2 The normal map

Talk III: Regularity of Nash-Kuiper curves

Vincent Borrelli

July 1, 2013

In the previous talk, we have seen that the sequencef₁, f₂, ...ofC^∞-maps generated by the Nash-Kuiper process C¹-converges toward a C¹-isometry f∞ provided that

X pδk−δk−1 <+∞.

In this talk we adress theC²-regularity off∞. To avoid some technicalities in the computations we still assume that the Nash-Kuiper process is applied to a smooth initial mapf₀:E/Z→E² such that:

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

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

and that the target speed function is r ≡1.We also assume that the Nks are chosen among even numbers. The general case, slightly more technical in nature, is left to the reader.

Proposition 1.– For every k∈N^∗, f_k is of constant speed r_k and radially symmetric. In particular,

f_k =IC(fk−1, h_k, N_k).

The functionsα_k are also constant and equal to J₀⁻¹ _r

k−1

rk

.

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

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

k−1

rk

is constant. Since N_k∈2N^∗, we have h_k(x+1

2,{N_k(x+1

2)}) =−h_k(x,{N_kx})

and consequently

Z 1 0

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

It ensues that

IC(fk−1, h_k, N_k) =IC(ff k−1, h_k, N_k)

and therefore f_k is of constant speed kf_k⁰(x)k =kh_k(x,{N x})k = r_k. It is also radially symmetric sincefk(x) =h(x,{N_kx}). .

1 C

-regularity

Proposition 1.– For every x∈E/Z,we have

f_k⁰⁰(x) = (−2πα_kN_ksin 2πN_kx+rk−1scalk−1(x))ir_ke^{iαkcos 2πNkx}tk−1(x) where scal_k denotes the signed curvature off_k.Moreover

r_kscal_k(x) =r0scal0(x)−2π

k

X

l=1

α_lN_lsin(2πN_lx).

Remark 1.– Recall that two regular C² curves differ by a rigid motion if and only if they have the same signed curvature function.

Remark 2.– Let µk(x) := scalk(x)kf_k⁰(x)kdx be the signed curvature measure off_k.We have

µ_k(x) =µ_k−1(x)−2πα_kN_ksin(2πN_kx)dx.

The convex integration process modifies the curvature mesure in the sim- plest way by a cosine term with frequencyN_k.

Remark 3.– More generally, ifh_k(x, s) =r_ke^iψk(s)tk−1, one can show that µk(x) =Nkψ⁰_k(Nkx) +µk−1(x).

Thus, if we require to modify the curvature mesure by a single term of frequencyNk, we have to choose

ψ_k⁰(x) :=α_ksin(2πx+ phase difference).

This is the reason why in Talk II we have chosen ψ_k(x) =α_kcos 2πx.

Proof.– We have f_k⁰⁰(x) = ∂

∂x r_ke^{iαkcos 2πNkx}tk−1(x)

= ∂

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

= r_k rk−1

∂

∂x cos(α_kcos 2πN_kx)f_k−1⁰ (x) + sin(α_kcos 2πN_kx)if_k−1⁰ (x)

= −2iπα_kN_ksin(2πN_kx)r_ke^{iαkcos 2πNkx}tk−1(x) + r_k

rk−1

cos(α_kcos 2πN_kx)f_k−1⁰⁰ (x) + sin(α_kcos 2πN_kx)if_k−1⁰⁰ (x) Sincefk−1 is of constant speedrk−1 we have

f_k−1⁰⁰ (x) =rk−1scalk−1(x)if_k−1⁰ (x) therefore

f_k⁰⁰(x) = −2iπα_kN_ksin(2πN_kx)r_ke^{iαkcos 2πNkx}tk−1(x) +rkrk−1scalk−1(x)ie^{iαkcos 2πNkx}tk−1(x) . Finally,

f_k⁰⁰(x) = (−2πα_kN_ksin 2πN_kx+rk−1scalk−1(x))ir_ke^{iαkcos 2πNkx}tk−1(x).

Becausef_k is of constant arc length we also have

f_k⁰⁰(x) =r_kscal_k(x)if_k⁰(x) =r_kscal_k(x)ir_ke^{iαkcos 2πNkx}tk−1(x).

From this we deduce

r_kscal_k(x) =r_k−1scal_k−1(x)−2πα_kN_ksin(2πN_kx) and by induction

r_kscal_k(x) =r₀scal₀(x)−2π

k

X

l=1

α_lN_lsin(2πN_lx).

Lemma 2 (Amplitude Lemma).– We have

α_k∼ q

2(1−r²₀)p

δ_k−δk−1

where ∼denotes the equivalence of sequences.

Proof.– By definition αk = J₀⁻¹(^rk−1_r

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 α_k∼2 r

1− rk−1

r_k .

Sincer²₀+ (1−r₀²) = 1,we have

r²_k=r²₀+δ_k(1−r₀²) = 1 + (δ_k−1)(1−r²₀) so

r²_k−r²_k−1= (δ_k−δk−1)(1−r₀²) and

1−r_k−1²

r_k² = (δ_k−δk−1)(1−r²₀)

1−(1−δk)(1−r²₀) ∼(δ_k−δk−1)(1−r²₀).

In an other hand 1−r²_k−1

r²_k =

1−rk−1

r_k 1 +rk−1

r_k

∼2

1−rk−1

r_k

.

Thus

1−rk−1

r_k

∼ 1

2(δ_k−δk−1)(1−r₀²).

and

αk∼2 r

1−rk−1

r_k ∼ q

2(1−r₀²)p

δk−δk−1.

Proposition 2.– If P

k∈N^∗

pδ_k−δk−1N_k<+∞ then f∞ is C².

Proof.– Since we already know that the sequence (f_k)k∈NC¹ converges, it is enough to prove that (f_k⁰⁰)_k∈

Nis a Cauchy sequence. From f_k⁰⁰(x) =r_kscal_k(x)if_k⁰(x)

we deduce

kf_k⁰⁰(x)−f_k−1⁰⁰ (x)k_C0 ≤ kr_kscalk(x)f_k⁰(x)−rk−1scalk−1(x)f_k−1⁰ (x)k_C0

≤ kr_k−1scalk−1(x)f_k⁰(x)−rk−1scalk−1(x)f_k−1⁰ (x)k_C0

+|r_kscal_k(x)−rk−1scalk−1(x)|kf_k⁰(x)k_C0

≤ rk−1|scal_k−1(x)|kf_k⁰(x)−f_k−1⁰ (x)k_C0

+r_k|r_kscal_k(x)−rk−1scalk−1(x)|.

Since

r_kscal_k(x) =r0scal0(x)−2π

k

X

l=1

α_lN_lsin(2πN_lx) we have

|r_kscal_k(x)−rk−1scalk−1(x)| ≤2πα_kN_k and

rk|scal_k(x)| ≤r0|scal₀(x)|+ 2π X

l∈N^∗

αlNl.

In particular ther_k|scal_k(x)|are uniformly bounded by M :=kr₀scal₀(x)k_C0+ 2π X

k∈N^∗

α_kN_k.

Note thatM <+∞.Indeedα_k∼p

2(1−r²₀)p

δ_k−δk−1 therefore X

k∈N^∗

pδ_k−δk−1N_k<+∞ =⇒ X

k∈N^∗

α_kN_k <+∞.

We deduce

kf_k⁰⁰(x)−f_k−1⁰⁰ (x)k_C0 ≤Mkf_k⁰(x)−f_k−1⁰ (x)k_C0+ 2πα_kN_k. Letp < q, we thus have

kf_q⁰⁰(x)−f_p⁰⁰(x)k_C0 ≤ M

q

X

k=p

pδ_k−δk−1+ 2π

q

X

k=p

α_kN_k

≤ M

∞

X

k=p

pδk−δk−1+ 2π

∞

X

k=p

αkNk.

Hence f_k⁰⁰

k∈N is a Cauchy sequence.

2 The normal map

2.1 Analogy with a Riesz product

Theorem 1.– Let n_k be the normal map of f_k.We have

∀x∈E/Z, nk(x) =e^{iαkcos(2πNkx)}nk−1(x)

where α_k 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 mapn∞off∞has the following expression

∀x∈E/Z, n∞(x) =

+∞

Y

k=1

e^{iαkcos(2πNkx)}

! n0(x).

Proof.– This is straightforward from Lemma 3, Talk II and the fact that

n_k=it_k.

Theorem 1 puts into light some resemblance of n∞ with a Riesz product, that is, an infinite product

p(x) :=

∞

Y

j=1

(1 +α_jcos(2πN_jx)),

where (α_j)j∈N is a sequence of real numbers such that for every j ∈ N^∗,

|α_j| ≤1,and

∀j∈N^∗, Nj+1

N_j ≥3 +q for some fixedq >0. In particular, if

p(x) = 1 +

∞

X

ν=1

γνcos(2πνx)

is the Fourier expansion of p, then γ_Nj =α_j and γ_ν = 0 if ν is not of the form Nj1 ±Nj2 ±...±Njk, j1 > j2 > ... > jk [3]. Riesz products are well known to have a fractal structure. Precisely, their Riesz measures p(x)dx have a fractionnary Hausdorff dimension [4].

An interesting case of a Riesz structure occurs forαj =a^j and Nj =b^j for some positive numbers a, b with a < 1 and ab > 1. Indeed, in that case, A∞:=P

jα_jcos(2πN_jx) is the well-known Weierstrass function:

A∞(x) =X

j

a^jcos(2πb^jx).

Graph of a Weierstrass function with a= 0.5 andb= 4.

Although its exact value is conjectural, the Hausdorff dimension of its graph is larger than one [2]. It follows that the Hausdorff dimension of the graph of

n∞=





+∞

Y

j=1

e^{iajcos(2πbjx)}



n0(x).

is also larger than one.

2.2 Spectrum

The normal map n∞ can be thought of as a 1-periodic map from R toC.

Let

∀x∈E/Z, n_k(x) =X

p∈Z

a_p(k)e^2iπpx

denotes the Fourier series expansion of the normal mapnk. We derive from Theorem 1 the following inductive formula.

Fourier series expansion of nk.–We have

∀p∈Z, ap(k) =X

n∈Z

un(k)ap−nN_k(k−1) where u_n(k) =iⁿJ_n(α_k).

In the above formula, Jn denotes the Bessel function of order n(see [1] or [5]):

α7−→J_n(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 (a_p(k))p∈Z from the spectrum (a_p(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) = iⁿJ_n(α_k).Since

|J_n(α_k)| ↓0

the weight is decreasing withn(see the figure below).

A schematic picture of the various spectra(ap(k))_p∈Z withNk =b^k. Lemma (Jacobi-Anger identity).– For every x∈R+, we have

e^ixcosθ=

+∞

X

n=−∞

iⁿJn(x)e^inθ.

Proof of the Jacobi-Anger identity. – Since θ 7−→ e^ixcosθ is a C^∞ periodic function, it admits an expansion in Fourier series:

e^ixcosθ=

n=∞

X

n=−∞

c_n(x)e^inθ with

c_n(x) := 1 2π

Z 2π 0

e^ixcosθe^−inθdθ.

The change of variableθ−→π−θ shows that c_n(x) = 1

2π Z 2π

0

e^{ixcosθ−inθ}dθ

= 1

2π Z 2π

0

cos (xcosθ−nθ) dθ ifn is even

= 1

2π Z 2π

0

isin (xcosθ−nθ) dθ ifn is odd

Now, by arguments similar to the ones of Lemma 1 Talk II, we obtain

c_n(x) =iⁿJ_n(x).

Proof of the Fourier series expansion of n_k.– From the Jacobi-Anger identity

e^ixcosθ=

+∞

X

n=−∞

iⁿJ_n(x)e^inθ we deduce

e^{iαkcos(2πNkx)}=

+∞

X

n=−∞

iⁿJ_n(α_k)e^2iπnNkx =

+∞

X

n=−∞

u_n(k)e^2iπnNkx. Since the Fourier coefficients of a product of two fonctions are given by the discrete convolution product of their coefficients, the product

n_k(x) =e^{iαkcos(2πNkx)}nk−1(x) can be written

n_k(x) =

+∞

X

n=−∞

un(k)e^2iπnNkx

! _+∞

X

p=−∞

ap(k−1)e^2iπpx

!

=

+∞

X

p=−∞

+∞

X

n=−∞

un(k)ap−nNk(k−1)

! e^2iπpx.

Therefore

a_p(k) =

+∞

X

n=−∞

u_n(k)ap−nN_k(k−1).

References

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

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

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

[4] F. R. Keogh, Riesz products, Proc. London Math. Soc. 14A (1965), 174-182.

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