Influencia del ambiente familiar en el hábito de lectura

Most of the work in the proof will be common to both parts of the theorem. We will be referring to the±of the second part with the understanding that in the case of the first part, we could just set±₌1.

We start off with¥₌≤, taken from Theorem 3.2.5, and will assume throughout that

k!_kL2_(B1)∑¥, with the understanding that the upper bound¥will be lowered at finitely many stages during the proof. For our weak solutionuto (4.4) corresponding to!, we will assume, without loss of generality, that ¯u=_|B11|

R

B1u=0.

To begin with we use Rivière’s decomposition of!(Theorem 3.2.5) in order to rewrite the equation (4.4) (equations (4.10) and (4.11) below). Theorem 3.2.5 gives us A ₂ W1,2(B1,GLm(R))\L1(B1,GLm(R)), B 2W₀1,2(B1,g lm(R)) andC =C(m)<1so that rA_°A!_=r?B (4.8) and

krA_kL2_(B1)+krBk_L2_(B1)+kdist(A,SO(m)k_L1_(B1)∑Ck!k_L2_(B1). (4.9) (We have actually replacedAbyA°1_{here, compared to how it appears in the state-} ment of Theorem 3.2.5.)

Now by (4.8) we have

div(Aru) = rA.ru+A¢u

= rA._ru_°A!._ru_°A f

and trivially

curl(A_ru)_=r?A._ru. (4.11) We note here that the above equations only hold in a weak sense, and more care should be taken in their calculation. We illustrate this for (4.10): A prioridiv(A_ru) is a distri- bution, so for¡₂C_c1(B1) we have

div(A_ru)[¡] _{= °} Z B1 A_ru._r¡ = Z B1 (_rA._ru)¡_{° r}(¡A)._ru = Z B1

(_rA._ru)¡_°(A!._ru)¡_°A f¡ sinceuweakly solves (4.4)

=

Z B1

(r?B.ru°A f)¡=(r?B.ru°A f)[¡].

We will now essentially carry out a Hodge decomposition ofA_ruinB1using the ex- pressions (4.10) and (4.11). We first extend all the quantities arising above to functions onR2.

LetE x:W1,2(B1)!W1,2(R2)\W₀1,2(B2) be a bounded extension operator. Denote ˜

u=E x(u)2W1,2(R2,Rm) and note that since we are assumingR_B1u=0, by the Poincaré inequality and by standard properties ofE xwe have

ku˜kW1,2_(R2₎∑Ckuk_W1,2_(B1)∑Ckruk_L2_(B1) andu₌u˜inB1.

For A, first let ˆA = A°_|B1_1|RB1A and ˜A =E x( ˆA)2W

1,2_(R2_{,g l}

m(R)). Noting that R

B1Aˆ=0 and using the same argument as foruwe have

kA˜kW1,2_(R2₎∑CkrAk_L2_(B1)

here we have used that_rA˜_=rAˆ_=rAinB1. Notice also that ˜A_ru˜₊≥_|B1_1|R_B1A¥_ru˜₌ A_ruinB1.

We extendf andBby zero (without relabelling), so by Remark 2.7.6, f 2h1(R2) with

kfkh1_(R2₎∑Ckfk_LlogL(B1). Now we define

D:₌N[_r?B._ru˜], E:₌N[_r?A˜._ru˜],

F :_=°N[A f],

whereN is the Newtonian potential (see section 2.2.3). Note that the quantity A f is well defined on the whole ofR2by the definition off. Finally let

H:₌A˜_ru˜₊ µ ₁ |B1| Z B1 A ∂ ru˜_{° r}D_{° r}F_{° r}?E. The first thing to notice aboutHis that

H=Aru° rD° rF° r?E (4.12) inB1. Hence we have

div(H)₌div(A_ru)_°¢(D₊F)₌div(A_ru)_{° r}?B._ru˜₊A f ₌0

weakly inB1, and a similar calculation shows curl(H)₌0 weakly inB1. (Again care must be taken in checking these.) ThereforeHis harmonic inB1(i.e. corresponds to a harmonic 1-form).

Supposer₂(0,1]. (For some estimates later it will need to be less than1₂.)

Without loss of generality, we may assume that ±₂(0,1]. (Recall that when ad- dressing the first part of the theorem, we are just setting±₌1.) For¥small enough, depending on±, we may assume (by (4.9)) that A is close to a special-orthogonal ma- trix in the sense that bothAandA°1_{change the length of any vector by at most a factor} of 1₊±. Therefore

kru_k2_L2_(Br)∑(1+±)2kAruk2_L2_(Br)

∑(1+3±)kAruk2_L2_(Br)

∑(1₊4±)_kH_k2_L2_(Br)+C(krDk2_L2_(Br)+krFk2_L2_(Br)+krEk2_L2_(Br)),

(4.13)

whereC is dependent on±. In order to obtain the inequalities of Theorem 4.2.2 we estimate_kH_kL2_(Br),krDk_L2_(Br),krFk_L2_(Br)andkrEk_L2_(Br).

First we consider _rD_=rN[_r?_{B.ru˜] andrE =rN[r}?_A˜_{.ru˜]. Notice that by the} work of Coifman-Lions-Meyer-Semmes [CLMS93] and the fact that

is a bounded linear operator (see Remark 2.7.4) we have, krD_kL2_(B1)+krEk_L2_(B1) ∑ C°krDk_L2,1_(B1)+krEk_L2,1_(B1)¢ = C°krN[r?B.ru˜]kL2,1_(B1)+krN[r?A˜.ru˜]k_L2,1_(B1)¢ ∑ C°_kr?B._ru˜_kH1_(R2₎+kr?A˜.ru˜kH1_(R2₎¢ ∑ C°_krB_kL2_(B1)+krAk_L2_(B1)¢kruk_L2_(B1) ∑ Ck!_kL2_(B1)kruk_L2_(B1) (4.14) where we have also used the continuous embeddingL2,1,_!L2and the estimate from Theorem 3.2.5.

ForrF =°rN[A f] we use (Theorem 2.2.1) that the Riesz potentialrN:L1(B1)! L2,1(B1) is a bounded operator; alsorN:h1(R2)!L2,1(B1) is bounded (Remark 2.7.4). We will also use the following: L2,1_{is the dual ofL}2,1_{; if f 2LlogL(B}

1) then for any g ₂L1_{,g f 2LlogL(B}

1) and kg fkLlogL(B1)∑ kgkL1kfkLlogL(B1) and finally we use the continuous embeddingLlogL(B1),!h1(R2). We have

krF_k2_L2_(B1) ∑ CkrFkL2,1_(B1)krFk_L2,1_(B1)

∑ CkA fkL1_(B1)kA fk_h1_(R2₎

∑ C_kf_kL1_(B1)kA fk_LlogL(B1)

∑ C_kf_kL1_(B1)kfk_LlogL(B1). (4.15) Also, using merely the boundedness ofrN:L1(B1)!L2,1(B1),!L1(B1), we have

krFkL1_(B1)∑Ckfk_L1_(B1). (4.16) From here, we proceed differently in order to prove the two different parts of the theorem. For the first part, we now estimate_kH_kL1_(B2/3)and apply standard estimates for harmonic functions in order to estimatekHkL2_(Br): Using estimates (4.14) and (4.16), we have

kHkL1_(B2/3) ∑ C°kruk_L1_(B2/3)+krDk_L1_(B2/3)+krEk_L1_(B2/3)+krFk_L1_(B2/3)¢

∑ C°_ku_kL1_(B1)+kfk_L1_(B1)+k!k_L2_(B1)kruk_L2_(B1)¢ (4.17) where we have also used the estimate

which follows from Propositions 2.3.1 and 2.4.2.

Since H is harmonic we have pointwise estimates onH and its derivatives on the interior ofB2/3in terms of_kH_kL1_(B2/3), and in particular

kH_kL1_(B1/2)+krHk_L1_(B1/2) ∑ CkHk_L1_(B2/3)

∑ C(_ku_kL1_(B1)+kfk_L1_(B1)+k!k_L2_(B1)kruk_L2_(B1)) by (4.17). Therefore if we considerr2(0,1₂], then

kHk2_L2_(Br) ∑ ºr2kHk2_L1_(Br) ∑ Cr2≥_ku_k2_L1_(B1)+kfk2_L1_(B1)+k!k2_L2_(B1)kruk2_L2_(B1) ¥ , (4.18) and krH_kL1_(Br) ∑ ºr2krHk_L1_(Br) ∑ Cr2°kukL1_(B1)+kfk_L1_(B1)+k!k_L2_(B1)kruk_L2_(B1)¢. (4.19) Now, looking back at inequality (4.13) and using (4.14), (4.15) and (4.18) we have

kru_k2_L2_(Br) ∑ C ≥

k!_k2_L2_(B1)kruk2_L2_(B1)+r2kuk2_L1_(B1)+kfkL1_(B1)kfk_LlogL(B1) ¥

which is the first inequality that we seek from the first part of the theorem.

In order to get the second estimate (4.6) of the first part of the theorem, we return to the Hodge decomposition (4.12) which tells us that

ru₌A°1(H_+rD_+rF_+r?E) inB1, and therefore r2u_=rA°1.(H_+rD_+rF_+r?E)₊A°1(_rH_+r2D_+r2F_+rr?E), with kr2ukL1_(Br) ∑ krA°1.(H+rD+rF+r?E)k_L1_(Br) (=I) + kA°1(_rH_+r2D_+r2F_+rr?E)_kL1_(Br). (=II) (4.20)

Using (4.9), (4.14), (4.15) and (4.18) we have (assuming also¥_<1) I _{∑ kr}A°1_kL2_(Br)kH+rD+rE+r?Fk_L2_(Br)

∑ C°rk!_kL2_(B1)kuk_L1_(B1)+k!k_L2_(B1)kruk_L2_(B1)+kfk_LlogL(B1)¢. (4.21) Now we use the fact that the operatorsr2N :h1(R2)!L1(B1) (Remark 2.6.15) and

r2N:H1(R2)!L1(R2) (Theorem 2.6.4) are bounded and the estimates (4.9) and (4.19) to conclude

II ∑ kA°1kL1_(Br)krH+r2D+r2F+rr?Ek_L1_(Br)

∑ C°r2_ku_kL1_(B1)+k!k_L2_(B1)kruk_L2_(B1)+kfk_LlogL(B1)¢. (4.22) Since we have assumed without loss of generality thatR_B1u₌0, by an application of the Poincaré inequality and looking back at (4.20), (4.21) and (4.22) we have

kr2u_kL1_(Br)∑C(k!k_L2_(B1)kruk_L2_(B1)+r2kuk_L1_(B1)+kfk_LlogL(B1)), as desired.

For the second part of the theorem, we return to a generalr ₂(0,1]. We will now controlH using the standard decay estimate

kH_k2_L2_(Br)∑r2kHk2_L2_(B1) which holds since_|H_|2_{is subharmonic.}

Then using (4.13) and (4.14) and (4.15) again, we find that

kru_k2_L2_(Br)∑(1+4±)kHk2_L2_(Br)+C(krDk2_L2_(Br)+krFk2_L2_(Br)+krEk2_L2_(Br)) ∑(1+4±)r2kHk2_L2_(B1)+C(krDk2_L2_(B1)+krFk2_L2_(B1)+krEk2_L2_(B1)) ∑(1₊5±)r2_kA_ru_kL22_(B1)+C(krDk2_L2_(B1)+krFk2_L2_(B1)+krEk2_L2_(B1)) ∑(1+5±)(1+±)2r2kruk2_L2_(B1)+C(krDk_L22_(B1)+krFk2_L2_(B1)+krEk2_L2_(B1)) ∑(1₊100±)r2_kru_k2_L2_(B1)+C ≥ k!_k2_L2_(B1)kruk2_L2_(B1)+kfkL1_(B1)kfk_LlogL(B1) ¥ . Thus, by repeating the argument with ±reduced by a factor of 100, we conclude the proof.