¤ 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