<< . .

. 86
( : 97)

. . >>

and in the real analytic case look at the proof of (50.12).
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]


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
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
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
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 ∈

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 .

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

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

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) =

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

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
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.
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 :

ds |s=0
exp (X)(Y )(x) = exp (X + sY )(x)
d 2tπ
ds |s=0
= Fl (1, x) = Y (x + k ) dt,

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

‚s |s=0 (X + sY )(Fl
= (t, x))
= Y (Fl (t, x)) + X (. . . )
= Y (x + t 2π ) + 0,

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



‚s |s=0 Fl (t, x) = Y (x + „ ) d„.

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-
tinuous (for the c∞ -topology).

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

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-
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 k
t’c (x) = (. . . ; 1 xk ; . . . ) = (?, . . . , ?; 2 xk ; ?, . . . ),
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 ’ 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.
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
the same inequality for the given k, and hence k∈N »’1 (»k (B)) is bounded as well.
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 |.

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

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)),
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 ).

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
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
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

∀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
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

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
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 :=

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

= 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 ,

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 ) =
k∈N k∈N

e(n+d)k xk = (xk )k
¤ n+d ,


<< . .

. 86
( : 97)

. . >>