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( : 97)



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1
¤ C( x
f (x) = f (0) + f (tx)x dt + x x r ).
n n+r n+2r
2
n
0


¤x ¤x ¤ 1 we are done.
Since x and x
n+r n+2r r 2r


Proof of (51.18). The idea comes from the 1-dimensional situation, where f (x) =
f (y) implies by the mean value theorem that there exists an r ∈ [x, y] := {tx + (1 ’
t)y : 0 ¤ t ¤ 1} with f (r) = f (x)’f (y) = 0.
x’y

51.20. Sublemma. There exists a δ > 0 such that for xj 2r ¤ δ we have
x1 ’ x0 0 ¤ C f (x1 ) ’ f (x0 ) 0 . In particular, we have that f is injective on
{x : x 2r ¤ δ}.

Proof. Using the Taylor formula
1
(1 ’ t)f (x0 + t(x1 ’ x0 ))(x1 ’ x0 )2 dt
f (x1 ) = f (x0 ) + f (x0 )(x1 ’ x0 ) +
0


and Ψ(x0 ) —¦ f (x0 ) = Id, we obtain that x1 ’ x0 = Ψ(x0 )(k), where

1
(1 ’ t)f (x0 + t(x1 ’ x0 )) (x1 ’ x0 )2 dt.
k := f (x1 ) ’ f (x0 ) ’
0

51.20
51.21 51. The Nash-Moser inverse function theorem 565

¤ 1 we can use the tame estimates of f and interpolation to get
For xj 2r


f (x0 + t(x1 ’ x0 ))(x1 ’ x0 )2 ¤
n
2
¤C x1 ’ x0 x1 ’ x0 x1 ’ x0
+ ( x1 + x0 n+2r )
n+r r n+2r r

¤C x1 ’ x0 x1 ’ x0 x1 ’ x0 x1 ’ x0
+ ( x1 + x0 n+2r )
n+2r 0 n+2r 2r 0

¤ C ( x1 x1 ’ x0 x1 ’ x0
+ x0 n+2r ) + ( x1 + x0 n+2r )2δ
n+2r 0 n+2r 0

¤ C( x1 x1 ’ x0 0 .
+ x0 n+2r )
n+2r


Using the tame estimate

¤ C k 0 (1 + x0 ¤ C k 0,
Ψ(x0 )k 2r )
0


we thus get

x1 ’ x0 ¤C k ¤
= Ψ(x0 )k
0 0 0

+ 1 C( x1
¤C f (x1 ) ’ f (x0 ) x1 ’ x0
+ x0 2r )
0 2r 0
2
+ x1 ’ x0 2 )
¤ C ( f (x1 ) ’ f (x0 ) 0 r
¤ C ( f (x1 ) ’ f (x0 ) + x1 ’ x0 · x1 ’ x0 0 ).
0 2r


Now use x1 ’ x0 ¤ x1 ¤ 2δ to obtain
+ x0
2r 2r 2r


x1 ’ x0 ¤ C( f (x1 ) ’ f (x0 ) + 2δ x1 ’ x0 0 ).
0 0

1
Taking δ < yields the result.
2C


¤ δ with δ as before. Then for n ≥ 0 we have
51.21. Corollary. Let xj 2r


x1 ’ x0 ¤C f (x1 ) ’ f (x0 ) f (x1 ) ’ f (x0 )
+ ( x1 + x0 n+2r ) .
n n n+2r 0



Proof. As before we have

f (x0 + t(x1 ’ x0 ))(x1 ’ x0 )2 ¤ C( x1 x1 ’ x0 0 .
+ x0 n+2r )
n n+2r


Since Ψ is tame we obtain now

x1 ’ x0 = Ψ(x0 ) f (x1 ) ’ f (x0 )
n
1
(1 ’ t)f (x0 + t(x1 ’ x0 ))(x1 ’ x0 )2

n
0

¤ Ψ(x0 ) f (x1 ) ’ f (x0 ) +
n
1
(1 ’ t)f (x0 + t(x1 ’ x0 ))(x1 ’ x0 )2
+ Ψ(x0 )
n
0

51.21
566 Chapter X. Further Applications 51.21


¤C f (x1 ) ’ f (x0 ) · f (x1 ) ’ f (x0 )
+ x0 +
n n+2r 0
1
(1 ’ t)f (x0 + t(x1 ’ x0 ))(x1 ’ x0 )2
+C +
n
0
1
(1 ’ t)f (x0 + t(x1 ’ x0 ))(x1 ’ x0 )2
·
+ x0 n+2r
0
0

¤C f (x1 ) ’ f (x0 ) f (x1 ) ’ f (x0 )
+ ( x1 + x0 n+2r )
n n+2r 0

x1 ’ x0
+ ( x1 + x0 n+2r )
n+2r 0

¤C f (x1 )’f (x0 ) 0


· ( x1 x1 ’ x0
+ ( x1 + x0 n+2r ) + x0 2r )
n+2r 2r 0

¤2δ C f (x1 )’f (x0 ) 0


¤C f (x1 ) ’ f (x0 ) f (x1 ) ’ f (x0 )
+ ( x1 + x0 n+2r ) .
n n+2r 0




Proof of (51.19). As in (51.18) we may assume that the initial condition is f :
0 ’ 0 and that E = Σ(B) and F = Σ(C).
The idea of the proof is to solve the equation f (x) = y via a di¬erential equation
for a curve t ’ x(t) whose image under f joins 0 and y a¬nely. More precisely we
consider the parameterization t ’ h(t) y of the segment [0, y], where h(t) := 1’e’ct
is a smooth increasing function with h(0) = 0 and limt’+∞ h(t) = 1. Di¬erentiation
of f (x(t)) = h(t) y yields f (x(t)) · x (t) = h (t) y and (if f (x) is invertible) that
x (t) = c Ψ(x(t)) · e’ct y. Substituting e’ct y = (1 ’ h(t)) y = y ’ f (x(t)) gives

x (t) = c Ψ(x(t)) · (y ’ f (x(t))).

In Fr´chet spaces (like Σ(B)) we cannot guarantee that this di¬erential equation
e
with initial condition x(0) = 0 has a solution. The subspaces Bt := {(xk )k ∈
Σ(B) : xk = 0 for k > t} however are Banach spaces (isomorphic to ¬nite products
of B), and they are direct summands with the obvious projections. So the idea is
to modify the di¬erential equation in such a way that for ¬nite t it factors over Bt
and to prove that the solution of the modi¬ed equation still converges for t ’ ∞
to a solution x∞ of f (x∞ ) = y. Since t is a non-discrete parameter we have to
consider the spaces Bt as a continuous family of Banach spaces, and so we have to
¬nd a family (σt )t∈R of projections (called smoothing operators). For this we take
a smooth function σ : R ’ [0, 1] with σ(t) = 0 for t ¤ 0 and σ(t) = 1 for t ≥ 1.
Then we set σt (x)(k) := σ(t ’ k) · x(k).
We have to show that σt ’ Id, more precisely we

¤ cn,m e(n’m)t x
Claim. For n ≥ m there exists a cn,m such that σt x and
n m
(1 ’ σt )x m ¤ cn,m e(m’n)t x n .
nk
xk . Since (σt x)k ¤
Recall that x n := ke xk for all t and k and

51.21
51.22 51. The Nash-Moser inverse function theorem 567

(σt x)k = 0 for t ¤ k and ((1 ’ σt )x)k = 0 for t ≥ k + 1 we have

enk (σt x)k
σt x =
n
k

enk xk ¤ e(n’m)k emk xk ¤ e(n’m)t x
¤ m,
k¤t k¤t

emk xk ¤ e(m’n)k enk xk ¤ en’m e(m’n)t x k .
(1 ’ σt )x ¤
m
k≥t’1 k≥t’1



Now we modify our di¬erential equation by projecting the arguments of the de¬ning
function to Bt , i.e.

x (t) = c Ψ(σt (x(t))) · (σt (y ’ f (x(t)))) with x(0) = 0.

Thus, our modi¬ed di¬erential equation factors for ¬nite t over some Banach space.
The following sublemma now provides us with local solutions

51.22. Sublemma. If a function f : F ⊇ U ’ F factors via smooth maps over a
Banach space E “ i.e., f = g —¦ h, where h : F ⊇ U ’ W ⊆ E and g : E ⊇ W ’ F
are smooth maps “ then the di¬erential equation y (t) = f (y(t)) has locally unique
solutions depending continuously (smoothly) on the initial condition y0 ∈ U .


RR f j w F
j
h h
F
yh R hh
Thg
R
h h
h
wE
x
R
Proof. Suppose y is a solution of the di¬erential equation y = f —¦ y with initial
t
condition y(0) = y0 , or equivalently y(t) = y0 + 0 f (y(s)) ds. The idea is to consider
the curve x := h —¦ y in the Banach space E. Thus,
t t
x(t) = h y0 + g(h(y(s))) ds = h y0 + g(x(s)) ds .
0 0

Now conversely, if x is a solution of this integral equation, then t ’ y(t) := y0 +
t
g(x(s)) ds is a solution of the original integral equation and hence also of the
0
t
di¬erential equation, since x(t) = h(y0 + 0 g(x(s)) ds) = h(y(t)), and so y(t) =
t t t
y0 + 0 g(x(s)) ds = y0 + 0 g(h(y(s))) ds = y0 + 0 f (y(s)) ds.
In order to show that x exists, we consider the map
t
k:x’ t ’ h y0 + g(x(s)) ds
0

and show that it is a contraction.
Since h is smooth we can ¬nd a seminorm on F , a C > 0 and an · > 0 such
q
that
h(y1 ) ’ h(y0 ) ¤ C y1 ’ y0 ¤ ·.
for all yj
q q


51.22
568 Chapter X. Further Applications 51.22

Furthermore, since g is smooth we ¬nd a constant C > 0 and θ > 0 such that

g(x1 ) ’ g(x0 ) ¤ C x1 ’ x0 for all xj ¤ θ.
q


¤ C and that θ ¤ 1. So we
Since we may assume that h(0) = 0, that g(0) q
obtain

h(y) ¤ C y ¤ · and g(x) ¤ 2C for all x ¤ θ.
for all y
q q q


˜ ˜
Let U := {y0 ∈ F : y0 q ¤ δ}, let V := {x ∈ C([0, µ], E) : x(t) ¤ θ for all t},
˜˜
and let k : F — C([0, µ], E) ⊇ U — V ’ C([0, µ], E) be given by

t
k(y0 , x)(t) := h y0 + g(x(s)) ds .
0

˜
Then k is continuous with values in V and is a C 2 µ-contraction with respect to x.
t
In fact, y0 + 0 g(x(s)) ds q ¤ y0 q + µ sup{ g(x(s)) q : s} ¤ δ + 2C µ ¤ · for
su¬ciently small δ and µ. So k(y0 , x)(t) ¤ C · ¤ θ for su¬ciently small ·. Hence,
˜
k(y0 , x) ∈ V . Furthermore,

t
k(y0 , x1 )(t) ’ k(y0 , x0 )(t) ¤ C g(x1 (s)) ’ g(x0 (s)) ds
0 q
¤ C µ sup{ g(x1 (s)) ’ g(x0 (s)) : s}
q
¤ C µ C sup{ x1 (s) ’ x0 (s) : s}.

Thus, by Banach™s ¬xed point theorem, we have a unique solution x of x = k(y0 , x)
˜
for every y0 ∈ U and x depends continuously on y0 . As said before, it follows that
t
y := y0 + 0 g(x(s)) ds solves the original di¬erential equation.

Let k(t) := y ’ f (x(t)) be the error we make at time t. For our original di¬erential
equation the error was k(t) = e’ct y, and hence k was the solution of the homo-
geneous linear di¬erential equation k (t) = ’c k(t). In the modi¬ed situation we
have

Claim. If x(t) is a solution, k(t) := y ’ f (x(t)), and g(t) is de¬ned by (f (σt · xt ) ’
f (xt )) x (t) then
k (t) + c σt · k(t) = g(t).

From x (t) = c Ψ(σt (x(t))) · (σt (y ’ f (x(t)))) = c Ψ(σt (x(t))) · (σt k(t)) we conclude
d
that f (σt xt ) · x (t) = c σt kt , and by the chain rule dt f (x(t)) = f (xt ) · x (t) we get

k (t) + cσt k(t) = (f (σt x(t)) ’ f (x(t))) · x (t) =: g(t).


In order to estimate that error k, we now consider a general inhomogeneous linear
di¬erential equation:

51.22
51.22 51. The Nash-Moser inverse function theorem 569

Sublemma. If the solution k of the di¬erential equation k (t) + c σt k(t) = g(t)
exists on [0, T ] then for all p ≥ 0 and 0 < q < c we have

T T
qt
eqt g(t)
dt ¤ C k(0)
e k(t) +C + g(t) dt
p p+q p p+q
0 0



Proof. If k is a solution of the equation above then the coordinates kj are solutions
of the ordinary inhomogeneous linear di¬erential equation

kj (t) + cσ(t ’ j)kj (t) = gj (t).

The solution kj of that equation can be obtained as usual by solving ¬rst the ho-
mogeneous equation via separation of variables and applying then the method of
variation of the constant. For this recall that for a general 1-dimensional inhomo-
geneous linear di¬erential equation k (t) + σ(t)k(t) = g(t) of order 1, one integrates
t
the homogeneous equation dkh(t) = ’σ(t)dt, i.e., log(kh (t)) = C ’ 0 σ(„ )d„ or
(t)
kh
t
kh (t) is a multiple of e’ 0 σ(„ )d„ . For vector valued equations this is no longer
t
a solution, since σ(t) does not commute with 0 σ(„ )d„ in general, and hence
t t
d ’ 0 σ(„ )d„
need not be ’σ(t) · e’ 0 σ(„ )d„ . But in our case, where σ is just
e
dt
a multiplication operator it is still true. For the inhomogeneous equation one
makes the ansatz k(t) := kh (t) C(t), which is a solution of the inhomogeneous
equation g(t) = k (t) + σ(t)k(t) = kh (t) C(t) + kh (t) C (t) + σ(t) kh (t) C(t) =
’σ(t) kh (t) C(t) + kh (t) C (t) + σ(t) kh (t) C(t) = kh (t) C (t) if and only if C (t) =
t
t ρ
kh (t)’1 g(t) = e 0 σ(„ )d„ g(t) or C(t) = C(0) + 0 e 0 σ(„ ) g(t)d„ dρ. So the solution
of the inhomogeneous equation is given by
t
t ρ
’ σ(„ )d„ σ(„ )d„
k(t) = kh (t) C(t) = e C(0) + g(ρ) e dρ .
0 0

0

<< . .

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( : 97)



. . >>