Poder de negociación de los consumidores

For our second proof of the weak version of Ladyzhenskaya’s inequality, we will use some of the standard theory of interpolation spaces. We recall here the basic facts we require: for full details, see the books of Bennett & Sharpley (1988), _§5.1, and Bergh & L¨ofstr¨om (1976), _§3.1.

Let (X0, X1) be a compatible couple of Banach spaces (that is, there is a Haus-

dorff topological vector spaceX such thatX0 and X1 embed continuously into X).

TheK-functional is defined for each f _∈X0+X1 andt >0 by

K(f, t) =K(f, t;X0, X1) := inf{kf0kX0+tkf1kX1 :f =f0+f1}

where the infimum is taken over all representations f =f0+f1 of f withf0 ∈X0

andf1 ∈X1.

Suppose 0 < θ < 1 and 1 _≤ q < _∞, or 0 _≤ θ _≤ 1 and q = _∞. We define the interpolation space (X0, X1)θ,q as the space of all f ∈ X0 +X1 for which the

functional kf_kθ,q :=        Z ∞ 0 [t−θK(f, t)]q 1/q 1_≤q <_∞ sup 0<t<∞ t−θ_{K(f, t) q=∞}

is finite. A very useful property of interpolation spaces is the estimate on the norms:

kfkθ,q ≤ckfk_X1−₀θkfkθX1 (3.15)

(see Bergh & L¨ofstr¨om (1976), _§3.5, p. 49). As a simple example of interpolation, note that

(L1(Rn), L∞(Rn))1−1/p,q =Lp,q(Rn)

if 1 < p < _∞ and 1 _≤ q _{≤ ∞} (see Bennett & Sharpley (1988), Chapter 5, The- orem 1.9). In fact, this equality remains true with L∞ replaced with BMO (see Bennett & Sharpley (1988), Chapter 5, Theorem 8.11):

(L1(Rn),BMO(Rn))1−1/p,q =Lp,q(Rn). (3.16)

The so-called reiteration theorem allows us to interpolate between interpolation spaces: it says that when we interpolate between two interpolation spaces of the same couple (X0, X1), we get another interpolation space in the same family.

Theorem 3.5 (Reiteration Theorem). Let (X0, X1) be a compatible couple of Ba-

andA1= (X0, X1)θ1,q1. If 0< θ <1 and 1≤q≤ ∞, then

(A0, A1)θ,q = (X0, X1)θ0_,q

providing θ0 _{= (1−θ)θ}

0+θθ1.

The proof may be found in Bennett & Sharpley (1988), Chapter 5, Theorem 2.4, or Bergh & L¨ofstr¨om (1976), Theorem 3.5.3.

3.2.1 Interpolation: Lorentz Version

Using the machinery of interpolation spaces, we can now give a very short proof of our generalised Ladyzhenskaya inequality (3.6).

Lemma 3.6 (Interpolation). Let 1 < q < p < _∞, and 1 _≤ r _{≤ ∞}. For any f _∈Lq,∞₍ Rn)∩BMO(Rn), kfkLp,r ≤ckfkq/p Lq,∞kfk 1−q/p BMO.

Proof. Using (3.16), we have

Lq,∞(Rn) = (L1(Rn),BMO(Rn))1−1/q,∞ provided that 1 < q < _∞. Set B := (L1₍

Rn),BMO(Rn))1,∞, and note that by

(3.15) we have_kf_kB ≤CkfkBMO. By the Reiteration Theorem (Theorem 3.5), we

obtain

Lp,r₍

Rn) = (Lq,∞(Rn),B)α,r

withq < p <∞, provided that α= 1−q/p. Thus, using (3.15), we obtain

kf_kLp,r ≤ckfkq/p_Lq,∞kfk 1−q/p B ≤ckfk q/p Lq,∞kfk 1−q/p BMO, as required.

Using the embedding ˙H1 _{⊂BMO in two dimensions (see Section 2.3), for f ∈}

L2,∞₍

R2)∩H˙1(R2), settingn= 2,p= 4 and q= 2 in Lemma 3.8 we obtain (3.4):

kf_kL4 ≤ckfk1/2

L2,∞k∇fk

1/2

L2 .

When Ω is a bounded Lipschitz domain inR2, we may extend a functionf ∈H01(Ω)

3.2.2 Interpolation: Weak Lp _Version

In the case p =r, we now give another proof of (3.6) which avoids the use of any Lorentz spaces (although it makes the proof a little more involved).

Lemma 3.7 (Weak interpolation). For any f ∈ Lq,∞₍

Rn)∩BMO(Rn), and any

q < p <_∞,

kf_kLp,∞ ≤ckfkq/p Lq,∞kfk

1−q/p

BMO.

Proof. Using (3.16), we have Lq,∞₍

Rn) = (L1(Rn),BMO(Rn))1−1/q,∞ provided that 1< q <_∞. SetB:= (L1₍

Rn),BMO(Rn))1,∞; note that by (3.15) we

havekfkB≤CkfkBMO. By the Reiteration Theorem (Theorem 3.5), we obtain

Lp,∞(Rn) = (Lq,∞(Rn),B)α,∞

withq < p <∞, provided thatα solves 1− 1

p = (1−α)(1−

1

q) +α·1, or in other

words thatα= 1₋q/p. Thus, using (3.15), we obtain

kf_kLp,∞ ≤ckfkq/p Lq,∞kfk 1−q/p B ≤ckfk q/p Lq,∞kfk 1−q/p BMO, as required.

By combining this with Lemma 2.2, we once again obtain (3.6).

Lemma 3.8 (Strong interpolation). For any f _∈ Lq,∞₍

Rn)∩BMO(Rn), and any

q < p <_∞,

kfkLp≤ckfkq/p Lq,∞kfk

1−q/p

BMO.

Proof. Givenp > q, choose any r and ssuch that q < r < p < s <_∞. Then

kf_kLp ≤ckfk1−α

Lr,∞kfkα_Ls,∞,

where 1−_rα+α_s = 1_p. Applying Lemma 3.7 to the two factors on the right, we obtain

kfkLp ≤c(ckfkq/r Lq,∞kfk 1−q/r BMO) 1−α_(ckfkq/s Lq,∞kfk 1−q/s BMO) α ≤ckfkL(1q,−∞α)q/r+αq/skfk (1−α)(1−q/r)+α(1−q/s) BMO =ckfkq/pLq,∞kfk 1−q/p BMO, as required.