Now we start to choose the eigenvectors smoothly, under the stronger assumption

in the smooth case, and in the real analytic case. Let us consider again eigenvalues

{»i (t) : 1 ¤ i ¤ N } contained in the interior of a smooth curve γ for t in an open

interval I. Then Vt := P (t, γ)(H) is the ¬ber of a smooth (real analytic) vector

bundle of dimension N over I. We choose a smooth framing of this bundle, and

use then the proof of theorem (50.14) to choose smooth (real analytic) sub vec-

tor bundles whose ¬bers over t are the eigenspaces of the eigenvalues with their

generic multiplicity. By the same arguments as in (50.14) we then get global vec-

tor sub bundles with ¬bers the eigenspaces of the eigenvalues with their generic

multiplicities, and thus smooth (real analytic) eigenvectors for all eigenvalues.

50.17. Result. ([Rellich, 1969, page 43], see also [Kato, 1976, II, 6.8]). Let A(t)

be a C 1 -curve of (¬nite dimensional) symmetric matrices. Then the eigenvalues

can be chosen C 1 in t, on the whole parameter interval.

This result is best possible for the degree of continuous di¬erentiability, as is shown

by the example in [Alekseevsky, Kriegl, Losik, Michor, 1996, 7.4]

50.17

553

51. The Nash-Moser Inverse Function Theorem

This section treats the hard implicit function theorem of Nash and Moser following

[Hamilton, 1982], in full generality and in condensed form, but with all details. The

main di¬culty of the proof of the hard implicit function theorem is the following:

By trying to use the Newton iteration procedure for a nonlinear partial di¬erential

equation one quickly ¬nds out that ˜loss of derivatives™ occurs and one cannot reach

the situation, where the Banach ¬xed point theorem is directly applicable. Using

smoothing operators after each iteration step one can estimate higher derivatives

by lower ones and ¬nally apply the ¬xed point theorem.

The core of this presentation is the following: one proves the theorem in a Fr´chet

e

space of exponentially decreasing sequences in a Banach space, where the smooth-

ing operators take a very simple form: essentially just cutting the sequences at some

index. The statement carries over to certain direct summands which respect ˜boun-

ded losses of derivatives™, and one can organize these estimates into the concept

of tame mappings and thus apply the result to more general situations. However

checking that the mappings and also the inverses of their linearizations in a certain

problem are tame mappings (a priori estimates) is usually very di¬cult. We do not

give any applications, in view of our remarks before.

51.1. Remark. Let f : E ⊇ U ’ V ⊆ E be a di¬eomorphisms. Then di¬erenti-

ation of f ’1 —¦ f = Id and f —¦ f ’1 = Id at x and f (x) yields using the chain-rule,

that f (x) is invertible with inverse (f ’1 ) (f (x)) and hence x ’ f (x)’1 is smooth

as well.

The inverse function theorem for Banach spaces assumes the invertibility of the

derivative only at one point. Openness of GL(E) in L(E) implies then local in-

vertibility and smoothness of inv : GL(E) ’ GL(E) implies the smoothness of

x ’ f (x)’1 .

Beyond Banach spaces we do not have openness of GL(E) in L(E) as the following

example shows.

51.2. Example. Let E := C ∞ (R, R) and P : E ’ E be given by P (f )(t) :=

f (t)’t f (t) f (t). Since multiplication with smooth functions and taking derivatives

are continuous linear maps, P is a polynomial of degree 2. Its derivative is given

by

P (f )(h)(t) = h(t) ’ t h(t) f (t) ’ t f (t) h (t).

In particular, the derivative P (0) is the identity, hence invertible. However, at the

1

constant functions fn = n the derivative P (fn ) is not injective, since hk (t) := tk

1 k

are in the kernel: P (fn )(hk )(t) = tk ’ t · 0 · tk ’ t · n · k · tk’1 = tk · (1 ’ n ).

Let us give an even more natural and geometric example:

51.3. Example. Let M be a compact smooth manifold. For Di¬(M ) we have

shown that the 1-parameter subgroup of Di¬(M ) with initial tangent vector X ∈

51.3

554 Chapter X. Further Applications 51.3

TId Di¬(M ) = X(M ) is given by the ¬‚ow FlX of X, see (43.1). Thus, the exponen-

tial mapping Exp : TId Di¬(M ) ’ Di¬(M ) is given by X ’ FlX .1

The derivative T0 exp : Te G = T0 (Te G) ’ Texp(0) (G) = Te G at 0 of the exponential

mapping exp : g = Te G ’ G is given by

FltX (1, e) = FlX (t, e) = Xe .

d d d

dt |t=0 dt |t=0 dt |t=0

T0 exp(X) := exp(tX) =

Thus, T0 exp = Idg . In ¬nite dimensions the inverse function theorem now implies

that exp : g ’ G is a local di¬eomorphism.

What is the corresponding situation for G = Di¬(M )? We take the simplest

compact manifold (without boundary), namely M = S 1 = R/2π Z. Since the

natural quotient mapping p : R ’ R/2πZ = S 1 is a covering map we can lift

˜

each di¬eomorphism f : S 1 ’ S 1 to a di¬eomorphism f : R ’ R. This lift

˜

is uniquely determined by its initial value f (0) ∈ p’1 ([0]) = 2πZ. A smooth

˜

mapping f : R ’ R projects to a smooth mapping f : S 1 ’ S 1 if and only if

˜ ˜

f (t + 2π) ∈ f (t) + 2πZ. Since 2πZ is discrete, f (t + 2π) ’ f (t) has to be 2πn for

˜

some n ∈ Z not depending on t. In order that a di¬eomorphism f : R ’ R factors

to a di¬eomorphism f : S 1 ’ S 1 the constant n has to be +1 or ’1. So we ¬nally

obtain an isomorphism {f ∈ Di¬(R) : f (t + 2π) ’ f (t) = ±1}/2πZ ∼ Di¬(S 1 ). In

=

particular, we have di¬eomorphisms Rθ given by translations with θ ∈ S 1 (In the

picture S 1 ⊆ C these are just the rotations by with angle θ).

Claim. Let f ∈ Di¬(S 1 ) be ¬xed point free and in the image of exp. Then f is

conjugate to some translation Rθ .

We have to construct a di¬eomorphism g : S 1 ’ S 1 such that f = g ’1 —¦ Rθ —¦ g.

Since p : R ’ R/2πZ = S 1 is a covering map it induces an isomorphism Tt p :

R ’ Tp(t) S 1 . In the picture S 1 ⊆ C this isomorphism is given by s ’ s p(t)⊥ ,

where p(t)⊥ is the normal vector obtained from p(t) ∈ S 1 via rotation by π/2.

Thus, the vector ¬elds on S 1 can be identi¬ed with the smooth functions S 1 ’ R

or, by composing with p : R ’ S 1 with the 2π-periodic functions X : R ’ R.

Let us ¬rst remark that the constant vector ¬eld X θ ∈ X(S 1 ), s ’ θ has as ¬‚ow

θ θ

FlX : (t, •) ’ • + t · θ. Hence exp(X θ ) = FlX = Rθ .

1

Xθ

FlX

Let f = exp(X) and suppose g —¦ f = Rθ —¦ g. Then g —¦ —¦g for t = 1.

= Flt

t

Let us assume that this is true for all t. Then di¬erentiating at t = 0 yields

T g(Xx ) = Xg(x) for all x ∈ S 1 . If we consider g as di¬eomorphism R ’ R this

θ

means that g (t) · X(t) = θ for all t ∈ R. Since f was assumed to be ¬xed point free

the vector ¬eld X is nowhere vanishing. Otherwise, there would be a stationary

tθ

point x ∈ S 1 . So the condition on g is equivalent to g(t) = g(0) + 0 X(s) ds. We

take this as de¬nition of g, where g(0) := 0, and where θ will be chosen such that

t+2π ds

g factors to an (orientation preserving) di¬eomorphism on S 1 , i.e. θ t X(s) =

2π

ds

g(t + 2π) ’ g(t) = 1. Since X is 2π-periodic this is true for θ = 1/ 0 X(s) . Since

the ¬‚ow of a transformed vector ¬eld is nothing else but the transformed ¬‚ow we

θ

obtain that g(FlX (t, x)) = FlX (t, g(x)), and hence g —¦ f = Rθ —¦ g.

51.3

51.4 51. The Nash-Moser inverse function theorem 555

In order to show that exp : X(S 1 ) ’ Di¬(S 1 ) is not locally surjective, it hence

su¬ces to ¬nd ¬xed point free di¬eomorphisms f arbitrarily close to the identity

which are not conjugate to translations. For this consider the translations R2π/n

and modify them inside the interval (0, 2π ) such that the resulting di¬eomorphism

n

π 3π k

f satis¬es f ( n ) ∈ n + 2πZ. Then f maps 0 to 2π, and thus the induced di¬eo-

/

morphism on S 1 has [0] as ¬xed point. If f would be conjugate to a translation,

the same would be true for f k , hence the translation would have a ¬xed point and

hence would have to be the identity. So f k must be the identity on S 1 , which is

impossible, since f ( n ) ∈ 3π + 2πZ.

π

/n

Let us ¬nd out the reason for this break-down of the inverse function theorem. For

this we calculate the derivative of exp at the constant vector ¬eld X := X 2π/k :

d

ds |s=0

exp (X)(Y )(x) = exp (X + sY )(x)

1

X+sY

d 2tπ

ds |s=0

= Fl (1, x) = Y (x + k ) dt,

0

where we have di¬erentiated the de¬ning equation for FlX+sY to obtain

FlX+sY (t, x) = |s=0 ‚t FlX+sY (t, x)

‚‚ ‚ ‚

‚t ‚s |s=0 ‚s

X+sY

‚

‚s |s=0 (X + sY )(Fl

= (t, x))

X

= Y (Fl (t, x)) + X (. . . )

= Y (x + t 2π ) + 0,

k

and the initial condition FlX+sY (0, x) = x gives

t

2π

X+sY

‚

‚s |s=0 Fl (t, x) = Y (x + „ ) d„.

k

0

If we take x ’ sin(kx) as Y then exp (X)(Y ) = 0, so exp (X) is not injective, and

since X can be chosen arbitrarily near to 0 we have that exp is not locally injective.

So we may conclude that a necessary assumption for an inverse function theorem

beyond Banach spaces is the invertibility of f (x) not only for one point x but for

a whole neighborhood.

For Banach spaces one then uses that x ’ f (x)’1 is continuous (or even smooth),

which follows directly from the smoothness of inv : GL(E) ’ GL(E), see (51.1).

However, for Fr´chet spaces the following example shows that inv is not even con-

e

tinuous (for the c∞ -topology).

51.4. Example. Let s be the Fr´chet space of all fast falling sequences, i.e.

e

s := {(xk )k ∈ RN : (xk )k n := sup{(1 + k)n |xk | : k ∈ N} < ∞ for all n ∈ N}.

Next we consider a curve c : R ’ GL(s) de¬ned by

c(t)((xk )k ) := ((1 ’ h0 (t))x0 , . . . , (1 ’ hk (t))xk , . . . ),

51.4

556 Chapter X. Further Applications 51.4

where hk (t) := (1 ’ 2’k ) h(kt) for an h ∈ C ∞ (R, R) which will be chosen appropri-

ately.

Then c(t) ∈ GL(s) provided h(0) = 0 and supp h is compact, since then the factors

1 ’ hk (t) are equal to 1 for almost all k. The inverse is given by multiplying with

1/(1 ’ hk (t)), which exists provided h(R) ⊆ [0, 1].

Let us show next that inv —¦c : R ’ GL(s) ⊆ L(s) is not even continuous. For this

take x ∈ s, and consider

1

1 ’1 k

t’c (x) = (. . . ; 1 xk ; . . . ) = (?, . . . , ?; 2 xk ; ?, . . . ),

k

1 ’ hk ( k )

provided h(1) = 1. Let x be de¬ned by xk := 2’k , then c( k )’1 (x)’c(0)’1 (x)

1

≥

0

1 ’ 2’k ’ 0.

It remains to show that c : R ’ GL(s) is continuous or even smooth. Since

smoothness of a curve depends only on the bounded sets, and boundedness in

GL(E) ⊆ L(E, E) can be tested pointwise because of the uniform boundedness

theorem (5.18), it is enough to show that evx —¦s : R ’ GL(s) ’ s is smooth.

Boundedness in a locally convex space can be tested by the continuous linear func-

tionals, so it would be enough to show that » —¦ evx —¦c : R ’ GL(s) ’ s ’ R is

smooth for all » ∈ s— . We want to use the particular functionals given by the coor-

dinate projections »k : (xk )k ’ xk . These, however, do not generate the bornology,

but if B ⊆ s is bounded, then so is k∈N »’1 (»k (B)). In fact, let B be bounded.

k

Then for every n ∈ N there exists a constant Cn such that (1 + k)n |xk | ¤ Cn for all

k and all x = (xk )k ∈ B. Then every y ∈ »’1 (»k (x)) (i.e., »k (y) = »k (x)) satis¬es

k

the same inequality for the given k, and hence k∈N »’1 (»k (B)) is bounded as well.

k

Obviously, »k —¦evx —¦c is smooth with derivatives (»k —¦evx —¦c)(p) (t) = (1’hk )(p) (t)xk .

Let cp (t) be the sequence with these coordinates. We claim that cp has values in s

and is (locally) bounded. So take an n ∈ N and consider

cp (t) = sup(1 + k)n |(1 ’ hk )(p) (t)xk |.

n

k

We have (1’hk )(p) (t) = 1(p) ’(1’2’k )k p h(p) (kt), and hence this factor is bounded

by 1+k p h(p) ∞ . Since (1+k n )(1+ h(p) ∞ k p )|xk | is by assumption on x bounded

we have that supt cp (t) n < ∞.

Now it is a general argument, that if we are given locally bounded curves cp : R ’ s

such that »k —¦ c0 is smooth with derivatives (»k —¦ c0 )(p) = »k —¦ cp , then c0 is smooth

with derivatives cp .

In fact, we consider for c = c0 the following expression

c(t) ’ c(0) »k (c(t)) ’ »k (c(0))

1 1

’ c1 (0) ’ »k (c1 (0)) ,

»k =

t t t t

which is by the classical mean value theorem contained in { 1 »k (c2 (s)) : s ∈

2

12

[0, t]}. Thus, taking for B the bounded set { 2 c (s) : s ∈ [0, 1]}, we conclude

51.4

51.7 51. The Nash-Moser inverse function theorem 557

that (c(t) ’ c(0))/t ’ c1 (0) /t is contained in the bounded set k∈N »’1 (»k (B)),

k

c(t)’c(0) 1 k 0

’ c (0). Doing the same for c = c shows that c is smooth

and hence t

with derivatives ck .

From this we conclude that in order to obtain an inverse function theorem we

have to assume beside local invertibility of the derivative also that x ’ f (x)’1 is

smooth. That this is still not enough is shown by the following example:

51.5. Example. Let E := C ∞ (R, R) and consider the map exp— : E ’ E given by

exp— (f )(t) := exp(f (t)). Then one can show that exp— is smooth. Its (directional)

derivative is given by

(f +sh)(t)

= h(t) · ef (t) ,

‚

‚s |s=0 e

(exp— ) (f )(h)(t) =

so (exp— ) (f ) is multiplication by exp— (f ). The inverse of (exp— ) (f ) is the multi-

plication operator with exp1 (f ) = exp— (’f ), and hence f ’ (exp— ) (f )’1 is smooth

—

as well. But the image of exp— consists of positive functions only, whereas the curve

c : t ’ (s ’ 1 ’ ts) is a smooth curve in E = C ∞ (R, R) through exp— (0) = 1, and

c(t) is not positive for all t = 0 (take s := 1 ).

t

So we will need additional assumptions. The idea of the proof is to use that a

Fr´chet space is built up from Banach spaces as projective limit, to solve the inverse

e

function theorem for the building blocks, and to try to approximate in that way an

inverse to the original function. In order to guarantee that such a process converges,

we need (a priori) estimates for the seminorms, and hence we have to ¬x the basis

of seminorms on our spaces.

51.6. De¬nition. A Fr´chet space is called graded, if it is provided with a ¬xed

e

increasing basis of its continuous seminorms. A linear map T between graded

Fr´chet spaces (E, (pk )k ) and (F, (qk )k ) is called tame of degree d and base b if

e

∀n ≥ b ∃Cn ∈ R ∀x ∈ E : qn (T x) ¤ Cn pn+d (x).

Recall that T is continuous if and only if

∀n ∃m ∃Cn ∈ R ∀x ∈ E : qn (T x) ¤ Cn pm (x).

Two gradings are called tame equivalent of degree r and base b if and only if the

identity is tame of degree r and base b in both directions.

51.7. Examples. Let M be a compact manifold. Then C ∞ (M, R) is a graded

Fr´chet space, where we consider as k-th norm the supremum of all derivatives

e

of order less or equal to k. In order that this de¬nition makes sense, we can

embed M as closed submanifold into some Rn . Choosing a tubular neighborhood

Rn ⊇ U ’ M we obtain an extension operator p— : C ∞ (M, R) ’ C ∞ (U, R), and

on the latter space the operator norms of derivatives f k (x) for f ∈ C ∞ (U, R) make

sense.

51.7

558 Chapter X. Further Applications 51.8

Another way to give sense to the de¬nition is to consider the vector bundle J k (M, R)

of k-jets of functions f : M ’ R. Its ¬ber over x ∈ M consists of all “Taylor-

polynomials” of functions f ∈ C ∞ (M, R). We obtain an injection of C ∞ (M, R)

into the space of sections of J k (M, R) by associating to f ∈ C ∞ (M, R) the section

having the Taylor-polynomial of f at a point x ∈ M . So it remains to de¬ne a

norm pk on the space C ∞ (M ← J k (M, R)) of sections. This is just the supremum

norm, if we consider some metric on the vector bundle J k (M, R) ’ M .

Another method of choosing seminorms would be to take a ¬nite atlas and a par-

tition of unity subordinated to the charts and use the supremum norms of the

derivatives of the chart representations.

A second example of a graded Fr´chet space, closely related to the ¬rst one, is the

e

space s(E) of fast falling sequences in a Banach space E, i.e.

:= sup{(1 + k)n xk | : k ∈ N} < ∞ for all n ∈ N}.

s(E) := {(xk )k ∈ E N : (xk )k n

A modi¬cation of this is the space Σ(E) of very fast falling sequences in a Banach

space E, i.e.

enk xk < ∞ for all n ∈ N}.

Σ(E) := {(xk )k ∈ E N : (xk )k :=

n

k∈N

51.8. Examples.

(1). Let T : s(E) ’ s(E) be the multiplication operator with a polynomial p, i.e.,

T ((xk )k ) := (p(k)xk )k .

We claim that T is tame of degree d := deg(p) and base 0. For this we estimate as

follows:

= sup{(1 + k)n p(k) xk : k ∈ N}

T ((xk )k ) n

¤ Cn sup{(1 + k)n+d xk : k ∈ N} = Cn (xk )k n+d ,

|p(k)|

where d is the degree of p and Cn := sup{ (1+k)d : k ∈ N}. Note that Cn < ∞,

since k ’ (1 + k)d is not vanishing on N, and the limit of the quotient for k ’ ∞

is the coe¬cient of p of degree d.

This shows that s(E) is tamely equivalent to the same space, where the seminorms

are replaced by k (1 + k)n xk . In fact, the sums are larger than the suprema.

Conversely, k (1+k)n xk ¤ k (1+k)’2 (1+k)n+2 x k ¤ ’2

k (1+k) x n+2 ,

showing that the identity in the reverse direction is tame of degree 2 and base 0.

(2). Let T : Σ(E) ’ Σ(E) be the multiplication operator with an exponential

function, i.e., T ((xk )k ) := (ak xk )k .

We claim that T is tame of some degree and base 0. For this we estimate as follows:

enk ak xk = e(n+log(a))k xk

T ((xk )k ) =

n

k∈N k∈N

e(n+d)k xk = (xk )k

¤ n+d ,

k∈N

51.8