in infinite time of Schrödinger equation: multidimensional case

0.21 MB·English

by  Vahagn Nersesyan

Global exact controllability in infinite time of
Schr¨odinger equation: multidimensional case
2
1
0
2 Vahagn Nersesyan, Hayk Nersisyan

n
a
J
Abstract. We prove that the multidimensional Schr¨odinger equation is exactly
7
controllable in infinite time near any point which is a finite linear combination of
1
eigenfunctions of the Schr¨odinger operator. We prove that, generically with respect

] to the potential, the linearized system is controllable in infinite time. Applying the
P
inversemapping theorem, we provethe controllability of thenonlinear system.
A
.
h
t
a Contents
m
[ 1 Introduction 1

1
2 Main results 4
v
2.1 Well-posedness of Schr¨odinger equation . . . . . . . . . . . . . . 4
5
4 2.2 Exact controllability in infinite time . . . . . . . . . . . . . . . . 8
4
3 3 Proof of Theorem 2.8 9
1. 3.1 Controllability of linearized system . . . . . . . . . . . . . . . . . 9
0 3.2 Proof of Proposition 3.4 . . . . . . . . . . . . . . . . . . . . . . . 12
2 3.3 Application of the inverse mapping theorem . . . . . . . . . . . . 13
1
v: References 15
i
X
1 Introduction
r
a
This paper is concerned with the problem of controllability for the following
Schr¨odinger equation
iz˙ = ∆z+V(x)z+u(t)Q(x)z, x D, (1.1)
− ∈
z =0, (1.2)
∂D
|
z(0,x)=z (x), (1.3)
0
where D Rd,d 1 is a rectangle, V,Q C∞(D,R) are given functions, u is
⊂ ≥ ∈
thecontrol,andzisthestate. Weprovethat(1.1)-(1.3)isexactlycontrollablein
infinitetimenearanypointwhichisafinitelinearcombinationofeigenfunctions
1
oftheSchr¨odingeroperator,extendingtheresultsof[24]tothemultidimensional
case.
Recallthatinthepapers[6,8,10]itisprovedthatthe1DSchr¨odingerequa-
tion is exactly controllable in finite time in a neighborhood of any finite linear
combinationofeigenfunctionsofLaplacian. In[13,26,19],approximatecontrol-
lability in L2 is proved for multidimensional Schr¨odinger equation, generically
with respect to functions V,Q and domain D. In [20, 11, 23, 22, 21], stabiliza-
tionresultsandapproximatecontrollabilitypropertiesareproved. Inparticular,
combinationof the results of[23]with the abovementionedlocalexactcontrol-
lability properties gives globalexact controllability in finite time for 1D case in
the spaces H3+ε,ε >0. See also papers [28, 29, 3, 2, 1, 9] for controllability of
finite-dimensional systems and papers [16, 17, 5, 31, 14, 15] for controllability
properties of various Schr¨odinger systems.
Thelinearizationof(1.1)-(1.3)aroundthetrajectorye−iλk,Vtek,V withu=0
and z = e (e is an eigenfunction of the Schr¨odinger operator ∆+V
0 k,V k,V
−
corresponding to some eigenvalue λ ) is of the form
k,V
iz˙ = ∆z+V(x)z+u(t)Q(x)e−iλk,Vtek,V, x D, (1.4)
− ∈
z =0, (1.5)
∂D
|
z(0,x)=0. (1.6)
Writing this in the Duhamel form
T
z(T)=−iZ S(T −s)[u(s)Qe−iλk,Vsek,V]ds, (1.7)
0
where S(t)=eit(∆−V) is the free evolution, we see that (1.4)-(1.6) is equivalent
to the following moment problem for d := ieiλm,VT z(T),e
mk hQem,V,ek,Vih m,Vi
T
dmk =Z eiωmksu(s)ds, m≥1, ωmk =λm,V −λk,V. (1.8)
0
It is well known that a gap condition for the frequencies ω is necessary for
mk
the solvability of this moment problem when T < + (e.g., see [30]). The
∞
asymptotic formula for the eigenvalues λm,V Cdmd2 implies that there is no
∼
gap in the case d 3 (when d = 2, existence of a domain for which there is a
≥
gap between the eigenvalues is an open problem). Moreover,it follows from [4]
that there is a linear dependence between the exponentials: there is a non-zero
{cm}∈ℓ2 suchthat +m∞=1cmeiωmks =0fort∈[0,T]. Hence(1.4)-(1.6)isnon-
controllable in finitePtime T < + . The situation is different when T = + .
∞ ∞
Indeed, by Lemma 3.10 in [22], the exponentials are independent on [0,+ ),
∞
andmoreover,(1.4)-(1.6)iscontrollable,byTheorem2.6in[24]. In[24],weused
the controllabilityoflinearizedsystem(1.4)-(1.6) to provethe controllabilityof
nonlinear system only in the case d=1. In the multidimensional case, we were
able to prove the controllability of (1.4)-(1.6) in a more regular Sobolev space
than the one where nonlinear system (1.1)-(1.3) is well posed. We do not know
2
ifthisdifficultyoflossofregularitycanbetreatedusingtheNash–Moserinverse
function theorem in the spirit of [6]. More precisely, in the multidimensional
case, it is very difficult to prove that the inverse of the linearization satisfies
the estimates in the Nash–Moser theorem. In this paper, we find a space
H
(see (1.11) for the definition), where the nonlinear problem is well posed and
the linearized problem is controllable. Applying the inverse inverse function
theoreminthe space ,wegetcontrollabilityfor(1.1)-(1.3). Letusnoticethat
H
is a sufficiently large space of functions, it contains the Sobolev space H3d.
H
Thus, in particular, we prove controllability in H3d. The result of this paper
is optimal in the sense that it seems that the multidimensional Schr¨odinger
equation (1.1)-(1.3) is not exactly controllable in finite time.
Acknowledgments. TheauthorswouldliketothankJ.-PPuelforprovid-
ing them in privat communication [27] some results about regularity questions
for the Schr¨odinger equation.
Notation
In this paper, we use the following notation. Let us define the Banach spaces
+∞
ℓ2 := a C∞ : a 2 = a 2 <+ ,
{{ j}∈ k{ j}kℓ2 | j| ∞}
X
j=1
ℓ2 := a ℓ2 :a R ,
0 {{ j}∈ 1 ∈ }
ℓ∞ := aj C∞ : aj ℓ∞ =sup aj <+ ,
{{ }∈ k{ }k | | ∞}
j≥1
ℓ∞ := a ℓ∞ : lim a =0 ,
0 {{ j}∈ j→+∞ j }
ℓ∞ := a ℓ∞ :a R .
01 {{ j}∈ 0 1 ∈ }
We denote by Hs := Hs(D) the Sobolev space of order s 0. Consider the
Schr¨odinger operator ∆+V, V C∞(D,R) with ( ∆≥+V) := H1 H2.
− ∈ D − 0 ∩
Let λ and e be the sets of eigenvalues and normalized eigenfunctions
j,V j,V
{ } { }
of this operator. Let , and be the scalar product and the norm in
the space L2. Define ht·he·ispacekH·(skV) := D((−∆ + V)s2) endowed with the
norm s,V = (λj,V)2s ,ej,V ℓ2. When D is the rectangle (0,1)d and
V(x ,..k.,·xk) = V (kx )+..h.·+V (ixk ), V C∞([0,1],R), the eigenvalues and
1 d 1 1 d d k
∈
eigenfunctions of ∆+V on D are of the form
−
λ =λ +...+λ , (1.9)
j1,...,jd,V j1,V1 jd,Vd
e (x ,...,x )=e (x ) ... e (x ), (x ,...,x ) D, (1.10)
j1,...,jd,V 1 d j1,V1 1 · · jd,Vd d 1 d ∈
where λ and e are the eigenvalues and eigenfunctions of operator
{ j,Vk} { j,Vk}
3
d2 +V on (0,1). Define the spaces
−dx2 k
= z L2 :(j3 ... j3) z,e ℓ∞,
H { ∈ 1 · · d h j1,...,jd,Vi∈ 0
kzkH :=k(j13·...·jd3)hz,ej1,...,jd,Vikℓ∞ <+∞}, (1.11)
+∞
= z L2 : z := (j3 ... j3) z,e <+ . (1.12)
V { ∈ k kV 1 · · d |h j1,...,jd,Vi| ∞}
X
j1,...,jd=1
The eigenvalues and eigenfunctions of Dirichlet Laplacian on the interval (0,1)
are λ =k2π2 and e (x)=√2sin(kπx), x (0,1). It is well known that for
k,0 k,0
any V L2([0,1],R) ∈
∈
1
λ =k2π2+ V(x)dx+r , (1.13)
k,V Z k
0
C
ek,V ek,0 L∞ , (1.14)
k − k ≤ k
de de
k,V k,0
C, (1.15)
(cid:13) dx − dx (cid:13)L∞ ≤
(cid:13) (cid:13)
(cid:13) (cid:13)
where +∞r2 < + (e.g., see [25]). For a Banach space X, we shall denote
k=1 k ∞
by B P(a,r) the open ball of radius r > 0 centered at a X. The integer part
X
ofx R is denotedby [x]. We denote byC a constantw∈hosevaluemaychange
∈
from line to line.
2 Main results
2.1 Well-posedness of Schr¨odinger equation
We assume that V(x ,...,x ) = V (x )+...+V (x ),x [0,1] and V
1 d 1 1 d d k k
C∞([0,1],R),k =1,...,d. Let us consider the following Schr¨o∈dinger equation∈
iz˙ = ∆z+V(x)z+u(t)Q(x)z+v(t)Q(x)y, (2.1)
−
z =0, (2.2)
∂D
|
z(0,x)=z (x). (2.3)
0
The following lemma shows the well-posedness of this system in H2 .
(V)
Lemma2.1. Foranyz H2 ,u,v L1 ([0, ),R)andy L1([0, ),H2 )
0 ∈ (V) ∈ loc ∞ ∈ ∞ (V)
problem (2.1)-(2.3) has a unique solution z C([0, ),H2 ). Furthermore, if
∈ ∞ (V)
v =0, then for all t 0 we have
≥
z(t) = z . (2.4)
0
k k k k
See [12] for the proof. In [10] it is proved that this problem is well posed in
H3 for d=1, and in [27] the well-posedness in H3 is proved for d 1.
(V) (V) ≥
4
For any integer l 3, let m=m(l):=[l−1] and define the space
≥ 2
dku
Cm := u Cm([0, ),R): (0)=0,k [0,m]
0 { ∈ ∞ dtk ∈ }
endowed with the norm of Cm([0, ),R). The following lemma shows that
∞
problem (2.1)-(2.3) is well posed in higher Sobolev spaces when u,v and y are
more regular.
Lemma2.2. Foranyintegerl 3,anyz Hl ,anyy Wm,1([0, ),H2 )
≥ 0 ∈ (V) ∈ loc ∞ (V)
andanyu,v Cm thesolutionzinLemma2.1belongstothespaceC([0, ),Hl)
∈ 0 ∞ ∩
C1([0, ),Hl−2). Moreover, there is a constant C >0 such that
∞
kz(t)kHl +kzkWm,1([0,t],H(2V)) ≤C(kz0kl,V +kvkC0mkykWm,1([0,t],H(2V)))
eC(kukC0m+1)t. (2.5)
×
See Appendix of [6] for the proof.
Lemma 2.3. Denote by (, ):H2 L1 (R ,R) H2 the resolving oper-
Ut · · (V)× loc + → (V)
ator of (1.1), (1.2). Then (, ) is locally Lipschitz continuous: there is C >0
t
U · ·
such that
kUt(z0,u)−Ut(z0′,u′)kHl ≤C(kz0−z0′kl,V +ku−u′kC0mkz0′kl,V)eC(kukC0m+1)t.
(2.6)
Proof. Notice that z(t):= (z ,u) (z′,u′) is a solution of problem
Ut 0 −Ut 0
iz˙ = ∆z+V(x)z+u(t)Q(x)z+(u(t) u′(t))Q(x) (z′,u′),
− − Ut 0
z =0,
∂D
|
z(0,x)=z (x) z′(x).
0 − 0
Applying Lemma 2.2, we get
kz(t)kHl ≤C(kz0−z0′kl,V +ku−u′kC0mkU·(z0′,u′)kWm,1([0,t],H(2V)))eC(kukC0m+1)t,
(2.7)
kU·(z0′,u′)kWm,1([0,t],H(2V)) ≤Ckz0′kl,VeC(kukC0m+1)t. (2.8)
Replacing (2.8) into (2.7), we get (2.6).
Let us rewrite (1.1)-(1.3) in the Duhamel form
t
z(t)=S(t)z i S(t s)[u(s)Qz(s)]ds, (2.9)
0− Z −
0
whereS(t)=eit(∆−V) is the freeevolution. Letus takeanyw L1(R ,R)and
+
∈
estimate the following integral
t
G (z):= S( s)[w(s)Qz(s)]ds.
t Z −
0
5
We take controls from the weighted space space
:= u L1(R ,R):u()eB· L1(R ,R)
+ +
G { ∈ · ∈ }
endowed with the norm u = u()eB· , where the constant B >0 will be
G L1
k k k · k
chosen later. For B > C +1, where C is the constant in Lemma 2.2, we have
the following result.
Proposition 2.4. Let us take any l 4d, z Hl ,w and u Cm,
≥ 0 ∈ (V) ∈ G ∈ 0
and let z(t) := (z ,u). Then there are constants δ,C > 0 such that for any
t 0
U
u BCm(0,δ) and for any t>s 0
∈ 0 ≥
t
G (z) G (z) C z(τ) w(τ)dτ, (2.10)
k t − s kH ≤ Z k kHl| |
s
and the following integral converges in
H
+∞
G (z):= S( τ)[w(τ)Qz(τ)]dτ. (2.11)
∞ Z −
0
Proof. Using(2.5)withv =0,thedefinitionof ,andchoosingδ >0sufficiently
G
small, we see that
+∞
z(τ) w(τ)dτ <+ .
Z k kHl| | ∞
0
Combining this with (2.10), we prove the convergence of the integral in (2.11).
Let us prove (2.10). To simplify the notation, let us suppose that d = 2; the
proofofthegeneralcaseissimilar. LetV(x ,x )=V (x )+V (x ). Integration
1 2 1 1 2 2
by parts gives
1 ∂2
Qz(s),e e = ( +V )(Qz),e e
h j1,V1 j2,V2i λ h −∂x2 1 j1,V1 j2,V2i
j1,V1 1
1 ∂2 ∂2
= ( +V )(Qz),( +V )e e
λ2 h −∂x2 1 −∂x2 1 j1,V1 j2,V2i
j1,V1 1 1
= 1 1 ∂2 (Qz)e dx ∂ e x1=1
λ2j1,V1 Z0 ∂x21 j2,V2 2∂x1 j1,V1(cid:12)(cid:12)x1=0
1 ∂2
+ V ( +V )(Qz),e e
λ2 (cid:16)h 1 −∂x2 1 j1,V1 j2,V2i
j1,V1 1
∂ ∂2 ∂
+ ( +V )(Qz), e e
h∂x1 −∂x21 1 ∂x1 j1,V1 j2,V2i(cid:17)
=:I +J .
j j
Let us estimate I . Since ∂2 (Qz(s))= 0 for all x [0,1] and for x = 0 and
j ∂x21 1 ∈ 2
6
x =1, integration by parts in x implies
2 2
I = 1 1( ∂2 +V ) ∂2 (Qz) e dx ∂ e x1=1
j λ2j1,V1λj2,V2 Z0 −∂x22 2 (cid:16)∂x21 (cid:17) j2,V2 2∂x1 j1,V1(cid:12)(cid:12)x1=0
= 1 1( ∂2 +V ) ∂2 (Qz) ( ∂2 +V )e dx ∂ e x1=1
λ2j1,V1λ2j2,V2 Z0 −∂x22 2 (cid:16)∂x21 (cid:17) −∂x22 2 j2,V2 2∂x1 j1,V1(cid:12)(cid:12)x1=0
= 1 ( ∂2 +V ) ∂2 (Qz) ∂ e ∂ e x2=1 x1=1
λ2j1,V1λ2j2,V2 −∂x22 2 (cid:16)∂x21 (cid:17)∂x2 j2,V2∂x1 j1,V1(cid:12)(cid:12)x2=0(cid:12)(cid:12)x1=0
+ 1 1V ( ∂2 +V ) ∂2 (Qz) e dx ∂ e x1=1
λ2j1,V1λ2j2,V2 Z0 2 −∂x22 2 (cid:16)∂x21 (cid:17) j2,V2 2∂x1 j1,V1(cid:12)(cid:12)x1=0
+ 1 1 ∂ ( ∂2 +V ) ∂2 (Qz) ∂ e dx ∂ e x1=1
λ2j1,V1λ2j2,V2 Z0 ∂x2 −∂x22 2 (cid:16)∂x21 (cid:17)∂x2 j2,V2 2∂x1 j1,V1(cid:12)(cid:12)x1=0
=:I +I +I . (2.12)
j,1 j,2 j,3
Let us consider the term I :
j,1
2j j π2 ∂2 ∂2
1 2
I = ( +V ) (Qz) cos(j πx )cos(j πx )
j,1 (cid:16)λ2 λ2 −∂x2 2 (cid:16)∂x2 (cid:17) 1 1 2 2
j1,V1 j2,V2 2 1
+ 1 ( ∂2 +V ) ∂2(Qz) ∂2 (e e e e ) x2=1 x1=1.
λ2j1,V1λ2j2,V2 −∂x22 2 (cid:16)∂x21 (cid:17)∂x1∂x2 j1,V1 j2,V2− j1,0 j2,0(cid:17)(cid:12)(cid:12)x2=0(cid:12)(cid:12)x1=0
Using (1.13), (1.15) and the Sobolev embedding Hs ֒ L∞,s> d, we get
→ 2
t t
j1s,ju2p≥1(cid:12)(cid:12)j13j23Zs ei(λj1,V1+λj2,V2)τw(τ)Ij,1dτ(cid:12)(cid:12)≤CZs kz(τ)kHl|w(τ)|dτ.
(cid:12) (cid:12)
The Riemann–Lebesgue theorem and (1.15) imply that
t
j13j23Z ei(λj1,V1+λj2,V2)τw(τ)Ij,1dτ→0 as j1+j2→+∞.
s
Thus
t
j13j23Z ei(λj1,V1+λj2,V2)τw(τ)Ij,1dτ ∈ℓ∞0 .
s
The terms I ,I and J are treated exactly in the same way. We omit the
j,2 j,3 j
details. Thus we get that
t t
G (z) G (z) = S( τ)[w(τ)Qz(τ)]dτ C z(τ) w(τ)dτ.
k t − s kH kZ − kH ≤ Z k kHl| |
s s
Let Tn + be a sequence such that e−iλV,jTn 1 as n for any j 1
→ ∞ → →∞ ≥
(e.g., see Lemma 2.1 in [24]). Then
S(T )z z as n + in for any z and t 0. (2.13)
n
→ → ∞ H ∈H ≥
7
Indeed, since
+∞
S(t)z = e−iλj,Vt z,e e , (2.14)
j,V j,V
h i
X
j=1
we have
kS(Tn)z−zkH ≤ sup (j13·...·jd3)|e−iλj1,...,jd,VTn −1||hz,ej1,...,jd,Vi|
λj1,...,jd,V≤N
ε ε
+2 sup (j3 ... j3) z,e + =ε
1 · · d |h j1,...,jd,Vi|≤ 2 2
λj1,...,jd,V>N
for sufficiently large integers N,n 1.
≥
Let us take t = T in (2.9) and pass to the limit n . Using Proposi-
n
→∞
tion 2.4, the embedding Hl ֒ and (2.13), we obtain the following result.
(V) →H
Lemma 2.5. Let us take any l 4d and z Hl . There is a constant δ >0
≥ 0 ∈ (V)
such that for any u BCm(0,δ) the following limit exists in
∈ 0 ∩G H
lim (z ,u)=: (z ,u). (2.15)
n→+∞UTn 0 U∞ 0
2.2 Exact controllability in infinite time
Let l 4d be the integer in Proposition 2.4. Take any integer s l and let
≥ ≥
Hs(R ,R):= u Hs(R ,R):u(k)(0)=0,k=0,...,s 1 .
0 + { ∈ + − }
The set of admissible controls is the Banach space
:= Hs(R ,R) (2.16)
F G∩ 0 +
endowed with the norm u := u + u . Equality (2.4) implies that
F G Hs
k k k k k k
it suffices to consider the controllability properties of (1.1), (1.2) on the unit
sphere S in L2.
We prove the controllability of (1.1), (1.2) under below condition.
Condition 2.6. Suppose that the functions V,Q C∞(D,R) are such that
∈
(i) inf (p j ... p j )3Q >0,Q := Qe ,e ,
p1,j1,...,pd,jd≥1| 1 1· · d d pj| pj h p1,...,pd,V j1,...,jd,Vi
(ii) λ λ =λ λ for all i,j,p,q 1 such that i,j = p,q and
i,V j,V p,V q,V
− 6 − ≥ { }6 { }
i=j.
6
See[24]and[26,23,18]fortheproofofgenericityof(i)and(ii),respectively.
Let us set
:=span e . (2.17)
j,V
E { }
Below theorem is the main result of this paper.
8
Theorem 2.7. Under Condition 2.6, for any z˜ S there is σ >0 such that
∈ ∩E
problem (1.1), (1.2) is exactly controllable in infinite time in S B (z˜,σ), i.e.,
H
∩
for any z S B (z˜,σ) there is a control u such that limit (2.15) exists
1 H
∈ ∩ ∈F
in and z = (z˜,u).
1 ∞
H U
SeeSection3.3fortheproof. SincethespaceH3d iscontinuouslyembedded
(V)
into , we obtain
H
Theorem 2.8. Under Condition 2.6, for any z˜ S there is σ > 0 such
∈ ∩E
that for any z S B (z˜,σ) there is a control u such that limit (2.15)
1 ∈ ∩ H(3Vd) ∈F
exists in and z = (z˜,u).
1 ∞
H U
Remark 2.9. Asinthecased=1(seeTheorems3.7and3.8in[24])herealsoone
can prove controllability in higher Sobolev spaces with more regular controls,
and a global controllability property using a compactness argument.
3 Proof of Theorem 2.8
3.1 Controllability of linearized system
In this section, we study the controllability of the linearization of (1.1), (1.2)
around the trajectory (z˜,0),z˜ S :
t
U ∈ ∩E
iz˙ = ∆z+V(x)z+u(t)Q(x) (z˜,0), (3.1)
t
− U
z =0, (3.2)
∂D
|
z(0,x)=z . (3.3)
0
The controllability in infinite time of this system is proved in [24], Section 2.
Forthe proofofTheorem2.8we needtoshowcontrollabilityof(3.1)-(3.3)in
H
which is larger than the space considered in [24]. Hence a generalization of the
arguments of [24] is needed.
Let S be the unit sphere in L2. For y S, let T be the tangent space to S
y
∈
at y S:
∈ T = z L2 :Re z,y =0 .
y
{ ∈ h i }
By Lemma 2.1, for any z H2 and u L1 (R ,R), problem (3.1)-(3.3) has
0 ∈ (V) ∈ loc +
a unique solution z C(R ,H2 ). Let
∈ + (V)
R (, ):H2 L1([0,t],R) H2 ,
t · · (V)× → (V)
(z ,u) z(t)
0
→
be the resolving operator. Then R (z ,u) T for any z T H2 and
t 0 ∈ Ut(z˜,0) 0 ∈ z˜∩ (V)
t 0. Indeed,
≥
d
Re R , =Re R˙ , +Re R , ˙
t t t t t t
dt h U i h U i h U i
=Re i(∆ V)R iu(t)Q(x) , +Re R ,i(∆ V)
t t t t t
h − − U U i h − U i
=Re i(∆ V)R , +Re R ,i(∆ V) =0.
t t t t
h − U i h − U i
9
Since Re R , =Re z ,z˜ =0, we get R (z ,u) T .
h 0 U0i h 0 i t 0 ∈ Ut(z˜,0)
As (3.1)-(3.3) is a linear control problem, the controllability of system with
z = 0 is equivalent to that with any z T . Henceforth, we take z = 0 in
0 0 z˜ 0
∈
(3.3). Let us rewrite this problem in the Duhamel form
t
z(t)= i S(t s)u(s)Q(x) (z˜,0)ds. (3.4)
− Z − Us
0
LetT be the sequence definedin Section2.1. Forany u the following
n
→∞ ∈F
limit exists in
H
R (0,u):= lim z(T )= lim R (0,u). (3.5)
∞ n→+∞ n n→+∞ Tn
Using (2.14) and (3.4), we obtain
+∞ t
hz(t),em,Vi=−i e−iλm,Vthz˜,ek,ViQmkZ eiωmksu(s)ds, m≥1, (3.6)
Xk=1 0
where ω = λ λ and Q := Qe ,e . Let us take t = T in
mk m,V k,V mk m,V k,V n
− h i
(3.6) and pass to the limit as n + . The choice of the sequence T implies
n
→ ∞
that
+∞ +∞
hR∞(0,u),em,Vi=−i hz˜,ek,ViQmkZ eiωmksu(s)ds. (3.7)
Xk=1 0
Moreover,R (0,u) T . Indeed, using (3.5) andthe convergence (z˜,0) z˜
∞ ∈ z˜ UTn →
in , we get
H
Re R (0,u),z˜ = lim Re R (0,u), (z˜,0) =0.
h ∞ i n→∞ h Tn UTn i
Lemma 3.1. The mapping R (0, ) is linear continuous from to T .
∞ z˜
· F ∩H
Proof. By(2.24)in[24], thereis aconstantC >0suchthatforanym ,k 1,
j j
≥
j =1,...,d we have
(m ... m )3
1 d
· · Qe ,e C. (3.8)
(cid:12) (k1 ... kd)3 h k1,...,kd,V m1,...,md,Vi(cid:12)≤
(cid:12) · · (cid:12)
(cid:12) (cid:12)
Then (3.7), (3.8) and the Schwarz inequality imply that
R (0,u) = sup (m3 ... m3) R (0,u),e
k ∞ kH | 1· · d h ∞ m1,...,md,Vi|
m1,...,md≥1
+∞
C sup (m3 ... m3) z˜,e Qe ,e u(s)ds
≤ m1,...,md≥1(cid:12)(cid:12) 1· · d h m1,...,md,Vih m,V m,ViZ0 (cid:12)(cid:12)
(cid:12) (m ... m )3 +∞ (cid:12)
+Ckz˜kVm,k≥su1p,m6=k(cid:12)(cid:12) (k11··...··kdd)3 hQek1,...,kd,V,em1,...,md,ViZ0 eiωmksu(s)ds(cid:12)(cid:12)
C z˜ 2 u 2 <+(cid:12) , (cid:12)
≤ k kVk kF ∞
where is defined by (1.12).
V
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