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t
In particular we have k(0) = C(0), and using a(t, s) := e’ s σ(„ ) d„ we get k(t) =
t t
a(0, t)k(0) + 0 a(ρ, t)g(ρ) dρ. We set aj,s,t := exp(’c s σ(„ ’ j)d„ ). Then

t
kj (t) = aj,0,t kj (0) + aj,„,t gj („ )d„.
0


We claim that ect aj,s,t ¤ ec (ecs + ecj ) for 0 ¤ s ¤ t ¤ T .
For t ¤ j + 1 this follows from aj,s,t ¤ 1. Let now t > j + 1. If s ≥ j + 1 then
σ(„ ’ j) = 1 for „ ≥ s, and so aj,s,t = e’c(t’s) and ect aj,s,t = ecs . If otherwise
s ¤ j + 1 then
t
σ(„ ’ j) d„ = e’c(t’j’1) .
aj,s,t ¤ exp ’c
j+1


So ect aj,s,t ¤ ec(j+1) = ec ecj , and the claim holds.

51.22
570 Chapter X. Further Applications 51.22


Next we claim that for 0 < q < c we have s eqt aj,s,t dt ¤ C(eqj + eqs ).
For j ¤ s ≥ t we have aj,s,t ¤ Cec(s’t) by the previous claim and

∞ ∞ (q’c)t
cs e
eqt aj,s,t dt ¤ Cecs (q’c)t
¤ Ceqs
e dt = Ce
q’c
s s t=s

using q < c. For s < j we split the integral into two parts. From aj,s,t ¤ 1 we
j j
conclude that s eqt aj,s,t dt ¤ s eqt dt ¤ Ceqj , and from aj,s,t ¤ Cec(j’t) (by the
∞ ∞
previous claim) we conclude that j eqt aj,s,t dt ¤ Cecj j e(q’c)t dt ¤ Ceqj , which
proves the claim.
Now the main claim. We have
T T
qt
eqt epj kj (t) dt ¤
e k(t) dt =
p
0 0 j
T t
qt pj
¤ ee aj,0,t kj (0) + aj,s,t gj (s) ds dt.
0 0
j


The ¬rst summand is bounded by
T
pj
eqt aj,0,t dt kj (0) ¤ C e(p+q)j kj (0) ¤ C k(0)
e p+q
0
j j


and the second by
T T T
pj qt
epj (eqj + q qs ) gj (s) ds
gj (s) ds ¤
e e aj,s,t dt
0 s 0
j j
T
+ eqs g(s) p ds.
¤C g(s) p+q
0



We will next show that the domain of de¬nition is not ¬nite and that the limit
limt’+∞ x(t) =: x∞ exists and is a solution of f (x∞ ) = y. For this we need the

is su¬ciently small then for n ≥ 2r and
Sublemma. If x is a solution and y 2r
q ≥ 0 we have
T
dt ¤ Cn,q eqT y
¤
x(T ) x (t) n.
n+q n+q
0



Proof. Using the inequality for smoothing operators we get for n ≥ 0 and q ≥ 0
that

= cΨ(σt x(t)) · σt k(t) ¤ C σt kt
x (t) + C σt x(t) σt k(t)
n+q n+q n+q n+q+2r 0

¤ Ceqt kt + x(t) kt .
n n+2r 0


51.22
51.22 51. The Nash-Moser inverse function theorem 571

¤ 1. Later we will show that this is
We will from now on assume that x(t) 2r
automatically satis¬ed. Then

qt
¤ k(t) 0 (1 + x(t)
x (t) 2r )Ce ,
q

¤2
and hence
t t
eqs k(s)
¤ x (s) q ds ¤ C
x(t) ds.
q 0
0 0

Using the estimates for k given by the sublemma, we have to estimate norms of

g(t) = (f (σt x(t)) ’ f (x(t))) · x (t) = ’b(x(t), σt x(t))((1 ’ σt )x(t), x (t)),

1
where b(x0 , x1 )(h0 , h1 ) := 0 f (x0 + t(x1 ’ x0 ))(h0 , h1 ) dt. Obviously, b is smooth,
is bilinear with respect to (h0 , h1 ), and satis¬es a tame estimate of degree 2r with
respect to x0 , x1 and degree r in h0 and h1 with base 0, i.e.

¤C
b(x0 , x1 )(h0 , h1 ) h0 h1 + h0 h1
n n+r r r n+r

+ ( x0 + x1 n+2r ) h0 h1 .
n+2r r r


¤ 1 we obtain
From this and xt 2r


¤C (1 ’ σt )xt (1 ’ σt )xt
g(t) x (t) + x (t)
n n+r r r n+r

¤Ce’rt xt ¤Cert kt ¤Ce’(n+r)t xt ¤Ce(n+r)t kt
0
n+2r n+2r 0


(1 ’ σt )xt
+ ( xt + σt xt n+2r ) x (t)
n+2r r r

¤Ce’rt xt ¤Cert kt
¤ xt n+2r +C xt n+2r 0
2r

¤ C xt kt 0 .
n+2r


In order to estimate x(t) q , we need the following estimate using the sublemma
giving an estimate for the solution k (t) + cσ(t)k(t) = g(t) for c > 2r

T T
qt
eqt
dt ¤ C ¤
e kt k(0) + g(t) + g(t)
p p+q p p+q
0 0
¤ xt ¤ xt
kt kt
p+2r 0 p+q+2r 0

T t t
qt (p+2r)s
e(p+q+2r)s ks
¤C k(0) + eC e ks ds + C ds kt dt
p+q 0 0 0
0 0 0
t
¤Ceqt e(p+2r)s ks ds
0
0
T T
qt
e(p+2r)s ks 0 ds .
¤C dt ·
k(0) + e kt
p+q 0
0 0


This inequality is of recursive nature. In fact, for p = 0 and q = 2r it says
T
KT := 0 e2rt kt 0 dt ¤ C y 2r + CKT and hence KT (1 ’ CKT ) ¤ C y 2r . If
2

KT ¤ 2C then 1’CKT ≥ 1 and hence KT ¤ 2C y 2r . Thus, KT ∈ (2C y 2r , 2C ].
1 1
/
2

51.22
572 Chapter X. Further Applications 51.22

1
Therefore, choosing y 2r < 4C 2 makes this a nonempty interval, and continuity of
T ’ KT and K0 = 0 shows that
T
e2rt kt 0 dt = KT ¤ 2C y ¤ δ.
for all y
2r 2r
0

Let us now show that the requirement x(t) 2r ¤ 1 is automatically satis¬ed.
Suppose not, then there is a minimal t0 > 0 with x(t0 ) 2r ≥ 1 since x(0) = 0.
Thus, for 0 ¤ t < t0 we have x(t) < 1, and hence the above estimates hold on
the interval [0, t0 ]. From x (t) 2r ¤ Ce2rt k(t) 0 we obtain by integration that
T t
x(t) 2r ¤ 0 x (t) 2r ¤ 0 Ce2rt k(t) 0 dt ¤ 2C y 2r . Thus, if y 2r ¤ δ with
Cδ < 1 then x(t0 ) 2r < 1, a contradiction. Note that this shows at the same time
that the sublemma is valid for q = 0 and n = 2r.
Now we proceed to show that
T
eqs ks ds ¤ C y for q ≥ 2r.
0 q
0

In fact, q = 2r and q = 2r + 1 will be su¬cient, and hence we have a common C.
The above estimate for p = 0 gives
T T T
qt qt
e2rs ks
dt ¤ C dt ·
e ks k(0) + e kt ds .
0 q 0 0
0 0 0
y
¤C y 2r ¤C


Thus
T T
qt
eqt ks
(1 ’ c y · dt ¤ C y dt ¤ C y
2r ) e ks and
0 q 0 q
0 0

for q ≥ 2r.Now for q ≥ p + 2r ¤C y ¤C y
q p+2r

T T T
qt qt
e(p+2r)s ks 0 ds
dt ¤ C dt ·
e ks k(0) + e kt
p p+q 0
0 0 0
¤ C( y ·y ¤C y
+y 2r ) p+q .
p+q p+q

¤1

Now we prove the main claim by induction on n = p + 2r. For p = 0 we have shown
it already. Next for n + 1 = p + 1 + 2r: Using the inequality at the very beginning
of the proof of the sublemma for n replaced by p and q by 1 + 2r + q
T T
Ce(1+2r+q)t
dt ¤
x (t) kt + x(t) kt dt
p+1+2r+q p p+2r 0
0 0
T
qT
e(1+2r)t k(t) + e(1+2r)t x(t)
¤ Ce · kt dt
p p+2r 0
0
¤C y p+2r


¤ CeqT C y +C y y 2r+1 +
p+2r+1 p+2r

+y p+2r C y 2r+1

¤ CeqT y +y y .
p+2r+1 p+2r 2r+1


51.22
51.22 51. The Nash-Moser inverse function theorem 573

¤ δ ¤ 1 we get by interpolation that
For y 2r


¤C y ¤C y
y y y p+1+2r ,
p+2r 2r+1 p+2r+1 2r


which eliminates the last summand and completes the induction.
Claim. For su¬ciently small y 2r the solution x exists globally, limt’+∞ x(t) =:
x∞ exists and solves f (x∞ ) = y. Moreover, we have x∞ n ¤ cn y n .
t
Furthermore, x(t) n ¤ 0 x (s) n ds ¤ C y n for n > 2r and y 2r ¤ δ using the
main claim for q = 0.
T
Suppose x exists on [0, ω) with ω chosen maximally. Since 0 x (t) n dt ¤ C y n
ω
with C independent on T < ω we have 0 x (t) n dt ¤ C y n < ∞, and hence
ω s
limt’ω t x („ ) d„ = 0. Thus, we obtain x(s) ’ x(t) n ¤ t x (r) dr ’ 0 for
s, t ’ ω. Hence, x(ω) := limt ω x(t) exists and x(ω) 2r ¤ 1.
Thus, we can extend the solution in a neighborhood of ω, a contradiction to the
maximality of ω. Then by the same argument as before x∞ := limt ∞ x(t) exists
and x∞ 2r ¤ 1 and x∞ n ¤ C y n for all n ≥ 2r. Since x (t) = cΨ(σ(t)x(t)) ·
(σ(t)(y’f (x(t)))) we have that limt’∞ x (t) = cΨ(x∞ )(y’f (x∞ )) exists, and since

x (t) n dt ¤ C y n < ∞ we have that (x∞ )(y ’ f (x∞ )) = limt ∞ x (t) = 0.
0
So we get y ’ f (x∞ ) = 0, hence we have obtained an inverse.

Proof of the inverse function theorem. By what we have shown so far, we
know that f is locally bijective and f ’1 y n ¤ C y n for all n ≥ 2r. Furthermore,

f ’1 y1 ’ f ’1 y0 + ( f ’1 y1 + f ’1 y0
¤C y1 ’ y 0 y 1 ’ y0
n+2r ) ,
n n n+2r 0


which shows continuity and Lipschitzness of the inverse, and the inverse f ’1 is
tame and locally Lipschitz.
We next show that f ’1 is Gˆteaux-di¬erentiable with derivative as expected, i.e.
a

(f ’1 ) (y)(k) = f (f ’1 y) · k.

For this let c(t) := f ’1 (y + t k), x := c(0) = f ’1 (y) and := Ψ(x) := f (f ’1 (y))’1 .
Then c is locally Lipschitz, and we have to show that c(t)’c(0) ’ · k for t ’ 0.
t
Now
c(t) ’ c(0) c(t) ’ c(0)
’ (k) = ( —¦ f (x)) ’ (k)
t t
c(t) ’ c(0) f (c(t)) ’ f (c(0))

= f (x)
t t
1
c(t) ’ c(0)
f (x) ’ f (x + s(c(t) ’ x)) ds ·
= .
t
0
=:g(t,s)


Since t ’ c(t) is locally Lipschitz, the map (t, s) ’ f (x) ’ f (x + s(c(t) ’ x)) is
locally Lipschitz, and hence in particular continuous. Therefore, g(t, s) ’ g(0, s)

51.22
574 Chapter X. Further Applications 51.23

1 1 1
for t ’ 0 uniformly on all s ∈ [0, 1]. Thus, 0 g(t, s)ds ’ 0 g(0, s)ds = 0 ds = 0
0
in L(E, F ), and since c(t)’c(0) stays bounded, this proves the claim.
t
Thus, we have for the Gˆteaux-derivative of the inverse function the formula
a

(f ’1 ) = inv —¦f —¦ f ’1 = Ψ —¦ f ’1 .

Since Ψ and f ’1 are tame, so is (f ’1 ) . By induction, using the chain rule for
di¬erentiable and for tame maps we conclude that (f ’1 ) is a tame smooth map,
since Ψ was assumed to be so. Note that in order to apply the chain-rule it is
not enough to have Gˆteaux-di¬erentiability, but because of tameness (or local
a
Lipschitzness) of the derivative we have the appropriate type of di¬erentiability
automatical. In fact, we have to consider d(f ’1 ) := ((f ’1 ) )§ = Ψ§ —¦ (f ’1 — Id)
and apply induction to that.

Let us ¬nally show that it is enough to assume that Ψ is a tame continuous in order
to assure that it is a tame smooth map.

51.23. Lemma. Let ¦ : E ⊇ U ’ GL(F ) be a tame smooth map and Ψ, de¬ned
by Ψ(x) := ¦(x)’1 , be a continuous tame map. Then Ψ is a tame smooth map.

Proof. For smoothness it is enough to show smoothness along continuous curves,
so we may assume that E = R = U , and so ¦ is a curve denoted c. Then
c(t)’1 ’ c(s)’1 c(t) ’ c(s)
= ’ comp(inv(c(t)), , inv(c(s)))

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