Complements to the paper "Lifting, degree, and the Distributional Jacobian Revisited"

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Complements to the paper ”Lifting, degree, and the Distributional Jacobian Revisited”

Jean Bourgain, Haïm Brezis, Petru Mironescu

To cite this version:

Jean Bourgain, Haïm Brezis, Petru Mironescu. Complements to the paper ”Lifting, degree, and the

Distributional Jacobian Revisited”. 2004. �hal-00747668�

COMPLEMENTS TO THE PAPER “LIFTING, DEGREE AND THE DISTRIBUTIONAL JACOBIAN REVISITED”

JEAN BOURGAIN

, HAIM BREZIS

AND PETRU MIRONESCU

1. Existence of a degree and optimal estimates.

Let 0 < s < ∞, 1 ≤ p < ∞ and set X = W

(S

; S

). We say that there is a (topological) degree in X if

a) C

(S

; S

) is dense in X;

b) the mapping g 7→ deg g, defined on C

(S

; S

), extends by continuity to X . We recall the following result, which is part of the folklore:

Lemma 1.1. There is a degree in X if and only if sp ≥ N .

Proof. Property a) holds for each s and p. When s is not an integer and sp < N , this was proved in [15]. When s = 1 and p < N , this assertion can be found in [4];

the same argument holds when s ≥ 2 is an integer and sp < N .

When sp > N , property a) follows immediately from the embedding W

֒ → C

. Finally, property a) when sp = N is essentially established in [13].

We next turn to property b). When sp > N , it is easy to see that the usual Brouwer degree of the (continuous) maps in W

has the required properties. When sp = N , we have W

֒ → V M O; in this case, the degree of VMO maps (studied in [13]) is the desired extension. Finally, we prove that b) does not hold when sp < N . We fix a map g ∈ C

( R

; S

) such that g(x) ≡ P when |x| ≥ 1 and deg g = 1;

here, P is the North pole of S

. Let π : S

→ R

be the stereographic projection and set g

(x) = g(kπ(x)), x ∈ S

. Then deg g

= 1, ∀k. However, it is easy to see that g

→ P strongly in W

and, therefore, the degree is not preserved in the strong limit.

In view of Lemma 1.1, it is natural to ask whether, for sp ≥ N , there is a control of the form

(1.1) | deg g| ≤ F (|g|

), ∀g ∈ W

(S

; S

).

Typeset byAMS-TEX

It follows from Theorem 0.6 and the Sobolev embeddings that the answer is yes. Indeed, if sp ≥ N and (s, p) 6= (1, 1), then there is some q > N such that W

(S

; S

) ֒ → W

(S

; S

). On the other hand, if s = p = N = 1, we have the estimate

| deg g| =

¯ ¯

¯ ¯ 1 2iπ

Z

S¹

˙ g g

¯ ¯

¯ ¯ ≤ 1

2π |g|

.

We next examine the optimality of the estimates in Theorem 0.6, which we restate in a slightly more general form.

Theorem 1. Let 1 ≤ p < ∞ and g ∈ W

(S

; S

). Then (1.2) | deg g| ≤ C

|g|

.

Proof. When p > N , this is the content of Theorem 0.6. When p = N , estimate (1.2) is an immediate consequence of the Kronecker formula

(1.3) deg g = 1

|S

| Z

S^N

det(∇g).

Finally, when 1 ≤ p < N , (1.2) follows from (1.3) and the Gagliardo-Nirenberg type inequality

(1.4) |g|

≤ |g|

kgk

. Estimate (1.2) is optimal in the following sense:

Lemma 1.2. For 1 ≤ p < ∞, there is a sequence (g

) ⊂ W

(S

; S

) such that |g

|

→ ∞

and deg g

≥ C

|g

|

.

Proof. Let h : R

→ S

be such that h(x) ≡ P for |x| ≥ 1. Then, clearly,

|h ◦ π|

∼ |h|

. In view of this remark, it suffices to construct a sequence (h

) ⊂ W

( R

; S

) such that |h

|

→ ∞, h

≡ P for |x| ≥ 1, deg h

≥ C

|h

|

. Fix a map g ∈ C

( R

; S

) such that deg g = 1 and g(x) ≡ P for |x| ≥ 1. For k ≥ 1, we fix k distinct points a

, . . . , a

∈ B

. Let

g

(x) = P + X

j=1

(g − P )

µ x − a

λ

¶

, λ > 0.

It is easy to see that

|g

|

−→ k|g|

.

In addition, for sufficiently small λ, the map g

is S

-valued, has degree k, and

equals P for |x| ≥ 1. If we set, for sufficiently large λ

, h

= g

, then |h

|

∼

k and deg h

= k.

2. Existence of a distributional Jacobian.

As in the previous section, we discuss whether, given 0 < s < ∞, 1 ≤ p < ∞, there is a notion of a distributional Jacobian in W

(S

; S

). As noted in the discussion before Theorem 0.8, the answer is yes in W

(S

; S

), and therefore also in W

(S

; S

) if sp ≥ N (via the Sobolev embeddings). On the other hand, there is no natural notion of distribution Jacobian if sp < N . Indeed, in this case C

(S

; S

) is dense in W

(S

; S

) (this follows from [4] and [5]). Let g : S

→ S

, g(x

, x

) = x

|x

| , for which Det(∇g) 6= 0. Consider a sequence (g

) ⊂ C

(S

; S

) such that g

→ g in W

. If a natural Det(∇) would exist, this would yield

0 = lim

k

Det (∇g

) = Det (∇g) 6= 0, impossible .

However, the answer given by Theorem 0.8 is not completely satisfactory. Indeed, the perfect analogs of a), b) in Section 1 are, for 0 < s < ∞, 1 ≤ p < ∞ such that N ≤ sp < N + 1:

a

) that the class R

= {g ∈ W

(S

; S

); g ∈ C

except a finite set, g ∈ W

} is dense in W

(S

; S

);

b

) that Det(∇) extends by continuity from R

to W

.

The proof of Theorem 0.8 combined with the Sobolev embeddings shows that b

) holds, provided a

) holds. However, we established a

) only for 0 < s < 1; when s = 1, a

) holds also, see [4]. It is plausible that a

) holds for any s.

Concerning the estimate

(2.1) k Det (∇g)k

≤ C|g|

, g ∈ W

(S

; S

), p > N, Theorem 2.4 implies its optimality.

3. The closure of C

(S

; S

).

As we have already noted, C

(S

; S

) in dense in W

(S

; S

) if sp < N or sp ≥ N + 1. It is easy to see that this is not true if N ≤ sp < N + 1.

We mention the following straightforward generalization of a result due to Bethuel [19] when p = N .

Theorem 2. Let N < p < ∞. For g ∈ W

(S

; S

), the following are equivalent:

a) g ∈ C

(S

; S

)

N/p,p

; b) Det (∇g) = 0.

A proof is presented in [20].

It is plausible that the assumptions s < 1, sp = N are irrelevant here.

Question. Let 0 < s < ∞, 1 ≤ p < ∞ be such that N ≤ sp < N + 1. It is true that

g ∈ C

(S

; S

)

N/p,p

⇔ Det (∇g) = 0?

4. An Alternative proof of Theorem 0.1.

In this section, we present another argument that yields the estimate

(4.1) |ϕ|

≤ C

(|e

|

+ |e

|

), 1 < p < ∞, ϕ ∈ W

(I).

We start with some preliminary results.

4.1. Basic estimates.

If ϕ ∈ L

(I ), with I ⊂ R interval, we set ϕ

= 1

|I|

Z

I

ϕ.

Lemma 4.1. Let ϕ ∈ C

((−ρ, ρ)). Assume that

(4.2) |e

|

≤ C

and

(4.3) |ϕ|

+ |ϕ|

≤ C

. Then

(4.4) |ϕ

− ϕ

| ≤ C(1 + C

)(1 + C

).

Proof. We start by introducing some notations. For 0 < l

< l

≤ ρ, set f (l

) = ϕ

− ϕ

,

h(l

, l

) = ϕ

− ϕ

.

Let C

= 2 + 2C

. If |f (ρ)| ≤ 10

C

, there is nothing to prove. Otherwise, assume, e.g., f (ρ) > 0 and set t = f (ρ)

C

> 10

. Let J =

· t 2

¸

− 1. For j = 1, . . . , J, we will construct inductively 0 < ρ

< . . . < ρ

< ρ such that, for j = 1, . . . , J − 1,

(4.5) f(ρ

) − f (ρ

) ∈ [C

, 2C

];

(4.6) ρ

≥ 2ρ

;

(4.7) dist (h(ρ

, ρ

), 2π Z ) > 1 2 .

Assume the ρ

’ s constructed, for the moment. By Corollary A.5, it follows that

|e

|

j,ρj+1)∪(−ρj+1,−ρj))

≥ C, j = 1, . . . , J − 1, and thus

|e

|

≥ C(J − 1) ≥ C

f (ρ), from which the conclusion of the lemma follows.

It remains to construct the ρ

’s. Let ρ

be the first l > 0 such that f(l) = C

. Assuming ρ

, . . . , ρ

constructed (j < J ), let a be the largest l > 0 such that f (l) = f (ρ

) + C

and let b be the smallest l > a such that f(l) = f (ρ

) + 2C

.

We claim that

(4.8) a ≥ 4ρ

(4.9) for at least one l ∈ [a, b], it holds that dist (h(ρ

, l), 2π Z ) > 1 2 .

Properties (4.5) - (4.7) follow immediately from (4.8) - (4.9); it suffices to take ρ

= l, where l ∈ [a, b] is such that (4.9) holds. It is also clear from our construc- tion that the ρ

’s exist up to j = J .

Proof of (4.8).

By Lemma A.1, we have

C

= |f(a) − f(ρ

)| ≤|ϕ

− ϕ

)| + |ϕ

− ϕ

| ≤ ρ

a

¡ |ϕ|

+ |ϕ|

¢

≤ ρ

a C

, and (4.8) follows from our choice of C

.

Proof of (4.9). Argue by contradiction and assume that dist (h(ρ

, l), 2π Z ) ≤ 1

2 , ∀ l ∈ [a, b].

Since l 7→ h(ρ

, l) is continuous, there is some fixed d ∈ Z such that

|h(ρ

, l) − 2πd| ≤ 1

2 , ∀ l ∈ [a, b].

In particular, |h(ρ

, a) − h(ρ

, b)| ≤ 1. By Lemma A.2, we have h(ρ

, a) = a

a − ρ

f (a) − ρ

a − ρ

f (ρ

) = f (ρ

) + C

a a − ρ

, and similarly

h(ρ

, b) = f(ρ

) + 2C

b b − ρ

. Thus

C

¯ ¯

¯ ¯ 2 b

b − ρ

− a a − ρ

¯ ¯

¯ ¯ ≤ 1.

Since 2 b

b − ρ

− a

a − ρ

≥ 1 2

a

a − ρ

, we find that C

a

a − ρ

≤ 2, which is impossible, since C

≥ 2.

Corollary 4.2. Let J, K be two adjacent intervals and ϕ ∈ C

(J ∪ K ). Assume that

(4.10) |e

|

≤ C

and

(4.11) |ϕ|

+ |ϕ|

≤ C

. Then

(4.12) |ϕ|

≤ C (1 + C

)(1 + C

).

Proof. Let L ⊂ J ∪ K be an interval. We have to prove that 1

|L|

Z

L

|ϕ − ϕ

| ≤ C(1 + C

)(1 + C

).

If L ⊂ J or L ⊂ K, this is clear. Otherwise, assume, e. g., L = (−a, b), with (−a, 0) ⊂ J and (0, b) ⊂ K . By Lemma A.3, we have

1

|L|

Z

L

|ϕ − ϕ

| ≤ 3(|ϕ|

+ |ϕ|

+ |ϕ

− ϕ

|), and the conclusion follows from Lemma 4.1.

We will also need the following variant of Lemma 4.1

Lemma 4.3. Let 0 < ρ

≤ 1

4 ρ and ϕ ∈ C

((ρ

, ρ) ∪ (−ρ, −ρ

)).

Assume that

(4.13) |e

|

≤ C

and

(4.14) |ϕ|

≤ C.

Then

(4.15) |(ϕ

− ϕ

) − (ϕ

− ϕ

) ¯

¯ ≤ C(1 + C

)(1 + C

).

Proof. Let C

= 2 + 3C

. We may assume, e. g., that tC

= ¡

ϕ

− ϕ

¢

− ¡

ϕ

− ϕ

¢

≥ 10

C

. Let J =

· t 4

¸

− 1. We construct inductively ρ

, j = 1, . . . , J, as follows: set ρ

= 4ρ

. Assume ρ

, . . . , ρ

already constructed such that

(4.16) ¡

ϕ

− ρ

¢

− ¡

ϕ

− ϕ

¢ ∈ [C

, 2C

];

(4.17) ρ

≥ 2ρ

;

(4.18) dist ¡

ϕ

− ϕ

, 2π Z ¢

> 1 2 , k = 1, . . . , j.

Let a be the largest l > ρ

such that

¡ ϕ

− ϕ

¢

− ¡

ϕ

− ϕ

¢

= C

, and let b be the smallest l > a such that

¡ ϕ

− ϕ

¢

− ¡

ϕ

− ϕ

¢

= 2C

. As in the proof of Lemma 4.1, we claim that

(4.19) a ≥ 2ρ

;

(4.20) there is some l ∈ [a, b] such that dist ¡

ϕ

− ϕ

, 2π Z ¢

> 1 2 .

Proof of (4.19). We have, by Lemma A.1, C

= ¯

¯ (ϕ

− ϕ

) − (ϕ

− ϕ

) ¯

¯

≤ ¯

¯ϕ

− ϕ

¯

¯ + ¯

¯ϕ

− ϕ

¯

¯

≤ a − ρ

ρ

− ρ

|ϕ|

+ a − ρ

ρ

− ρ

|ϕ|

≤ a − ρ

ρ

− ρ C

, so that

a ≥ C

C

ρ

− C

+ C

C

ρ

≥ 2ρ

;

the last inequality follows from the inequalities C

≥ 3C

and ρ

≥ 4ρ

.

Proof of (4.20). Argue by contradiction. As in the proof of Lemma 4.1, it follows that

¯ ¯ ¡

ϕ

− ϕ

¢

− ¡

ϕ

− ϕ

¢ ¯

¯ ≤ 1.

Starting from the identity

ϕ

= a − ρ

a − ρ

ϕ

− ρ

− ρ

a − ρ

ϕ

, we obtain, as in the proof of Lemma 4.1, that

C

¯ ¯

¯ ¯ 2 b − ρ

b − ρ

− a − ρ

a − ρ

¯ ¯

¯ ¯ ≤ 1.

As in the proof of Lemma 4.1, this implies that C

a − ρ

a − ρ

≤ 2, which is impossible, since C

≥ 2.

The remaining part of the proof of Lemma 4.3 is identical to the one of Lemma 4.1 and will be omitted.

Proof of Theorem 0.1. We may assume that ϕ ∈ C

∩ W

. As explained in the main paper, when |e

|

is sufficiently small, (4.1) follows from the inequality

|g|

≤ C |g|

combined with

Lemma 4.4 ([14]). Let ϕ ∈VMO(I ). There are constants C > 0, δ > 0 such that (4.21) |ϕ|

≤ C|e

|

if |e

|

≤ δ.

Let γ = min(δ

, δ). It suffices to establish (4.1) when |e

|

≥ δ. Let N be the smallest integer ≥ |g|

γ . We consider a partition of I with N successive intervals I

, . . . , I

chosen such that |e

|

1∪···∪Ij)

= jγ, j = 1, . . . , N − 1. Thus

|e

|

j)

≤ γ, ∀ j.

It suffices to establish the estimate

(4.22) 1

|I | Z

I

|ϕ − ϕ

| ≤ C|e

|

. In view of lemmas A.6 and 4.4, we have

1

|I|

Z

I

|ϕ − ϕ

| ≤ Cγ + 1

|I |

X

j,k

|I

||I

||ϕ

− ϕ

|, so that (4.22) bounds to proving

(4.23) 1

|I|

X

j,k

|I

||I

||ϕ

− ϕ

| ≤ C|e

|

.

The remaining part of the proof is devoted to estimating the differences

|ϕ

− ϕ

|. Without any loss of generality, we will assume j = 1.

Lemma 4.5. Assume that |I

| = |I

| ≥

max |I

|. Let l

be such that |I

| = max{|I

|; 2 ≤ l ≤ k − 1} and set J = I

∪ · · · ∪ I

. Assume that |I

| ≤ 4|J |, and consider the following intervals

picture

(with | I ¯

| = | I ¯

| = | I ˜

| = | I ˜

| =

|I

|). Then

(4.24) |ϕ

− ϕ

| ≤ |ϕ

− ϕ

| + |ϕ

− ϕ

| + C µ

1 + log |I |

|I

| .

¶

Proof. We have

(4.25) |ϕ

−ϕ

| ≤ |ϕ

−ϕ

|+|ϕ

−ϕ

|+|ϕ

−ϕ

l0

|+|ϕ

l0

−ϕ

k

|+|ϕ

k

−ϕ

|.

By Lemma A.7, we have

(4.26) |ϕ

− ϕ

| + |ϕ

− ϕ

l0

| + |ϕ

k

− ϕ

| ≤ C µ

1 + log |I|

|I

¶

,

and the conclusion follows.

Lemma 4.6. Same hypotheses as above, except that we assume |I

| > 4|J |. Let I

, I

be as below.

picture

(with |I

| = |I

| = 4|J |). Then (4.27)

|ϕ

− ϕ

| ≤ |ϕ

− ϕ

| + |ϕ

l0

− ϕ

k

| + C µ

1 + log 4|J|

|I

| + |e

|

1\I₁^∗)∪(Ik\I_k^∗))

¶ .

Proof. We have

(4.28) |ϕ

− ϕ

| ≤ |ϕ

− ϕ

| + ¯

¯ (ϕ

− ϕ

) − (ϕ

− ϕ

) ¯

¯ . By Lemma 4.5, we have

(4.29) |ϕ

− ϕ

| ≤ |ϕ

− ϕ

| + |ϕ

l0

− ϕ

k

| + C µ

1 + log 4|J |

|I

|

¶ .

On the other hand, Lemma 4.3 (with (−ρ

, ρ

) replaced by J and (−ρ, ρ) replaced by I

∪ J ∪ I

) yields

(4.30) |(ϕ

− ϕ

) − (ϕ

− ϕ

k

)| ≤ C ³

1 + |e

|

1\I₁^∗)∪(Ik\I_k^∗))

´ , and the conclusion follows.

Corollary 4.7. If |I

| = |I

| ≥

max

2≤l≤k−1

|J

|, then (with l

as above)

(4.31)

|ϕ

− ϕ

| ≤|ϕ

− ϕ

| + |ϕ

− ϕ

|+

C µ

1 + log min{4|J |, |I

|}

|I

| + |e

|

1\I₁^∗)∪(Ik\I_k^∗))

¶ . (Here, I

\ I

and I

\ I

could be empty).

Lemma 4.8. Assume that |I

| = |I

| ≥

max

2≤l≤k−1

|J

|. Then (4.32) |ϕ

− ϕ

| ≤ C(k + |e

|

).

Proof. We start by applying Corollary 4.7. We note that, by construction, we may

apply again Corollary 4.7 to the consecutive intervals ¯ I

, I

, . . . , I

, I ¯

, respec-

tively to ˜ I

, I

, . . . I

, I ˜

; next we iterate this procedure.

We find that

(4.33) |ϕ

− ϕ

| ≤

2(k−1)

X

l=1

|ϕ

− ϕ

| + 2C(k − 1) + Σ

+ Σ

.

Here, J

, K

are adjacent intervals of equal length, each one contained into one of the original I

’s; Σ

is the sum of the logarithmic terms, while Σ

is the sum of the

| |

terms. Lemma 4.1 implies that

(4.34)

2(k−1)

X

l=1

|ϕ

− ϕ

| ≤ C (k − 1) ≤ Ck.

On the other hand, the | |

terms we consider appear on disjoint intervals, and thus

(4.35) Σ

≤ C|e

|

. Therefore,

(4.36) |ϕ

− ϕ

| ≤ Ck + C|e

|

+ Σ

.

Claim We have Σ

≤ Ck. In order to prove the claim, we give a formal description of how Σ

is computed.

Let Z

= {2, . . . , k − 1} and let s

∈ Z

be such that |I

| = max |I

| (with the notations used up to now, we have s

= l

). Let Z

= {2, . . . , s

− 1}, Z

= {s

+ 1, . . . , k − 1} if s

is closer to 2 than to k − 1; otherwise, let Z

= {s

+ 1, . . . , k − 1}, Z

= {2, . . . , s

− 1}. We have

|Z

| = 1 + |Z

| + |Z

|, |Z

| ≤ |Z

|.

Assuming Z

constructed, we proceed to constructing Z

and Z

as above.

More specifically, if Z

6= φ, we pick s

∈ Z

such that |I

| = max

s∈Zc

|I

|. We next write Z

\ {s

} = Z

⊔ Z

, with Z

, Z

intervals of integers and

|Z

| ≤ |Z

|. If Z

= ∅, we stop.

If Z

= {m, m + 1, . . . , n}, then the corresponding term in Σ

is of the form

(4.37) log min{|K |, 4 P

s∈Zc

|I

|}

|I

| ;

here, K is an interval contained in I

and of length ≤ min{|I

, |I

|}.

Assume c 6= ∅. If ˆ c is the predecessor of c, we have either m − 1 ∈ Z

, or n + 1 ∈ Z

, and thus |K | ≤ |I

|.

In conclusion,

(4.38) Σ

≤ C log min{4|J|, |I

|}

|I

| + C X

c6=∅

log 4 min{|I

|, P

s∈Zc

|I

|}

|I

|

| {z }

Rc

.

Setting R

= min{|J|, |I

|}

|I

| , the claim amounts to proving that

(4.39) X

c

log(4R

) ≤ Ck.

This is an immediate consequence of the two following Lemma 4.9. We have

(4.40) X

c

log R

≤ Ck + X

c

log |Z

|.

Lemma 4.10. We have

(4.41) X

c

log |Z

| ≤ Ck.

Proof of Lemma 4.9. Let t be the largest integer such that Z

1, 1, . . . , 1

| {z } t times

,

0) 6= φ.

Set a

= Z

, a

= Z

, . . . , a

= Z

1, 1, . . . , 1

| {z } t times

,

0) , x

= R

, x

= R

, . . . , x

= R (1, . . . , 1)

| {z } t times

.

Writing

Z

= (Z

∪ {s

}) ∪ (Z

∪ {s

}) ∪ (Z

∪ {s

}) ∪ . . . , we find that

R

≤ P

s∈Zφ

|I

|

|I

| = X

s∈Z₍₀₎∪{sφ}

|I

|

|I

| + X

s∈Z_(1,0)∪{s1}

|I

|

|I

| + X

s∈Z_(1,1,0)∪{s_(1,1,0)}

|I

|

|I

| +. . . ,

so that

R

≤ 2|Z

| + 2|Z

| |I

|

|I

| + 2|Z

| |I

|

|I

| + . . . . Since

|I

|

|I

| ≤ 1

R

, |I

|

|I

| = |I

|

|I

| · |I

|

|I

| ≤ 1

R

R

, . . . , we obtain

(4.42) 1

2 x

≤ a

+ a

x

+ a

x

x

+ · · · + a

x

x

. . . x

and similarly

(4.43)



 

 



 

 

 1

2 x

≤ a

+ a

x

+ · · · + a

x

. . . x

.. .

1

2 x

≤ a

+ a

x

.

Noting that x

= 1, we find from (4.42) - (4.43) by backward induction on j that

x

. . . x

≤ X

m=1

2

X

J⊂{j,...,t}

|J|=m

Y

l∈J

a

.

In particular, since a

≥ 1, ∀j, we obtain (4.44)

x

. . . x

≤ X

m=1

2

X

J⊂{0,...,t}|J|=m

Y

l∈J

a

≤ X

m=1

2

µ t + 1 m

¶

Y

0

a

≤ 3

Y

0

a

.

Similarly, for any fixed ¯ c we have

(4.45) Y

c contains only 1’s

R

≤ Y

ccontains only 1’s

(3|Z

|).

Since each c can be uniquely written as c = (¯ c, 0, ¯ ¯ c) where ¯ c ¯ contains only 1’s, by multiplying the inequalities of type (4.45) we find that

Y

c

R

≤ 3

Y

c

|Z

|,

from which the conclusion of the lemma follows.

Proof of Lemma 4.10. Let, for l ≥ 0, S

= {c ; |Z

| ∈ [2

, 2

)}. We claim that [c 6= c

, c, c

∈ S

] ⇒ Z

∩ Z

= φ.

Argue by contradiction and assume that Z

∩ Z

6= φ.

Then, for example, we have Z

⊂

6=

Z

, so that Z

⊂ Z

, by construction.

Thus |Z

| ≤ 1

2 |Z

| ≤ 1

2 |Z

|, which is impossible if c, c

∈ S

. Therefore, Y |Z

| =

[log₂k]+1

Y

l=1

Y

c∈Sl

|Z

| ≤

[log₂k]+1

Y

l=1

2

≤ Y

l≥1

2

= 2

,

where A = P

l≥1

l2

.

Lemma 4.11. Assume that |I

| ≥ |I

|, l = 2, . . . , k. Then (4.46) |ϕ

− ϕ

| ≤ C

µ

k + |e

|

1∪···∪Ik)

+ log |I

|

|I

|

¶ .

Proof. Let l

= 1 and define inductively l

such that |I

| = max

lj−1<l≤k

|I

|. Then

|ϕ

− ϕ

| ≤ X

j≥1

|ϕ

− ϕ

|.

Let ¯ I

be as follows:

picture

(such that | I ¯

| = |I

|). We may apply Lemma 4.7 to the sequence of intervals I ¯

, . . . , I

, and find that

(4.47) |ϕ

− ϕ

| ≤ C(s

− s

+ |e

|

sj−1∪···∪I_sj)

).

On the other hand, Lemma A.7 yields (4.48) |ϕ

− ϕ

| ≤ C

µ

1 + log |I

|

|I

|

¶ .

By summing up all the inequalities of type (4.47)-(4.48), we find that (4.46) holds.

Lemma 4.12. For each j, k we have (4.49) |ϕ

− ϕ

| ≤ C

µ

|k − j| + |e

|

+ log |I |

|I

||I

|

¶ .

Proof. Assume j = 1. If I

(or I

) is the largest among the intervals I

, . . . , I

, the conclusion follows from Lemma 4.11. Otherwise, let l ∈ {2, . . . , k − 1} be such that

|I

| ≥ |I

|, t = 1, . . . , k. By Lemma 4.11, we have (4.50) |ϕ

− ϕ

| ≤ C

µ

(l − 1) + |e

|

1∪···∪Il)

+ log |I

|

|I

|

¶

and

(4.51) |ϕ

− ϕ

| ≤ C µ

(k − l) + |e

|

l∪···∪Ik)

+ log |I

|

|I

|

¶ , from which the lemma follows.

Corollary 4.13. We have

(4.51) |ϕ

− ϕ

| ≤ C Ã

N + |e

|

+ |I | p |I

||J

|

! .

Proof of Theorem 0.1. We have to estimate the r.h.s. of (4.23). In view of Corollary 4.13, we have

1

|I|

X

j,k

|I

||I

||ϕ

− ϕ

| ≤C ³

N + |e

|

´ + C

|I | X

j,k

|I

|

|I

|

≤C ³

N + |e

|

´

≤ C|e

|

, since N ≤ C|e

|

.

5. An improvement of Theorem 0.1 and the answer to OP2 when N = 1.

If I ⊂ R is an interval and g : I → C , we set, for δ > 0, J (g, δ, I) =

Z Z

{(x,y)∈I²;|g(x)−g(y)|≥δ}

1

|x − y|

.

In this section, we prove the following generalization of Theorem 0.1.

Theorem 3. For sufficiently small δ > 0, we have

(5.1) |ϕ|

≤ C(δ + J (e

, δ, I )), ∀ ϕ ∈ C

(I; R ).

An immediate consequence is the following

Theorem 4. Let g ∈ C

(S

; S

). Then, for sufficiently small δ > 0, we have (5.2) | deg g| ≤ CJ (g, δ, S

).

This answer OP 2 when N = 1.

Proof of Theorem 4. By Lemma A.8, we have |g|

≤ δ + 2J (g, δ, S

). Recall that deg g = 0 provided |g|

is sufficiently small (see [13]). Thus (5.2) holds (for small δ > 0) provided J(g, δ, S

) is sufficiently small.

When J(g, δ, S

) is not to small, estimate (5.2) is obtained from (5.1) in the same way (0.6) follows from Theorem 0.1.

Proof of Theorem 3. The proof is the same as the one of Theorem 0.1, except that

| |

has to be replaced by J (g, δ, I). The only two places where | |

comes into the picture are the inequality

(5.3) |g|

≤ |g|

and Lemma A.4 (together with Corollary A.5). The substitute of (5.3) is Lemma A.8. The analog of Lemma A.4/Corollary A.5 are Lemma A.9/Corollary A.10 presented into the appendix.

Appendix. Elementary properties of averages.

Lemma A.1. Let J ⊂ K . Then |ϕ

− ϕ

| ≤ |K|

|J | |ϕ|

. Proof. We have

|ϕ

−ϕ

| = 1

|J |

¯ ¯

¯ ¯ Z

J

(ϕ−ϕ

)| ≤ 1

|J | Z

K

|ϕ−ϕ

¯ ¯

¯ ¯ = |K |

|J | 1

|K|

Z

K

|ϕ−ϕ

| ≤ |K |

|J | |ϕ|

. The following identities are trivial:

Lemma A.2. Let J, K be two adjacent intervals. Then (A.1) ϕ

− ϕ

= |K |

|J| + |K | (ϕ

− ϕ

), ϕ

= |J| + |K |

|J | ϕ

− |K |

|J| ϕ

and

(A.2) ϕ

= |K |

|J| + |K | ϕ

+ |J|

|J | + |K| ϕ

.