A SIMPLE NASH-MOSER IMPLICIT FUNCTION
THEOREM IN WEIGHTED BANACH SPACES
Sanghyun Cho†
Abstract. We prove a simplified version of the Nash-Moser implicit function theorem in
weighted Banach spaces. We relax the conditions so that the linearized equation has an approximate inverse in different weighted Banach spaces in each recurrence step.
During the last several decades, “Nash-Moser implicit function theorem” helped to resolve
several difficult problems of solvability for nonlinear problems, especially, nonlinear partial
differential equations [6,11].
Usually nonlinear partial differential equations (or nonlinear problems in general) can be
tranformed into solving the problem :
φ(u) = 0
where φ involves the variables x, the unknown function u(x) and its derivatives up to the
order m.
To prove implicit function theorem in infinite dimensional spaces (as spaces of functions
usually are), we first linearize the equation, and then solve the linear equation so that
we get the recursive solutions with appropriate recurrence estimates. The simplest one is
known as Picard’s iterative scheme. However when φ involves the derivatives of u up to
order m, the Picard’s scheme can not be convergent unless the linearized equation gets m
derivatives (as does elliptic equations). To overcome this difficulty, Nash[11] and Moser [10]
proposed another scheme involving smoothing operators so that the solution of the linearized
equation could be estimated inductively in each Sobolev space of order s. Later Hörmander
proposed improved schemes [6,7,8] to get optimal results with respect to the regularity of
the solution. However, these schemes are too complicated and rather frightening for the
uninitiated reader.
In [12], Saint Raymond established a simplified version of C ∞ existence theorem so that
the number of derivatives that are used (provided that it is finite) does not matter, and this
method is useful when we are working on C ∞ category [1,2,3]. Raymond used the scheme
proposed by Moser [10] which consists alternately in using Newton’s scheme and Nash’s
smoothing operators closer and closer to the identity :
vk = −ψ(uk )φ(uk ), uk+1 = uk + Sk vk ,
2000 Mathematics Subject Classification. 47J07.
Key words and phrases. Nash-Moser, Banach spaces, Sobolev spaces.
(†)Partially supported by R01-1999-00001 from KOSEF
Typeset by AMS-TEX
1
2
S. CHO
and proved the convergence in C ∞ category. Here Sk is a smoothing operator and ψ(u) is
the right inverse of (∂φ/∂u)(u), that is,
φ0 (u)ψ(u) = I,
(1)
as an operator (I=identity). Also we need an estimate for v = −ψ(u)φ(u), so called “tame
estimates”;
(2)
|v|s ≤ Cs (kφ(u)ks+d + |u|s+d kφ(u)kd ),
in Hamilton [4].
In some cases, however, we have to deal with the case that a linearized equation has right
inverse with error terms of second order, c.f.,[1,2,3], or, in each recurrence step, we have
to solve a linearized equation with different weights in each weighted Sobolev space. For
example, when we try to prove an embedding problem of a Cauchy-Riemann structure, the
(approximate) linearized equation becomes an inhomogeneous Cauchy-Riemann equation
on compact pseudoconvex almost complex manifolds (close to being integrable). In this
case, we can not get an elliptic regularity for the solution up to the boundary. Therefore,
we have to use weighted estimates for ∂ [5,9] with different weights in each Sobolev space.
We then use these weighted estimates for ∂ (as in (5) below) in each recurrence step in the
process of Nash-Moser iteration.
In this paper we prove the Nash-Moser implicit function theorem in weighted Banach
spaces. We relax the conditions in (1) and (2) as mentioned above. That is, in each
k-th recurrence step, φ(u) + φ0 (u)(v) = 0 has a solution with an error that depends on
(k)
(kφ(u)k3d )1+ε , for ε > 0, where d is a positive integer, and can be estimated as
(3)
E(s)
|v|(k)
s ≤ Cs tk
(k)
(k)
(kφ(u)ks+d + |u|s+d kφ(u)kd ),
(k)
(k)
where {tk } is an increasing sequence of positive integers and the norms | · |s (or k · ks k),
defined in each recurrence spaces, are Sobolev norms of order s with weight e−tk λ , where
λ is a C ∞ function, and E(s) is an integer depending on s. We state the main theorem as
follows.
Theorem 5.1. Let {tk }k≥0 be a strictly increasing sequence of positive integers and suppose
that Bs²,k and Bs²,k , 0 ≤ s, k ∈ Z, 0 < ² ∈ R, are families of Banach spaces with the following
properties.
(i) For each fixed k and ², Bs²,k ⊂ Bt²,k and Bs²,k ⊂ Bt²,k if s > t.
(ii) For each fixed ² > 0, and for each fixed s, Bs²,k1 ⊂ Bs²,k2 and Bs²,k1 ⊂ Bs²,k2 if k2 ≥ k1 .
(k)
(k)
(iii) If | · |s,² and k · ks,² denote the norms on Bs²,k and Bs²,k , respectively, then
(k)
(k)
(k)
| · |(k)
s,² ≥ | · |t,² and k · ks,² ≥ k · kt,² if s > t.
²,k
(iv) For each fixed ² > 0 and k, set B∞
=
²
²,0
exists an open set U0 ⊂ B∞ with
T
s>0
²,k
Bs²,k and B∞
=
(0)
²,0
U0² = {u ∈ B∞
; |u − u²0 |3d,² < δ},
T
s>0
Bs²,k . Then there
NASH-MOSER IMPLICIT FUNCTION THEOREM
3
²,0
where d is a positive integer, u²0 ∈ B∞
, and δ is a given positive number. Further2
²
²,0
²,0
more, there is a C -map φ² : U0 −→ B∞
such that if u ∈ U0² and v, w ∈ B∞
,
then
(0)
−E(s,d)
kφ² (u)k(0)
(1 + |u|s+d,² ), s ≥ d,
s,² ≤ Cs ²
(0)
(0)
(0)
(0)
kφ0² (u)(v)k2d,² ≤ C1 ²−E(d) |v|3d,² ,
(0)
kφ00² (u)(v, w)k2d,² ≤ C2 ²−E(d) |v|3d,² |w|3d,² ,
Here E(s, d) and E(d) are polynomials in s and d.
(v) There is a positive number a > 0 with the following properties : for each fixed ² > 0
²,k
−→
and for each k, and for all u ∈ U0² , there exists a linear operator ψk² (u) : B∞
²,k
Bk+T (a,d)+2d , such that
(5.1)
³
´1+a
(k)
(k)
kφ² (u) − φ0² (u)ψk² (u)φ² (u)k2d,² ≤ C1 kφ² (u)k3d,²
,
and vk² := −ψk² (u)φ² (u) satisfies, for each s, d ≤ s ≤ k + T (a, d) + 2d, the following
estimates (so called “tame estimates”) :
(5.2)
E (s)
−E1 (s) 2
|vk² |(k)
tk
s,² ≤ Cs ²
(k)
(0)
(k)
(kφ² (u)ks+d,² + |u|s+d,² kφ² (u)kd,² ),
where E1 (s), E2 (s) are polynomials in s, Cs is a constant, and T (a, d) is an integer,
for example, the smallest integer bigger than or equal to 3d + 3 + 120(d+1)
(2d + 3).
S∞ ²,ka
²,k
and
(vi) For each ² > 0 and k, there are smoothing operators Sθ : s=0 Bs → B∞
S∞
²,k
²,k
S̃θ : s=0 Bs → B∞ for all θ > 1 such that for each real number s, t, there are a
constant Cs,t and integers E1 (s, t) and E2 (s, t) such that
(5.3)
E (s,t) s−t
|v|t,² , s ≥ t,
E (s,t) s−t
|v|²,t , s ≤ t,
−E1 (s,t) 2
tk
|Sθ v|(k)
s,² ≤ Cs,t ²
−E1 (s,t) 2
|v − Sθ v|(k)
tk
s,² ≤ Cs,t ²
θ
θ
(k)
(k)
and the similar estimates hold for S̃θ .
(0)
Then there exist an integer D and a small number b > 0 such that if kφ² (u0 )kD < b, for
some ² > 0 and for some u²0 ∈ U0² , then there exists an element u ∈ U0² such that φ² (u) = 0.
Remark 5.2. (a). In many cases, we approximate the non-linear problem up to second order
errors terms. Hence a = 1 in these cases.
(b). If λ is a smooth bounded function, we can modify λ so that 1 ≤ λ ≤ 1 + a4 . Let Bsk , Bsk
be the weighted Sobolev spaces of order s on a bounded domain Ω ⊂ Cn with weighted
norm
X Z
(k) 2
(5.4)
(kf ks ) =
|Dα f |2 e−tk λ dV, f ∈ Bsk ,
|α|≤s
then (5.3) holds with θ0 = e.
Ω
4
S. CHO
Theorem 1. Let {tk }k≥0 be a strictly increasing sequence of positive integers and suppose
that Bsk and Bsk , s, k ≥ 0, are families of Banach spaces with the following properties.
(i) For each fixed k, Bsk ⊂ Btk and Bsk ⊂ Btk if s > t.
(k)
(k)
(ii) If | · |s and k · ks denote the norms on Bsk and Bsk , respectively, then
(k)
| · |(k)
s ≥ | · |t
k
(iii) For each fixed k, if B∞
=
k
set Uk ⊂ B∞ with
(k)
and k · k(k)
s ≥ k · kt
T
s>0
if s > t.
T
k
Bsk and B∞
=
s>0
Bsk , then there exists an open
(k)
k
; |u − u0 |3d < δ},
Uk = {u ∈ B∞
0
, and δ is a given positive number. Furwhere d is a positive integer, u0 ∈ B∞
2
k
k
thermore, there is a C -map φ : Uk −→ B∞
such that if u ∈ Uk and v, w ∈ B∞
,
then
(k)
kφ(u)k(k)
s ≤ Cs (1 + |u|s+d ), s ≥ d
(k)
(k)
(k)
(k)
kφ0 (u)(v)k2d ≤ C1 |v|3d ,
(k)
kφ00 (u)(v, w)k2d ≤ C2 |v|3d |w|3d ,
(when one deals with (nonlinear) partial differential equations of order m, these
estimates classically hold for d > m + n/2).
(iv) There is a positive number ε > 0 with the following properties : for each k, and for
k
k
−→ Bk+T
all u ∈ Uk , there exists a linear operator ψk (u) : B∞
(²,d)+2d , such that
(4)
(k)
(k)
kφ(u) − φ0 (u)ψk (u)φ(u)k2d ≤ C1 (kφ(u)k3d )1+ε ,
and vk := −ψk (u)φ(u) satisfies, for each s, d ≤ s ≤ k + T (², d) + 2d, the following
estimates (so called “tame estimate”) :
E(s)
|vk |(k)
s ≤ Cs tk
(5)
(k)
(k)
(k)
(kφ(u)ks+d + |u|s+d kφ(u)kd ),
where E(s) is a polynomial in s, Cs is a constant, and T (², d) is an integer, for
example, the smallest integer bigger than or equal to 3d + 3 + 120(d+1)
(2d + 3).
²
(v) There is θ0 > 1 such that
(6)
(1+ 4² )(tk −tk+1 )
(k+1)
| · |(k)
≤ θ0k
s ≤ | · |s
(1+ 4² )(tk −tk+1 )
(k+1)
k · k(k)
≤ θ0k
s ≤ k · ks
θ0
θ0
t −tk+1
| · |(k)
s ,
t −tk+1
k · k(k)
s ,
where ² > 0 is the number in (iv) satisfying (4), and hence Bsk ⊂ Bsk+1 , Bsk ⊂ Bsk+1 ,
for each s ≥ 0.
NASH-MOSER IMPLICIT FUNCTION THEOREM
5
S∞
S∞
k
(vi) For each k, there are smoothing operators Sθ : s=0 Bsk → B∞
and S̃θ : s=0 Bsk →
k
B∞
for all θ > 1 such that for each real number s, t, there are a constant Cs,t and
an integer E(s, t) such that
E(s,t) s−t
|v|t , s ≥ t,
E(s,t) s−t
|v|t , s ≤ t,
|Sθ v|(k)
s ≤ Cs,t tk
|v − Sθ v|(k)
s ≤ Cs,t tk
θ
θ
(k)
(k)
and the similar estimates hold for S̃θ .
(0)
Then there exist an integer B and a small number b > 0 such that if kφ(u0 )kB < b,
for some u0 ∈ U0 , then there exists an element u ∈ U0 such that φ(u) = 0.
(k)
Remark 2. (a). Since kφ(u)k3d ≤ 1, we may assume that 0 < ² ≤ 1.
(b). In many cases, we approximate the non-linear problem up to second order error terms,
and hence ² = 1 in these cases.
(c). If λ is a smooth bounded function, we can modify λ so that 1 ≤ λ ≤ 1 + 4² . Let Bsk be
the weighted Sobolev space of order s on a bounded domain Ω ⊂ Cn with weighted norm :
2
(kf k(k)
s )
=
X Z
|α|≤s
Ω
|Dα f |2 e−tk λ dV, f ∈ Bsk .
Then (6) holds with θ0 = e.
By choosing a subsequence if necessary, we may assume that the sequence {tk }k≥0 satisfies
(7)
tk+1 ≥
3
tk , k ≥ 0,
2
and t0 is sufficiently large. Also, we will use a sequence of real numbers {θk } defined
inductively as follows :
θk = θ0tk , k ≥ 1,
(8)
and will use the corresponding smoothing operators Sθk . In the sequel, we set
τ0 = (1 + 2²/3) > 1,
and τk = 1 +
(10 − k)
², k = 1, 2, . . . , 5,
15
and hence 1 < τ5 < τ4 < τ3 < τ2 < τ1 < τ0 . For a convenience, we set
(9)
T := T (², d) := [[3d + 3 +
120(d + 1)
(2d + 3)]],
²
where [[Γ]] denotes the smallest integer bigger than or equal to Γ. We first prove the following
Lemma which is a crucial step in the proof of Theorem 1.
6
S. CHO
Lemma 3. With the same assumption as in the theorem and with the smoothing operators
Sθk of the remark, the sequences
vk = −ψk (uk )φ(uk ), uk+1 = uk + Sθk+1 vk ,
(0)
are well defined if kφ(u0 )k2d ≤ θ0−2t0 for sufficiently large t0 ; more precisely, there exist
constants (Ut )t≥d , and V (independent of k) such that for k ≥ 0,
(i)k
(ii)k
(iii)k
(k)
(k)
|uk − u0 |3d < δ and kφ(uk )k2d ≤ θk−τ0 ,
(k)
|vk |3d+3 ≤ V θk−τ3 ,
(k+1)
2d+1/2
(1 + |uk+1 |s+2d ) ≤ Us θk+1
(k)
(1 + |uk |s+2d ), d ≤ s ≤ k + T + 2d.
Proof. Since the property (i)k implies that the sequence uk and vk are well defined, it is
sufficient to prove (i)k , (ii)k and (iii)k inductively. The property (i)0 is true by assumption.
Proof of (ii)k+1 . The tame estimate (5) gives, for every s, d ≤ s ≤ k + T + 2d, that
³
´
E(s)
(k)
(k)
(k)
(10)
|vk |(k)
≤
C
t
kφ(u
)k
+
|u
|
kφ(u
)k
.
s
k
k
k
s
k
s+d
s+d
2d
For s = d, and using (i)k and (10), we have
³
´
(k)
E(d)
(k)
(k)
(k)
(11)
|vk |d ≤ Cd tk
1 + |uk − u0 |2d + |u0 |2d kφ(uk )k2d ≤ V0 θk−τ1 ,
E(d)
because tk θkτ1 −τ0 is bounded.
Let T be the number defined in (9) and set N = 4(2d+1). From the tame estimate (5) and
(k)
the properties of kφ(uk )ks stated in (iii) of Theorem 1, it follows, for d ≤ s ≤ k + T + 2d,
that
³
´
E(s)
(k)
(k)
(k)
|vk |(k)
≤
C
t
kφ(u
)k
+
|u
|
kφ(u
)k
s
k
k
k
s
k
s+d
s+d
d
³
´
E(s)
(k)
(k)
(k)
(12)
≤ Cs t k
Cs+d (1 + |uk |s+2d ) + Cd (1 + |uk |2d )|uk |s+d
³
´
E(s)
(0)
(k)
≤ Cs t k
Cs+d + C2d (1 + δ + |u0 |3d ) · (1 + |uk |s+2d ).
The estimate
(13)
(j)
N − 41
(0)
(1 + |uj |T +2d ) ≤ (1 + |u0 |T +2d )θj
holds obviously for j = 0. Moreover, if it holds for some j < k, we obtain from (7), (13)
and (iii)j that
(j+1)
2d+1/2
(1 + |uj+1 |T +2d ) ≤ UT θj+1
(j)
(1 + |uj |T +2d )
−1/4
(0)
−1/4
(0)
2d+1− 41
≤ UT θj+1 (1 + |u0 |T +2d )θjN θj+1
N− 1
≤ UT θj+1 (1 + |u0 |T +2d )θj+1 4 ,
NASH-MOSER IMPLICIT FUNCTION THEOREM
7
2N/3
because θjN ≤ θj+1 and (2d + 1) + 2N/3 < N . Therefore, by induction for j ≤ k that, (13)
−t1 /4
holds provided t0 (and hence t1 ) is sufficiently large so that θ0
Thanks to (13), we may write (12) as :
(14)
E(s)
(k)
³
|vk |T ≤ CT tk
UT ≤ 1.
´
N− 1
(0)
(0)
CT +d + C2d (1 + δ + |u0 |3d ) · (1 + |u0 |T +2d )θk 4 ≤ V1 θkN ,
E(s) −1/4
because tk θk
is bounded, for d ≤ s ≤ k + T .
Combining (9), (11), (14) and the properties (vi) in Theorem 1 of the smoothing operaτ1 −τ2
tors, the interpolation formula, with θ̄k = θk(2d+3) , can be written as
(15)
(k)
(k)
(k)
|vk |3d+3 ≤ |Sθ̄k vk |3d+3 + |vk − Sθ̄k vk |3d+3
E(3d+3,d) 2d+3
(k)
E(3d+3,T ) 3d+3−T
(k)
θ̄k |vk |d + C3d+3,T tk
θ̄k
|vk |T
E(3d+3,d)+E(d) −τ2
E(3d+3,T ) −4(2d+2)+N
C3d+3,d V0 tk
θk + C3d+3,T V1 tk
θk
V θk−τ3 ,
≤ C3d+3,d tk
≤
≤
E(3d+3,T ) −4(2d+2)+N +τ
E(3d+3,d)+E(d)
3
because tk
θk
and tk
θk−τ2 +τ3 are bounded. This proves
(ii)k+1 .
(k+1)
(k)
Proof of (iii) Now we want to estimate |uk+1 |s+2d in terms of |vk |s . Since uk+1 =
uk + Sθk+1 vk , it follows, for d ≤ s ≤ k + T , that
(k+1)
(k+1)
(k+1)
(k)
E(s+2d,s) 2d
θk+1 |vk |(k)
s .
|uk+1 |s+2d ≤ |uk |s+2d + |Sθk+1 vk |s+2d ≤ |uk |s+2d + Cs+2d,s tk
Thus one obtains from (13) that
(k+1)
E(s+2d,s)+E(s) 2d
θk+1 (1
1 + |uk+1 |s+2d ≤ Ws tk
2d+1/2
≤ Us θk+1
(k)
+ |uk |s+2d )
(k)
(1 + |uk |s+2d ),
E(s+2d,s)+E(s) − 1
2
because tk
θk+1
, d ≤ s ≤ k + T + 2d, is bounded. This proves (iii)k with
constants Us does not depend on k. P
Proof of (i)k+1 . Since uk − u0 = j<k Sθj+1 vj , ∀t ∈ [0, 1], one can write, from (6) and
(ii)k , that
(k+1)
|uk + tSθk+1 vk − u0 |3d
≤
X
(k+1)
|Sθj+1 vj |3d
j≤k
t −tk+1 E(3d,3d)
tk
≤ C3d,3d θ0k
X
(j)
|vj |3d
j≤k
≤ C3d,3d V
−t
/3 E(3d,3d)
θ0 k+1 tk
X
j≤k
−tk+1 /4
≤ C3d,3d V Sθ0
,
θj−τ3
8
S. CHO
so that
P
−τ3
<∞
j≤∞ θj
−t1 /4
C3d,3d V Sθ0
< δ,
where S =
is a constant. By choosing t0 (and hence t1 ) sufficiently large
we have
(k+1)
|uk+1 − u0 |3d
=|
X
(k+1)
Sθj+1 vj |3d
< δ,
j<k+1
and this is the first part of (i)k+1 .
By virtue of Taylor’s formula, we can write ;
φ(uk+1 ) = φ(uk ) + φ0 (uk )Sθk+1 vk
Z 1
+
(1 − t)φ00 (uk + tSθk+1 vk )(Sθk+1 vk , Sθk+1 vk )dt
0
= φ1 + φ2 + φ3 ,
where
φ1 = φ(uk ) + φ0 (uk )vk
φ2 = φ0 (uk )(Sθk+1 vk − vk )
Z 1
φ3 =
(1 − t)φ00 (uk + tSθk+1 vk )(Sθk+1 vk , Sθk+1 vk )dt.
0
First, we estimate φ1 . For this, we use the following two estimates ;
(k)
(16)
−(1+2ε/3)
kφ(uk )k2d ≤ θk
= θk−τ0
(k)
and kφ(uk )kT +d ≤ AθkN .
Note that the second inequality comes from the properties (iii) in Theorem 1 and (14) with
(k+1)
(k+1)
(k+1)
s = T +d. From (4), we have kφ1 k2d
= kφ(uk )+φ0 (uk )(vk )k2d
≤ C1 (kφ(uk )k3d )1+ε .
2ε
Setting θ̄k = θk3d , it follows from (16) that
(k)
(k)
(k)
kφ(uk )k3d = kS̃θ̄k φ(uk )k3d + kφ(uk ) − S̃θ̄k φ(uk )k3d
E(3d,2d) d
(k)
θ̄k kφ(uk )k2d
≤ C3d,2d tk
E(3d,T +d) 3d−T −d
(k)
θ̄k
kφ(uk )kT +d
+ C3d,T +d tk
E(3d,2d) −1
E(3d,T +d)
θk + C3d,T +d tk
Aθk−1
E(3d,2d)
E(3d,T +d)
(C3d,2d tk
+ C3d,T +d tk
A)θk−1 .
= C3d,2d tk
=
Hence it follows from (6) that
(k+1)
kφ1 k2d
(k+1) 1+ε
≤ C1 (kφ(uk )k3d
(1+ε)(t −t
)
)
(k)
k
k+1
= C1 θ0
(kφ(uk )k3d )1+ε
³
´1+ε
(1+ε)(tk −tk+1 ) −(1+ε)
E(3d,2d)
E(3d,T +d)
≤ C1 C3d,2d tk
+ C3d,T +d tk
A
θ0
θk
³
´1+ε
−(t
)(1+ε)
E(3d,2d)
E(3d,T +d)
≤ C1 C3d,2d tk
+ C3d,T +d tk
A
θ0 k+1
≤
1 −(1+2ε/3)
θ
,
3 k+1
NASH-MOSER IMPLICIT FUNCTION THEOREM
9
by choosing t0 sufficiently large.
Next, let us estimate φ2 . From (6), (15) and the property (iii) of Theorem 1, we obtain
that
(k+1)
kφ2 k2d
t −tk+1
≤ θ0k
(k)
C1 |Sθk+1 − vk |3d
t −t
E(3d,3d+3) −3
(k)
θk+1 |vk |3d+3 θ0k k+1
E(3d,3d+3) −3 −τ3 tk −tk+1
θk+1 θk θ0
C1 C3d,3d+3 V tk
≤ C1 C3d,3d+3 tk
≤
≤
1 −(1+2ε/3)
θ
,
3 k+1
provided t0 is sufficiently large.
Finally, we estimate φ3 . By choosing t0 sufficiently large, we have, from (15), that
(k+1)
kφ3 k2d
2(t −t
(k+1)
)
(k)
≤ (|Sθk+1 vk |3d )2 ≤ θ0 k k+1 (|Sθk+1 vk |3d )2
³
´2
2(t −t
)
E(3d,3d)
(k)
≤ C3d,3d tk
|vk |3d
θ0 k k+1
³
´2
2(t −t
)
E(3d,3d)
≤ C3d,3d V tk
θ0 k k+1 θk−2τ3
≤
1 −(1+2ε/3)
θ
.
3 k+1
If we combine the estimates of φ1 , φ2 and φ3 , the second part of (i)k+1 follows. ¤
We define a constant M =
set
τ4 +N +1
τ3 −τ4
=
(17)
Λ(k) =
k + T + (2 + M )d − 1
.
M +1
15 6
² ( 15 ² + 4(2d + 1) + 1),
and for each integer k ≥ 0,
(k)
The proof of the property (ii)k of Lemma 3 can be modified to prove an estimate for |vk |s
for every s ≥ d.
Lemma 4. There exist constants (Vs )s≥d such that the sequence {vk } of Lemma 3 satisfies,
for every k ≥ 0 and d ≤ s ≤ Λ(k), that
max(E(s,d),E(s,D)) −τ4
θk ,
|vk |(k)
s ≤ Vs t k
where D = [[ (τ(s−d)
(τ4 + N + 1) + s]].
3 −τ4 )
Proof. Keeping the value N = 4(2d + 1), we obtain from (7) and (iii)k of Lemma 3 that
(k+1)
2d+1/2
−N
(1 + |uk+1 |s+2d )θk+1
≤ Us θk+1
(k)
−N
(1 + |uk |s+2d )θk+1
−1/2
(k)
−1/2
(k)
−N +2d+1
≤ Us θk+1 (1 + |uk |s+2d )θk+1
≤ Us θk+1 (1 + |uk |s+2d )θk−N ,
10
S. CHO
(k)
for d ≤ s ≤ k + T + 2d. Therefore for each fixed s, the sequence (1 + |uk |s+2d )θk−N is
bounded, and hence there exists a constant Ks such that
(k)
(1 + |uk |s+2d )θk−N ≤ Ks .
Substituting this into (13), we obtain, for d ≤ s ≤ k + T + 2d, that
(18)
(0)
E(s) N
θk
|vk |(k)
s ≤ Cs (Cs+d + C2d (1 + δ + |u0 |3d ))Ks tk
E(s) N
θk
= Ws0 tk
≤ Ws θkN +1 ,
where N = 4(2d + 1) does not depend on s. Now, if d ≤ s ≤ Λ(k), the definition of Λ(k) in
(17) shows that D ≤ k + T + 2d. Therefore for d ≤ s ≤ Λ(k), we rewrite our interpolation
(τ3 −τ4 )
formula with θ̄k = θk (s−d) ;
(k)
(k)
|vk |(k)
s ≤ |Sθ̄k vk |s + |vk − Sθ̄k vk |s
E(s,d) s−d
(k)
E(s,D) s−D
(k)
θ̄k |vk |d + Cs,D tk
θ̄k |vk |D
E(s,d)
E(s,D)
Cs,d tk
V θk−τ4 + Cs,D tk
WD θk−τ4
max(E(s,d),E(s,D)) −τ4
Vs t k
θk ,
≤ Cs,d tk
≤
≤
where we have used the estimate (18).
¤
Proof of the Theorem 1 Let uk and vk be as in Lemma 3. From lemma 4 we have,
for any j ≥ 0 and d ≤ s ≤ Λ(j), that
E(s,s)
|Sθj+1 vj |(j)
s ≤ Cs,s tj
E(s,s)+max(E(s,d),E(s,D)) −τ4
θj
|vj |(j)
s ≤ Cs,s Vs tj
≤ As θj−τ5 .
By (6) one thus obtains, for d ≤ s ≤ Λ(j), that
(1+²/4)(tj −t0 )
|Sθj+1 vj |(0)
s ≤ θ0
(19)
|Sθj+1 vj |(j)
s
(1+²/4)tj −τ5
θj
≤ Cs,s As θ0
−²/12
= Cs,s As θj
because (1 + ²/4) − τ5 = −²/12. By virtue of (17), we also have
Λ(j) < s if and only if j < Λ−1 (s) = s(M + 1) − (2 + M )d − T + 1.
P
Now for each fixed s ≥ 0, the sequence uk = u0 + j<k Sθj+1 vj is convergent in Bs0
because
X
X
X
|
Sθj+1 vj |(0)
Sθj+1 vj |(0)
Sθj+1 vj |(0)
s ≤|
s +|
s < ∞,
(20)
j<Λ−1 (s)
j≥Λ−1 (s)
where the first sum in the right is finite sum by (20), and the second sum in the right is
0
finite by (19). Moreover, the limit u ∈ B∞
of the sequence uk satisfies
Z 1
(0)
(0)
(0)
kφ(u)k2d ≤ kφ(uk )k2d + k
φ0 (uk + t(u − uk ))(u − uk )dtk2d
0
≤
(0)
kφ(uk )k2d
(0)
+ C1 |u − uk |3d ,
and by (6), one obtains that
(0)
(1+²/4)(tk −t0 )
kφ(uk )k2d ≤ θ0
(k)
−5²/12
kφ(uk )k2d ≤ θk
,
for all k. Therefore, by taking limit for k = ∞, it follows that φ(u) = 0 and this proves
Theorem 1 with B = 2d and b = θ0−2t0 .
NASH-MOSER IMPLICIT FUNCTION THEOREM
11
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Department of Mathematics, Sogang University, Seoul, 121-742 Korea
E-mail address: (†):[email protected]