247A Notes on Lorentz spaces Definition 1. For 1 ≤ p < ∞ and f : R ...

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2 Proposition 3. Given f ∈ L p,q , we write f = fm where Then fm(x) := f(x)χ {x:2 m ≤|f(x)| λ}| 2 m 2 m−1 λ λ q q/p dλ |En| λ n≥m 1/pq |En| . n≥m To obtain a lower bound, we keep only the summand n = m; for an upper bound, we use the concavity of fractional powers. This yields 2 m |Em| 1/p q f ℓm ∗ Lp,q 2 m |En| 1/p (5) . m≤n As 2 m χEm L p = 2m |Em| 1/p , we have our desired lower bound. To obtain the upper bound, we use the triangle inequality in ℓq (Z): ∞ RHS(5) = 2 −k 2 m+k ∞ χEm+kLp ≤ 2 −k k=0 ℓ q m k=0 ℓ q m 2 m χEmLp q ℓm This completes the proof of the upper bound. Lemma 4. Given 1 ≤ q < ∞ and a finite set A ⊂ 2Z , q A ≤ A q ≤ 2 max A∈A A q ≤ 2 q A q where all sums are over A ∈ A. More generally, for any subset A of a geometric series and any 0 < q < ∞, q A ≈ A q where the implicit constants depend on q and the step size of the geometric series. Proposition 5. For 1 ∞ and 1 ≤ q ≤ ∞, (6) sup | fg| : g ∗ L p′ ,q ′ ≤ 1 ≈ f ∗ L p,q. Indeed, LHS(6) defines a norm on Lp,q . Note that by (6), this norm is equivalent to our quasi-norm. Moreover, under this norm, Lp,q is a Banach space and when q = ∞, the dual Banach space is Lp′ ,q ′ , under the natural pairing. Remark. When p = 1 (and q = 1), the RHS(6) is typically infinite; indeed, E |f| may well be infinite even for some set E of finite measure. In fact, there there cannot be a norm on Lp,q equivalent to our quasi-norm. For example, the impossibility of finding an equivalent norm for L1,∞ (R) can be deduced by computing N |x − n| −1 ∗ N ≈ N log(N) and |x − n| −1 ∗ L1,∞ ≈ N. n=0 L 1,∞ n=0
Proof. Because the quasi-norm is positively homogeneous, we need only verify (6) in the case that f and g have quasi-norm comparable to one. We may also assume that f = 2nχFn and g = 2mχEm. By the normalization just mentioned, n (7) 2 |Fn| 1/pq m ≈ 1 ≈ 2 |Em| 1/p′q ′ n Combining the above with Lemma 4 to obtain (8) 2 n A 1/p q ≈ A∈2 Z n:|Fn|≈A and similarly for g. Now we compute: |fg| dx = n,m ≤ A,B∈2 Z ≤ A,B∈2 Z 2 n 2 m |Fn ∩ Em| n:|Fn|∼A n:|Fn|∼A A∈2 Z m n:|Fn|≈A 2 n · min(A, B) · 2 n A 1/p A · min B 2 n |Fn| 1/p q ≈ 1. p ′ , B A 2 m m:|Em|∼B 1 1 p · m:|Em|∼B 2 m B 1/p′ Notice that this has the structure of a bilinear form: two vectors (indexed over 2 Z ) with a matrix sitting between them. Moreover, by Schur’s test, the matrix is a bounded operator on ℓ q (2 Z ). Thus, |fg| dx 2 n:|Fn|∼A n A 1/p ℓq (A∈2Z ) · 2 m:|Em|∼B m B 1/p′ ℓq ′ (B∈2Z ) by (8) and the corresponding statement for g. This completes proof of the part of (6). We turn now to the opposite inequality. Given f = 2 n χFn ∈ L p,q , we choose Then g = n fg = n 2 n |Fn| 1 p 2 n |Fn| 1 p It remains to show that g ∗ ∗ gLp′ ,q ′ ′ q ≈ A∈2 Z A q′ q−1 1 − |Fn| p ′ χFn = q−1 2 n 1 1− |Fn| p ′ = L p′ ,q n∈N(A) n n 2 n(q−1) |Fn| q−p p χFn . 2 n |Fn| 1 q p ′ 1. By Proposition 3, q |Fn| ′ /p ′ ≈ 1 ≈ f ∗ Lp,q ≈ 1. . where n ∈ N(A) ⇔ 2 n(q−1) |Fn| q−p p ≈ A. Notice that for each A, the sum in n is over part of a geometric series; indeed, Thus Lemma 4 applies and yields ∗ gLp′ ,q ′ ′ q ≈ A q′ n ∈ N(A) =⇒ |Fn| ⇐⇒ A p p(q−1) q−p −n 2 q−p . A∈2 Z n∈N(A) |Fn| q′ /p ′ ≈ n 2 nq |Fn| q/p ≈ 1. This provides the needed bound on g and so completes the proof of (6). The fact that LHS(6) is indeed a norm is a purely abstract statement about vector spaces and (separating) linear functionals. The proof that L p,q is complete in this norm differs little from the usual Riesz–Fischer argument. Let ℓ be a continuous linear functional on L p,q . By definition, |ℓ(χE)| |E| 1/p and so the measure E ↦→ ℓ(χE) is absolutely continuous with respect to Lebesgue measure and so is represented by some locally L 1 function g. This is the Radon– Nikodym Theorem. By linearity this representation of the functional extends to 3
Page 1: 247A Notes
Page 5: Proof of Theorem 7. By the duality

247A Notes on Lorentz spaces Definition 1. For 1 ≤ p < ∞ and f : R ...

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