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51.10 51. The Nash-Moser inverse function theorem 559

where d is any integer greater or equal to log(a). Note however, that T is not well
de¬ned on s(E) for a > 1, and this is the reason to consider the space Σ(E).
Note furthermore, that as before one shows that one could equally well replace the
sum by the corresponding supremum in the de¬nition of Σ(E), one only has to use
that k∈N e’k = 1’1/e < ∞.
1


(3). As a similar example we consider a linear di¬erential operator D of degree
d, i.e., a local operator (the values Df depend at x only on the germ of f at x)
|±|
which is locally given in the form Df = |±|¤d g± · ‚ ± f , with smooth coe¬cient
‚x

functions g± ∈ C (M, R) on a compact manifold M .
Then D : C ∞ (M, R) ’ C ∞ (M, R) is tame of degree d and base 0. In fact, by the
product rule we can write the k-th derivative of Df as linear combination of partial
derivatives of the g± and derivatives of order up to k + d of f .
(4). Now we give an example of a non-tame linear map. For this consider T :
C ∞ ([0, 1], R) ’ C ∞ ([’1, 1], R) given by T f (t) := f (t2 ). It was shown in the proof

of (25.2) that the image of T consists exactly of the space Ceven ([’1, 1], R) of even
functions. Since (T f )(n) (t) = f (n) (t2 )(2t)n + 0<2k¤n cn f (n’k) (t2 )tn’2k with some
k
n
ck ∈ Z, we have that T is tame of order 0 and degree 0. But the inverse is not
tame since (T f )(2n) (0) is proportional to f (n) (0), hence in order to estimate the
n-th derivative of T ’1 g we need the 2n-th derivative of g.

51.9. De¬nition. A graded Fr´chet space F is called tame if there exists some
e
Banach space E such that F is a tame direct summand in Σ(E), i.e. there are tame
linear mappings i : F ’ Σ(E) and p : Σ(E) ’ F with p —¦ i = IdF .

Our next aim is to show that instead of Σ(E) we can equally well use s(E). For
this we consider a measured space (X, µ) and a measurable positive weight function
w : X ’ R and de¬ne

L1 (X, µ, w) := f ∈ L1 (X, µ) : f en w(x) |f (x)| dµ(x) < ∞ .
:=
n
Σ
X



51.10. Proposition. Every space L1 (X, µ, w) is a tame Fr´chet space.
e
Σ


Proof. Let Xk := {x ∈ X : k ¤ w(x) < k + 1}. Then the Xk form a countable
disjoint covering of X by measurable sets. Let χk be the characteristic function of
Xk , and let R : L1 (X, µ, w) ’ ΣL1 (X, µ) and L : ΣL1 (X, µ) ’ L1 (X, µ, w) be
Σ Σ
de¬ned by Rf := (χk · f )k and L((fk )k ) := k χk · fk . Then obviously L —¦ R = Id.
The linear map R is well-de¬ned and tame of degree 0 and base 0, since


enk χk f enk |f | dµ ¤
Rf = =
n 1
Xk
k k

ew(x)n |f (x)| dµ(x) = ew(x)n |f (x)| dµ(x) = f
¤ n.
Xk X
k

51.10
560 Chapter X. Further Applications 51.13

Finally, L is a well-de¬ned linear map, which is tame of degree 0 and base 0, since

en w(x) en w(x) |fk (x)| dµ(x)
L((fk )k ) = χk fk (x) dµ(x) =
n
X Xk
k k

en(k+1) |fk (x)| dµ(x) ¤ en(k+1) |fk (x)| dµ(x)
¤
Xk Xk
k k

= en enk fk = en (fk )k n.
1
k

51.11. Corollary. For every Banach space E the space s(E) is a tame Fr´chet
e
space.

Proof. This result follows immediately from the proposition (51.10) above, if one
replaces L1 (X, µ, w) by the vector valued function space L1 (X, µ, w; E) and simi-
Σ Σ
1 1
larly the space L (X, µ) by the Banach space L (X, µ; E).

Now let us show the converse direction:

51.12. Proposition. For every Banach space E the space Σ(E) is a tame direct
summand of s(E).

Proof. We de¬ne R : Σ(E) ’ s(E) and L : s(E) ’ Σ(E) by R((xk )k ) := (yk )k ,
where y[ek ] := xk and 0 otherwise, and L((yk )k ) := (y[ek ] )k . The map R is well-
de¬ned, linear and tame, since (yk )k n := k (1 + k)n yk = j (1 + [ej ])n xj ¤
jn n
j (2e ) xj = 2 (xj )j n . The map L is well-de¬ned, linear and tame, since
kn kn kn
y[ek ] ¤ ¤
(xk )k n := ke xk = ke k (1 + [e ]) y[ek ] j (1 +
j)n yj = y n . Obviously, L —¦ R = Id.

51.13. De¬nition. A non-linear map f : E ⊇ U ’ F between graded Fr´chet e
spaces is called tame of degree r and base b if it is continuous and every point in U
has a neighborhood V such that
∀n ≥ b ∃Cn ∀x ∈ V : f (x) ¤ Cn (1 + x n+r ).
n


Remark. Every continuous map from a graded Fr´chet space into a Banach space
e
is tame.
For ¬xed x0 ∈ U choose a constant C > f (x0 ) and let V := {x : f (x) < C}.
Then V is an open neighborhood of x0 , and for all n and all x ∈ V we have
f (x) n = f (x) ¤ C ¤ C(1 + x n ).
Every continuous map from a ¬nite dimensional space into a graded Fr´chet space
e
is tame.
Choose a compact neighborhood V of x0 . Let Cn := max{ f (x) n : x ∈ V }. Then
f (x) n ¤ Cn ¤ Cn (1 + x ).
It is easily checked that the composite of tame linear maps is tame. In fact
¤ C(1 + g(x)
f (g(x)) n+r )
n

¤ C(1 + C(1 + x ¤ C(1 + x
n+r+s )) n+r+s )

for all x in an appropriately chosen neighborhood and n ≥ bf and n + r ≥ bg .

51.13
51.15 51. The Nash-Moser inverse function theorem 561

51.14. Proposition. The de¬nition of tameness of degree r is coherent with the
one for linear maps, but the base may change.

Proof. Let ¬rst f be linear and tame as non-linear map. In particular, we have
locally around 0
f (x) n ¤ C(1 + x n+r ) for all n ≥ b.
If we increase b, we may assume that the 0-neighborhood is of the form {x :
x b+r ¤ µ} for some µ > 0. For y = 0 let x := y µb+r y, i.e., x b+r = µ.
Thus, f (x) n ¤ C(1 + x n+r ). By linearity of f , we get

y y y
b+r b+r b+r
¤C
f (y) = f (x) +x
n n n+r
µ µ µ
y b+r
=C +y .
n+r
µ
¤x for b ¤ n we get
Since y b+r n+r

1
¤C
f (y) +1 x n+r .
n
µ

Conversely, let f be a tame linear map. Then the inequality

¤C x ¤ C(1 + x for all n ≥ b
f (x) n+r )
n n+r


is true.

De¬nition. For functions f of two variables we will de¬ne tameness of bi-degree
(r, s) and base b if locally

∀n ≥ b ∃C ∀x, y : f (x, y) ¤ C(1 + x +y n+s );
n n+r


and similar for functions in several variables.

51.15. Lemma. Let f : U — E ’ F be linear in the second variable and tame of
b+r —
base b and degree (r, s) in a b+s -neighborhood. Then we have


∀n ≥ b ∃C : f (x)h ¤ C( h +x h b+s )
n n+s n+r


for all x in a b+r -neighborhood and all h.
If f : U — E1 — E2 is tame of base b and degree (r, s, t) in a — —
b+r b+s
b+t -neighborhood. Then we have


¤ C( h
f (x)(h, k) k +h k +x h k b+t )
n n+s b+t b+s n+t n+r b+s


for all x in a b + r-neighborhood and all h and k.
¯ µ
Proof. For arbitrary h let h := h. Then
h b+s


¯ ¯
f (x)h ¤ C(1 + x +h n+s ).
n+r


51.15
562 Chapter X. Further Applications 51.18

Therefore
h h µ
b+s b+s
¯ ¤
f (x)h = f (x)h C 1+ x + h
n n n+r n+s
µ µ h b+s
C h b+s C h b+s
¤ + x n+r + C h n+s
µ µ
1 C
¤C + 1 h n+s + x n+r h b+s .
µ µ
The second part is proved analogously.

51.16. Proposition. Interpolation formula for Σ(E).

·x ¤x ·x for 0 ¤ r ¤ n ¤ m.
x n m n’r m+r



Proof. Let us ¬rst consider the special case, where n = m and r = 1. Then
2
n’1 · ’x
x x =
n+1 n

e(n’1)k xk e(n+1)l xl ’ enk xk enl xl
=
k l k l

(e(n’1)k e(n+1)k ’ e2nk ) xk 2
=
k=l

(e(n’1)k e(n+1)l + e(n+1)k e(n’1)l ’ 2en(k+l) ) xk
+ xl .
k<l

In both subsummands the expression in brackets is positive, since

e(n’1)k e(n+1)l + e(n+1)k e(n’1)l ’ 2en(k+l) =
= 2en(k+l) (el’k + ek’l ’ 2) = 4en(k+l) (cosh(l ’ k) ’ 1) ≥ 0.

By transitivity, it is enough to show the general case for r = 1. Without loss of
generality we may assume x = 0. Then this case is equivalent to
xn x m+1
¤ for n ¤ m.
x n’1 xm
Again by transitivity it is enough to show this for m = n.

51.17. The Nash-Moser inverse function theorem. Let E and F be tame
Fr´chet spaces and let f : E ⊇ U ’ F be a tame smooth map. Suppose f has a
e
tame smooth family Ψ of inverses. Then f is locally bijective, and the inverse of f
is a tame smooth map.

The proof will take the rest of this section.

51.18. Proposition. Let E and F be tame Fr´chet spaces and let f : E ⊇ U ’ F
e
be a tame smooth map. Suppose f has a tame smooth family Ψ of linear left
inverses. Then f is locally injective.


51.18
51.19 51. The Nash-Moser inverse function theorem 563

51.19. Proposition. Let E and F be tame Fr´chet spaces and let f : E ⊇ U ’ F
e
be a smooth tame map. Suppose f has a tame smooth family Ψ of linear right
inverses. Then f is locally surjective (and locally has a smooth right inverse).

By a tame smooth mapping f we will for the moment understand an in¬nitely often
Gˆteaux di¬erentiable map, for which the derivatives f (n) (x) are multilinear and
a
are tame as maps U — E n ’ F .
By a tame smooth family of (one-sided) inverses of f we understand a family
(Ψ(x))x∈U : F ’ E of (one-sided) inverses of (f (x))x∈U , which gives a tame
smooth map Ψ§ : U — F ’ E.
Let us start with some preparatory remarks for the proofs. Contrary to good
manners the symbol C will almost never denote the same constant even not in the
same inequality. This constant may depend on the index of the norm n but not on
any argument of the norms.
For all three proofs we may assume that the initial values are f : 0 ’ 0 (apply
translations in the domain and the codomain).

Claim. We may assume that E = Σ(B) and F = Σ(C).
First for (51.18). In fact, E and F are direct summands in such spaces Σ(B) and
Σ(C). We extend f to a smooth tame mapping f : Σ(B) ⊇ U ’ Σ(B — C) ∼
˜ ˜ =
˜
Σ(B) — Σ(C), by setting U := p’1 (U ), where p : Σ(B) ’ E is the retraction,
˜
and f := (Id ’p, f —¦ p). Note that (Id ’p) preserves exactly that part which gets
˜
annihilated by f —¦ p. More precisely injectivity of f implies that of f . In fact,
f (x) = f (y) implies x = p(x), y = p(y), and hence (Id ’p)(x) = 0 = (Id ’p)(y), and
˜ ˜ ˜x ˜ ˜ ˜ ˜x
so f (x) = f (y). Since f (˜)(h) = ((Id ’p)(h), f (p(˜)) · p(h)), let Ψ(˜) := (Id ’p) —¦
x
˜x ˜x
pr1 +Ψ(p(˜))—¦pr2 . Then Ψ(˜)—¦ f (˜) = (Id ’p)—¦(Id ’p)+Ψ(p(˜))—¦f (p(˜))—¦p = Id.
x x x
Now for (51.19). Here we extend f to a smooth tame mapping f : Σ(B — C) ∼
˜ =
˜
˜ ˜
Σ(B) — Σ(C) ⊇ U ’ Σ(C), by setting U := p’1 (U ) — Σ(C) and f := (f —¦ p) •
(Id ’q), where p : Σ(B) ’ E and q : Σ(C) ’ F are the retractions. Since
˜x˜ ˜x˜
f (˜, y ) = f (p(˜)) —¦ p • (Id ’q) let Ψ(˜, y ) : Σ(C) ’ Σ(B) — Σ(C) be de¬ned
x
˜x˜ ˜ ˜ ˜ ˜x˜
by Ψ(˜, y )(k) := (Ψ(p(˜))(q(k)), (Id ’q)(k)), i.e. Ψ(˜, y ) := (Ψ(p(˜)) —¦ q, (Id ’q)).
x x
Then

˜x˜ ˜x˜
f (˜, y ) —¦ Ψ(˜, y ) = ((f —¦ p) (˜) • (Id ’q) (˜)) —¦ (Ψ(p(˜)) —¦ q, (Id ’q))
x y x
= f (p(˜)) —¦ p —¦ Ψ(p(˜)) —¦q + (Id ’q) —¦ (Id ’q)
x x
Ψ(p(˜))
x

= q + (Id ’2q + q 2 ) = Id .


Claim. We may assume that x ’ f (x), (x, h) ’ f (x)h, (x, h) ’ f (x)(h, h) and
(x, k) ’ Ψ(x)k satisfy tame estimates of degree 2r in x, of degree r in h and 0 in
k (for some r) and base 0 on the set { x 0 ¤ 1}.
Consider on Σ(B) the linear operators p which are de¬ned by ( p x)k := epk xk .
p
x n = x n+p . If f satis¬es f (x) n ¤ C(1 + x n+s ) on x a ¤ δ
Then

51.19
564 Chapter X. Further Applications 51.20

˜ ˜
for n ≥ b then f := q —¦ f —¦ ’p satis¬es f (x) m = f ( ’p x) m+q ¤ C(1 +
’p
x m+q+s ) = C(1 + x m+q+s’p ) on x a’p ¤ δ for m ≥ b ’ q.
Choosing q and p su¬ciently large, we may assume that f , f , f , and Ψ satisfy
tame estimates of base 0 (choose q large in comparison to b) on {x : x 0 ¤ δ}
(choose p large in comparison to a). Furthermore, we may achieve that (x, k) ’
Ψ(x)k is tame of order 0 (since by linearity we don™t need p for the neighborhood,
which is now global, but we have to choose it so that m + q + s ’ p ¤ m) in k (but
we cannot achieve that this is also true for f ). Now take r su¬ciently large such
that the degrees are dominated by 2r and r, and ¬nally replace f by x ’ f (cx) to
obtain δ = 1.

¤ 1 we have for all n ≥ 0 a Cn > 0 such that
Claim. On x 2r


¤ Cn x
f (x) n+2r ,
n

¤ Cn ( h
f (x)h +x h r ),
n n+r n+2r

¤ Cn ( h1
f (x)(h1 , h2 ) h2 + h1 h2 +x h1 h2 r ),
n n+r r r n+r n+2r r

¤ Cn ( k
Ψ(x)k +x k 0 ).
n n n+2r


The 2nd, 3rd and 4th inequality follow from the corresponding tameness and
(51.15), since the neighborhood is given by a norm with index higher then base
+ degree. For the ¬rst inequality one would expect f (x) n ¤ C(1 + x n+2r ), but
since f (0) = 0 one can drop the 1, which follows from integration of the second
estimate:
1

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