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The Nash-Kuiper process for curves

Vincent Borrelli, Sa¨ıd Jabrane, Francis Lazarus and Boris Thibert January 18, 2013

An isometric immersion of a Riemannian manifold into an Euclidean space is a C1 map f : (Mm, g) −→ Eq = (Rq,h., .i) such that fh., .i = g.

Such a map preserves the length of curves that is:

Length(f ◦γ) =Length(γ)

for every rectifiable curve γ : [a, b] −→ Mm. In a local coordinate system x= (x1, ..., xm) the isometric condition gives rise to a system ofsm= m(m+1)2 equations

1≤i≤j≤n,

∂f

∂xi

(x), ∂f

∂xj

(x)

=gx

∂xi

, ∂

∂xj

.

Thus generically, isometric maps are expected to exist –at least locally– if the target space has dimension greater thansm. In 1926, Janet [10] proved that any analytic Riemannian surface (M2, g) admit local isometric analytic immersions in Es2. Shortly after, this result was generalized by Cartan [5]

for analytic Riemannian manifold of dimensionm: local isometric analytic maps do exist if the dimension of the target Euclidean space is at leastsm. Thirty years after the result of Janet, J. Nash showed in an outstanding ar- ticle [15] that everyCkRiemannian manifold (3≤k≤+∞) can be mapped Ck isometrically into an Euclidean space Eq with q = 3sm+ 4m if Mm is compact andq = (m+1)(3sm+4m) if not. This result was then improved by Gromov [12] and G¨unther [9] who proved thatq= max{sm+2m, sm+m+5}

is enough for the compact case.

Another amazing result of J. Nash is the discovery that, in a C1 setting, the barrier formed by the Janet dimension can be completely destroyed: in the compact case, if a Riemannian manifold admits an immersion into some Eq, q ≥m+ 1, then it admits a C1 isometric immersion into the same Eu- clidean space (Nash [14] proved the caseq≥m+ 2 and Kuiper [13] the case

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q =m+ 1). As a consequence every compact riemannian surface admits a C1 isometric immersion inE3 but in general, for obvious curvature reasons, the immersion can not be enhanced to be C2.

Beyond the breaking of the dimensional barrier, there is another phenomenon which is utterly baffling in the Nash-Kuiper result: not only C1 isometric maps do exist but they are plentiful! In fact, there is a C1 isometric map near every strictly short map. A mapf0 : (Mm, g)−→ Eq is called strictly shortif it strictly shortens distances, that is, if the difference g−f0h., .i is a metric. The Nash-Kuiper approach reveals that if f0 is a strictly short embedding, then for every > 0 there exists a C1 isometric embedding f : (Mm, g)−→Eq such that

kf−f0kC0

wherek.kC0 denotes the supremum norm overMm(this manifold is assumed to be compact for the simplicity of the presentation). For instance, for every >0,there is a C1 isometric embedding of the unit sphere inside a ball of radius.

Recently [4], we have converted the Nash-Kuiper proof into an algorithm, using the Gromov convex integration theory ([12], [16], [7]). We have imple- mented this algorithm and produced numerical pictures of a C1 isometric embeddingf of the square flat torusE2/Z2 insideE3 that isC0 close to a strictly short embeddingf0 ofE2/Z2 as a torus of revolution. Our algorithm generates a sequence of maps

f0, f1,1, f1,2, f1,3, f2,1, f2,2, f2,3, ...

defined recursively thatC1 converges towardf.The geometry of the limit map consists merely of the behavior of its tangent planes or, equivalently, of the properties of its Gauss mapn:E2/Z−→ S2 ⊂E3. From the algo- rithm, one can extract a formal expression of that Gauss map as an infinite product ofcorrugation matricesapplied to the initial Gauss map off0.One major obstacle to the understanding of n lies in the inherent complexity of the coefficients of these corrugation matrices. The main theorem of [4]

(the Corrugation Theorem) describes their asymptotic behaviour.

In this article, we propose to study the normal map of isometric maps result- ing from a convex integration process in the simpler situation of isometric immersions of the circle E/Z into E2. In this case, the isometric problem

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in itself is totally trivial but the way the Nash-Kuiper process solves it, produces a sequence of curves

f0, f1, f2, ...

whose limit f has a non trivial geometry. Of course, in that one dimen- sional setting, some of the difficulties inherent to the dimension two vanish.

In particular, if the initial curve f0 :E/Z−→ E2 'C is parametrized with constant speed and is radially symmetric (see the definition below) all com- putations can be completely carried out and lead to an explicit formula for the normal mapn of the limit curve f.

Theorem 1.– Let nk be the normal map of fk.We have

∀x∈E/Z, nk(x) =ekcos(2πNkx)nk−1(x)

where αk∈]0,π2[ is the amplitude of the loop used in the convex integration to build fk−1 from fk and Nk ∈ 2N is the number of corrugations of fk

(precise definitions below). In particular, the normal mapnof f has the following expression

∀x∈E/Z, n(x) =

+∞

Y

k=1

ekcos(2πNkx)

! n0(x).

The above expression of the normal mapn is reminiscent of aRiesz prod- uct, that is a product of the form

h(x) =

+∞

Y

k=1

(1 +αkcos(2πNkx)).

It is a fact that an exponential growth ofNk, known as Hadamard’s lacunary condition, results in a fractional Hausdorff dimension of the Riesz measure1 µ:=h(x)dx[11].

The normal mapncan be thought of as a 1-periodic map fromRtoC.In

§3 we perform its Fourier series expansion. Its spectrum, whose structure is very similar to the spectrum of a Riesz product, is obtained as a limit of an

1Letdimsupµ(resp. diminfµ) denotes the supremum (resp. the infimum) of the Haus- dorff dimension of the Borel sets of positiveµ-measure. Ifd=dimsupµ=diminfµ then the measureµis said to have Hausdorff dimensiond.

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iterative process starting with the spectrum of the initial mapn0.Precisely, let

∀x∈E/Z, nk(x) =X

p∈Z

ap(k)e2iπpx

denotes the Fourier series expansion of the normal mapnk. We derive from the above theorem the following inductive formula (cf. Lemma 3):

Fourier series expansion of nk.–We have

∀p∈Z, ap(k) =X

n∈Z

un(k)ap−nNk(k−1) where un(k) =inJnk).

In the above formula, Jn denotes the Bessel function of order n(see [1] or [17]):

α7−→Jn(α) = 1 π

Z π

0

cos(nu−αsinu)du.

The Fourier expansion of nk gives the key to understand the construction of the spectrum (ap(k))p∈Z from the spectrum (ap(k−1))p∈Z. The k-th spectrum is obtained by collecting an infinite number of shifts of the previous spectrum. The n-th shift is of amplitude nNk and weighted by un(k) = inJnk).Since

|Jnk)| ↓0

the weight is decreasing withn(see the figure of§3).

In the Nash-Kuiper process there is a infinite number of degrees of freedom in the construction of the sequence (fk)k∈N.In particular, given any sequence of positive numbers (δk)k∈N increasing toward 1, the process produces a sequence such that

kfk0 −fk−10 kC0 ≤Ctep

δk−δk−1. Thus, if

X pδk−δk−1<+∞

the sequence (fk)k∈Nis C1 converging toward a C1 limitf.Moreover if X pδk−δk−1Nk <+∞

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thenfis C2 (see Proposition 5). Regarding the intermediary regularities, we prove the following:

Theorem 2.– Assume that

X pδk−δk−1 <+∞ and X p

δk−δk−1Nk= +∞.

Let 0< η <1 andSk :=Pk

l=1

l−δl−1Nl. If X(δk−δk−1)1−η2 Sηk <+∞

thenf is C1,η.

In the simplified one dimensional approach followed in this article, the se- quence (Nk)k∈N can be chosen freely. This is no longer possible in the general case: some constraints appear that force the Nks to be increasing.

The control of the growth of the Nks is then the key to understand the C1,η regularity of the limit map. In the original proof of Nash, the chosen sequence forδk was 1−2−(k+1).For such a choice, the numerical result we have obtained for the square flat torus seems to suggest that the sequence (Nk)k∈Nis exponentially growing (see also the theoretical arguments of [6]).

This gives the motivation for the following corollary.

Corollary 3.– Let 0< γ <1 and δk:= 1−e−γ(k+1). If there exists β >0 such that

∀k∈N, Nk≤N0eβk thenf is C1,η for any η >0 such that

η < γ 2β.

The question of the C1,η regularity of isometric maps resulting from the Nash-Kuiper process is addressed in [2], [3] and [6]. The optimalC1,η reg- ularity of an isometric immersion of a Riemannian surface in E3 is still an open question.

1 The convex integration process for curves

The convex integration process.–Letf0: [0,1]→R2 be aCmap and let

h: [0,1] −→ C(R/Z,C) x 7−→ h(x, .)

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be a C family of loops such that

∀x∈[0,1], Z 1

0

h(x, s)ds=f00(x).

LetN ∈Nthe any natural number. We define a newCmapf : [0,1]−→

R2 by the formula

∀x∈[0,1], f(x) :=f0(0) + Z x

0

h(s,{N s})ds

where{N s}denotes the fractional part ofN s.We call such a formula giving a new mapf from the data off0 andh aconvex integration. We sometimes write

f :=IC(f0, h, N).

The new mapf has a derivative whose image obviously lies inside the image ofh since

∀x∈[0,1], f0(x) =h(x,{N x}).

Moreover,f remainsC0 close tof0.Indeed, it can be shown that kf−f0kC0 =O

1 N

(see [4] for instance).

Curves with given speeds.– Letf0 : [0,1]−→E2'Cbe a regular curve (∀x∈[0,1], f00(x)6= 0) and let r: [0,1]−→R+ be anyC map such that

∀x∈[0,1], r(x)>kf00(x)k.

Leth defined by

h(x, s) :=r(x) (cos(α(x) cos 2πs)t0(x) + sin(α(x) cos 2πs)n0(x)) witht0 := kff000

0k,n0:=it0 and α(x)∈]0, κ[ is such that r(x)J0(α(x)) =kf00(x)k

where J0 denotes the Bessel function of the first kind and of order 0 and κ'2.4 denotes the first zero ofJ0.Since the Bessel functionJ0 is decreasing

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on the interval [0, κ] and J0(0) = 1, there is a unique α(x) that solves the above implicite equation. Note that

Z 1 0

h(x, s)ds=r(x)J0(α(x))t0(x)

therefore the above implicit condition on α(x) implies that the average of h(x, .) is f00(x). The map f obtained by convex integration from f0 and h has speedkf0kequal to the given functionrand is arbitrarilyC0 close tof0. Closed curves with given speeds.– Iff0is defined overE/Zrather than [0,1] the curvef obtained fromf0 andh by convex integration is not closed in general. This defect can be easily corrected by the following modification of the convex integration formula:

∀x∈[0,1], f(x) :=f0(0) + Z x

0

h(s,{N s})ds−x Z 1

0

h(s,{N s})ds.

For short we writef :=IC(ff 0, h, N).The C0 closeness implies that

Z 1 0

h(x, s)ds

=O 1

N

so that the correction can be made arbitrarily small. We still have kf − f0kC0 =O N1

but nowkf0k is only approximately equal tor(x),precisely

∀x∈E/Z, kf0(x)k=

h(x,{N x})− Z 1

0

h(s,{N s})ds

and therefore, for all x∈E/Z,we have |kf0(x)k −r(x)|=O N1 .

Nash and Kuiper process.– In the spirit of the Nash and Kuiper proof, the way to obtain a mapf :E/Z−→E2'C with speed the given function r is to produce a sequence of closed curves (fk)k∈N by iteratively applying the modifying convex integration formula so that to reduce step by step the isometric defaultr− kf00k.

Let (δk)k∈N be a sequence of increasing positive number converging to- ward 1, we set

∀k∈N,∀x∈E/Z, r2k(x) :=kf00(x)k2k r2(x)− kf00(x)k2 .

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Note that for everyx ∈E/Z, the sequence rk(x) is increasing towardr(x).

We definefk to be IC(ff k−1, hk, Nk) with

hk(x, s) :=rk(x)ek(x) cos 2πstk−1(x) whereαk(x) =J0−1kf0

k−1(x)k rk(x)

andtk−1is the normalized derivative offk−1. Eachfk has a speed which is approximatelyrk:

kfk0(x)k −rk(x) =O

1 Nk

.

Since the sequencerk(x) is strictly increasing for everyx∈E/Z,the number Nk can be chosen large enough such that

∀x∈E/Z, rk+1(x)>kfk(x)k.

This is crucial to definefk+1 asIC(ff k, hk+1, Nk+1).If the sequence (δk)k∈N

is chosen so that

X pδk−δk−1<+∞

and if (Nk)k∈Nis rapidly diverging then the sequencefk:=ICf(fk−1, hk, Nk) isC1 converging toward aC1 limitfwith speed kf0 k=r. This is proven further in the text in the particular case of closed curves with constant speed.

The general case, slightly more technical in nature, is left to the reader.

Closed curves with constant speed.– From now on, in order to get the most pleasant computations we consider the simplified case wherer≡1 and f0:E/Z→E2 is aC map such that:

• (Cond1) it is of constant speedr0:=kf00k<1

• (Cond2) it is radially symmetric, that is: f00(x+12) =−f00(x).

In all what follows, we will also assume that the Nash-Kuiper sequence of Cmaps derived from f0:

fk:=IC(ff k−1, hk, Nk), k∈N

is such that hk(x, s) = rkek(x) cos 2πstk−1(x) and Nk ∈ 2N. Note that (Cond1) implies that every functionrk =p

r20k(1−r02) is constant.

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Proposition 1.– For every k∈N, fk is of constant speed rk and radially symmetric. In particular,

fk =IC(fk−1, hk, Nk).

The functionsαk are also constant and equal to J0−1 r

k−1

rk

.

Proof.– By induction. Assume thatfk−1 satisfies (Cond1) and (Cond2).

In particular fk−1 is of constant speed rk−1 and thus the function αk = J0−1

rk−1

rk

is constant. Since Nk∈2N, we have hk(x+1

2,{Nk(x+1

2)}) =−hk(x,{Nkx}) and consequently

Z 1 0

hk(s,{N s})ds= 0.

It ensues that

IC(fk−1, hk, Nk) =IC(ff k−1, hk, Nk)

and therefore fk is of constant speed kfk0(x)k =khk(x,{N x})k = rk. It is also radially symmetric sincefk(x) =h(x,{Nkx}). .

· · ·

The convex integration process applied to circle (left f0, center f1,right f)

2 C

1

convergence

It turns out that the sequence (δk)k∈N mainly determines the sequence (αk)k∈N.

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Lemma 1 (Amplitude Lemma).– We have αk

q

2(1−r20)p

δk−δk−1

where ∼denotes the equivalence of sequences. We also have αk ≤ 1

r0

q

2(1−r02)p

δk−δk−1.

Proof.– By definition αk = J0−1(rk−1r

k ). Recall that the Taylor expansion ofJ0(α) up to order 2 is

w= 1− α2

4 +o(α2).

Lety= 1−w andX =α2, we havey= X4 +o(X) thus X= 4y+o(y) and soX∼4y. We finally get

α∼2√

1−w and αk∼2 r

1− rk−1

rk . Sincer20+ (1−r02) = 1,we have

r2k=r20k(1−r02) = 1 + (δk−1)(1−r20) so

r2k−r2k−1= (δk−δk−1)(1−r02) and

1−rk−12

rk2 = (δk−δk−1)(1−r20)

1−(1−δk)(1−r20) ∼(δk−δk−1)(1−r20).

In an other hand 1−r2k−1

r2k =

1−rk−1

rk 1 +rk−1

rk

∼2

1−rk−1

rk

.

Thus

1−rk−1

rk

∼ 1

2(δk−δk−1)(1−r02).

and

αk∼2 r

1−rk−1

rk

∼ q

2(1−r02)p

δk−δk−1.

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The Taylor expansion ofJ0 up to order 4 shows that w≤1−α2

4 +α4 64 =

1− α2

8 2

(because it is alternating) and hence α2k ≤8

1−

rrk−1

rk

. Thus

α2k ≤ 8

√rk

√rk−√ rk−1

≤ 8

√rk(√ rk+√

rk−1)(rk−rk−1)

≤ 8

√rk(√ rk+√

rk−1)(rk+rk−1) r2k−r2k−1

≤ 2

r02 rk2−rk−12 sincer0 < rk−1 < rk.We deduce

αk≤ 1 r0

q

2(r2k−r2k−1) = 1 r0

q

2(1−r02)p

δk−δk−1.

Let (Ak)k∈N be the sequence of functions defined by

∀x∈E/Z, Ak(x) :=

k

X

l=1

αlcos(2πNlx).

Lemma 2.– For every x∈E/Z,we have:

fk0(x) =eiAk(x)rk

r0f00(x).

Proof.– Letnk−1 :=itk−1.From

fk0(x) = rk(cos(αkcos(2πNkx))tk−1(x) + sin(αkcos(2πNkx))nk−1(x))

= rkekcos(2πNkx) 1 rk−1

fk−10 (x) we deduce by induction : fk0(x) =eiAk(x)rrk

0f00(x).

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Proposition 2.– If P p

δk−δk−1 <+∞ then

i) the sequence(Ak)k∈Nis normally converging andA:= limk→+∞Ak is continuous.

ii) the sequence (fk)k∈N is C1 converging toward f := limk→+∞fk

and

∀x∈R/Z, f0 (x) =eiA(x) 1 r0f00(x).

Proof.– From the Amplitude Lemma we deduce that Xαk<+∞

thus the sequence (Ak)k∈N is normally converging and A:= lim

k→+∞Ak is continuous. Moreover, from the relation

fk0(x) =eiAk(x)rk r0f00(x) we also deduce that fk0

k∈N is normally converging toward eiA(x)1

r0

f00(x).

Since (fk(0))k∈N obviously converges, we obtain that the sequence (fk)k∈N

isC1 converging towardf:= limk→+∞fk.

Corollary 1.– Let γ >0and δk:= 1−e−γ(k+1).Then sequence (δk)k∈N is increasing toward 1 andp

δk−δk−1∼√

δ0eγ2k.In particular, f isC1.

3 The normal map

From now on, we assume

X pδk−δk−1<+∞

so that the sequence (fk)k∈N is C1 converging toward its limit f. The following theorem is a straightforward consequence of the results of the pre- ceding section:

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Theorem 1.– Let nk be the normal map of fk.We have

∀x∈E/Z, nk(x) =ekcos(2πNkx)nk−1(x)

In particular, the normal mapn of f has the following expression

∀x∈E/Z, n(x) =eiA(x)n0(x).

We deduce from this theorem the following result about Fourier expansion ofnk.

Lemma 3 (Fourier expansion of nk).– For all k∈Nwe denote by

∀x∈E/Z, nk(x) =X

p∈Z

ap(k)e2iπpx the Fourier expansion ofnk. We have

∀p∈Z, ap(k) =X

n∈Z

un(k)ap−nNk(k−1)

where un(k) =inJnk) (Jn denotes the Bessel function of order n).

Proof.– From the Jacobi-Anger identity eizcosθ =

+∞

X

n=−∞

inJn(z)einθ we deduce

ekcos(2πNkx)=

+∞

X

n=−∞

inJnk)e2iπnNkx =

+∞

X

n=−∞

un(k)e2iπnNkx. Since the Fourier coefficients of a product of two fonctions are given by the discrete convolution product of their coefficients, the product

nk(x) =ekcos(2πNkx)nk−1(x) can be written

nk(x) =

+∞

X

n=−∞

un(k)e2iπnNkx

! +∞

X

p=−∞

ap(k−1)e2iπpx

!

=

+∞

X

p=−∞

+∞

X

n=−∞

un(k)ap−nNk(k−1)

! e2iπpx.

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Therefore

ap(k) =

+∞

X

n=−∞

un(k)ap−nNk(k−1).

A schematic picture of the various spectra (ap(k))p∈Z withNk =bk. Remark.– The analogy with Riesz products suggests that the Hausdorff dimension of the graph of the normal map n could be fractional. Note that the relevant part of this map is the 1-periodic function

R3x7−→A(x) =

+∞

X

k=1

αkcos(2πNkx)∈R.

In the simple case where αk = ak and Nk = bk with 0 < a < 1 < b , the mapA is a Weiestrass function2. Ifab >1,it is known that its graph has a fractional Hausdorff dimension. The exact value of this dimension is still an open question. It is believed to be equal to 2 +lnlnab (see [8]).

4 C

1,η

regularity

Proposition 3.– We have

kfk0 −fk−10 kC0 ≤Cte1

k−δk−1

2Cf. Lemma 1 and the lines above Corollary 3 in the introductory part of this article for a motivation for such choice forαk andNk.

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withCte1=p

7(1−r20).

Proof.– For every pointx∈E/Z, we have

kfk0 −fk−10 k2 =kfk0k2+kfk−10 k2−2kfk0kkfk−10 kcos(αkcos 2πNkx) sinceαkcos 2πNkxis the angle betweenfk0(x) andfk−10 (x).An upper bound for this angle isαk =J0−1(w) wherew=rk−1/rk∈]0,1[ since

rk =kfk0(x)k and rk−1=kfk−10 (x)k.

Recall that from the Amplitude Lemma we have the following inequality αk2

2 ≤4(1−√ w).

By using the upper boundαk, we obtain

kfk0 −fk−10 k2 ≤ rk2+rk−12 −2rk−1rkcosαk

≤ rk2−rk−12 + 2rk−1(rk−1−rkcosαk).

Since

cosαk≥1−αk2 2 we have

rk−1(rk−1−rkcosαk) ≤ rk−12 −rkrk−1+rk−1rkα2k 2

≤ rk−12 −rkrk−1+ 4rk−1rk

1−

rrk−1

rk

≤ rk−12 + 3rk−1rk−4rk−1

√rkrk−1

≤ rk−12 + 3r2k−4rk−1

q

rk−12 ( since rk−1< rk)

≤ 3(r2k−r2k−1).

Therefore

kfk0 −fk−10 k2 ≤7 kfk0k2− kfk−10 k2 . Now

kfk0k2− kfk−10 k2 = r2k−rk−12

= (δk−δk−1)(1−r02).

Finally

kfk0 −fk−10 kC0 ≤Cte1p

δk−δk−1

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withCte1 =p

7(1−r20).

For everyk ∈N, we denote by Mk(g) the supremum over E/Z of the k-th derivative g(k) of g : E/Z −→ C (if k = 0, it is understood that g(0) = g) and we definekgkCk to be the sumM0(g) +...+Mk(g).

Corollary 2.– We have

kfk−fk−1kC1 ≤2Cte1

k−δk−1

withCte1=p

7(1−r20).

Proof.– From the theorem we deduce by a mere integration kfk−fk−1kC0 ≤Cte1

k−δk−1

thus the result since

kfk−fk−1kC1 =kfk−fk−1kC0+M1(fk−fk−1).

Proposition 4.– For every x∈E/Z,we have

fk00(x) = (−2παkNksin 2πNkx+rk−1scalk−1(x))irkekcos 2πNkxtk−1(x) where scalk denotes the signed curvature offk.Moreover

rkscalk(x) =r0scal0(x)−2π

k

X

l=1

αlNlsin(2πNlx).

Proof.– We have fk00(x) = ∂

∂x rkekcos 2πNkxtk−1(x)

= ∂

∂x(rk(cos(αkcos 2πNkx)tk−1(x) + sin(αkcos 2πNkx)nk−1(x))

= rk rk−1

∂x cos(αkcos 2πNkx)fk−10 (x) + sin(αkcos 2πNkx)ifk−10 (x)

= −2iπαkNksin(2πNkx)rkekcos 2πNkxtk−1(x) + rk

rk−1

cos(αkcos 2πNkx)fk−100 (x) + sin(αkcos 2πNkx)ifk−100 (x) Sincefk−1 is of constant speedrk−1 we have

fk−100 (x) =rk−1scalk−1(x)ifk−10 (x)

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therefore

fk00(x) = −2iπαkNksin(2πNkx)rkekcos 2πNkxtk−1(x) +rkrk−1scalk−1(x)iekcos 2πNkxtk−1(x) . Finally,

fk00(x) = (−2παkNksin 2πNkx+rk−1scalk−1(x))irkekcos 2πNkxtk−1(x).

Becausefk is of constant arc length we also have

fk00(x) =rkscalk(x)ifk0(x) =rkscalk(x)irkekcos 2πNkxtk−1(x).

From this we deduce

rkscalk(x) =rk−1scalk−1(x)−2παkNksin(2πNkx) and by induction

rkscalk(x) =r0scal0(x)−2π

k

X

l=1

αlNlsin(2πNlx).

Proposition 5.– If P

k∈N

k−δk−1Nk<+∞ then f is C2.

Proof.– Since we already know that the sequence (fk)k∈NC1 converges, it is enough to prove that (fk00)k∈

Nis a Cauchy sequence. From fk00(x) =rkscalk(x)ifk0(x)

we deduce

kfk00(x)−fk−100 (x)kC0 ≤ krkscalk(x)fk0(x)−rk−1scalk−1(x)fk−10 (x)kC0

≤ krk−1scalk−1(x)fk0(x)−rk−1scalk−1(x)fk−10 (x)kC0

+|rkscalk(x)−rk−1scalk−1(x)|kfk0(x)kC0

≤ rk−1|scalk−1(x)|kfk0(x)−fk−10 (x)kC0

+rk|rkscalk(x)−rk−1scalk−1(x)|.

Since

rkscalk(x) =r0scal0(x)−2π

k

X

l=1

αlNlsin(2πNlx)

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we have

|rkscalk(x)−rk−1scalk−1(x)| ≤2παkNk and

rk|scalk(x)| ≤r0|scal0(x)|+ 2π X

l∈N

αlNl.

In particular therk|scalk(x)|are uniformly bounded by M :=kr0scal0(x)kC0+ 2π X

k∈N

αkNk.

Note thatM <+∞.Indeedαk∼p

2(1−r20)p

δk−δk−1 therefore X

k∈N

k−δk−1Nk<+∞ =⇒ X

k∈N

αkNk <+∞.

We deduce

kfk00(x)−fk−100 (x)kC0 ≤Mkfk0(x)−fk−10 (x)kC0+ 2παkNk. Letp < q, we thus have

kfq00(x)−fp00(x)kC0 ≤ M

q

X

k=p

k−δk−1+ 2π

q

X

k=p

αkNk

≤ M

X

k=p

k−δk−1+ 2π

X

k=p

αkNk.

Hence fk00

k∈N is a Cauchy sequence.

Theorem 2.– Assume that

X pδk−δk−1 <+∞ and X p

δk−δk−1Nk= +∞.

Let 0< η <1 andSk :=Pk l=1

l−δl−1Nl. If

X(δk−δk−1)1−η2 Sηk <+∞

thenf is C1,η.

Proof.– Let 0< η <1.We are going to use the interpolation inequality kfkC1,η ≤Ctekfk1−η

C1 kfkη

C2

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to show that (kfk−fk−1kC1,η)k∈N is a Cauchy sequence. From the above sections, we have

kfk−fk−1kC1 ≤2Cte1

k−δk−1

and

M2(fk−fk−1) ≤ M2(fk) +M2(fk−1)

≤ M0(rk scalk)M1(fk) +M0(rk−1 scalk−1)M1(fk)

≤ M0(scalk) +M0(scalk−1)

≤ 2M0(scal0) + 4π

k

X

l=1

αlNl.

From the Amplitude Lemma we deduce M2(fk−fk−1) ≤ 2M0(scal0) +

2(1−r20) r0

Pk

l=1

l−δl−1Nl

≤ 2M0(scal0) +

2(1−r20) r0 Sk. So

kfk−fk−1kC2 ≤2Cte1p

δk−δk−1+ 2M0(scal0) +4πp

2(1−r20) r0

Sk.

Since limk→+∞Sk= +∞, for klarge enough we have kfk−fk−1kC2 ≤Cte2Sk. for some constantCte2. We now have

kfk−fk−1k1−ηC1 kfk−fk−1kηC2 ≤Cte3k−δk−1)1−η2 Skη

withCte3 = (2Cte1)1−ηCteη2.

Corollary 3.– Let 0< γ <1 and δk:= 1−e−γ(k+1). If there exists β >0 such that

∀k∈N, Nk≤N0eβk thenf is C1,η for any η >0 such that

η < γ 2β.

(20)

Proof.– We have

δk−δk−10e−γk thus

Sk=

k

X

l=1

l−δl−1Nl≤p δ0N0

k

X

l=1

e(β−γ2)l<p

δ0N0eβ−γ21−e(β−γ2)(k+1) 1−eβ−γ2 Suppose first thatβ > γ

2.We then have : Sk≤Cte4e(β−γ2)k. Finally

k−δk−1)1−η2 Skη ≤Cte5e−γ1−η2 keη(β−γ2)k. Now

−γ1−η 2 +η

β−γ 2

<0 if and only if

η < γ 2β. Therefore, under that condition

X(δk−δk−1)1−η2 Sηk <+∞

hence the corollary in the case whereβ > γ

2.We left to the reader the easier caseβ ≤ γ

2.

References

[1] M. Abramowitz, I. Stegun, Handbook of Mathematical Functions, Dover, 1965.

[2] J. Borisov,C1,α-isometric immersions of Riemmannian spaces, Dokl.

Akad. Nauk. SSSR 163 (1965), 11-13, translation in Soviet. Math. Dokl.

6 (1965), 869-871.

[3] J. Borisov, Irregular C1,β-surfaces with an analytic metric, Siberian Math. J. 45, 1 (2004), 19-52.

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[4] V. Borrelli, S. Jabrane, F. Lazarus, B. Thibert, Flat tori in three-dimensional space and convex integration, PNAS 2012, vol 109, 7218-7223.

[5] E. Cartan,Sur la possibilit´e de plonger un espace riemannien donn´e dans un espace euclidien, Ann. Soc. Pol. Math., 6 (1927), 1-7.

[6] S. Conti, C. De Lellis and L. Szkelyhidi,H-principle and rigidity forC1,α-isometric embeddings, to appear in the Proceedings of the Abel Symposium 2010, arXiv:0905.0370v1.

[7] Y. Eliahsberg, N. Mishachev, Introduction to the h-principle, Graduate Studies in Mathematics, Vol 48, AMS, Providence, 2002.

[8] K. Falconer,Fractal Geometry, Wiley, 2003.

[9] M. G¨unther,On the pertubation problem associated to isometric em- beddings of Riemannian manifolds, Ann. Global Anal. Geom. 7(1989), 69-77.

[10] M. Janet, Sur la possibilit´e de plonger un espace riemannien donn´e dans un espace euclidien, Ann. Soc. Pol. Math., 5 (1926), 38-43.

[11] J.-P. Kahane, Jacques Peyri`ere et les produits de Riesz, arXiv.org/abs/1003.4600v1.

[12] M. Gromov, Partial Differential Relations, Ergebnisse der Mathe- matik und ihrer Grenzgebiete (3) [Results in Mathematics and Related Areas (3)], 9. Springer-Verlag, Berlin, 1986.

[13] N. Kuiper,OnC1-isometric imbeddings, Indag. Math., 17 (1955), 545- 556.

[14] J. Nash,C1-isometric imbeddings, Ann. of Math. (2), 60 (1954), 383- 396.

[15] J. Nash, The imbedding problem for Riemannian manifolds, Ann. of Math. (1), 63 (1956), 20-63.

[16] D. Spring,Convex Integration Theory, Bikhauser, 1998.

[17] G. N. Watson, A Treatise on the Theory of Bessel Functions, Cam- bridge University Press, 1995.

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