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247A Notes on Lorentz Spaces Definition 1. for 1 ≤ P < ∞ and F : R D → C We Define (1) Weak(Rd) := Sup Λ∣∣{X

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247A Notes on Lorentz Spaces Definition 1. for 1 ≤ P < ∞ and F : R D → C We Define (1) Weak(Rd) := Sup Λ∣∣{X 247A Notes on Lorentz spaces Definition 1. For 1 ≤ p < 1 and f : Rd ! C we define ∗ 1=p (1) kfkLp ( d) := sup λ fx : jf(x)j > λg weak R λ>0 and the weak Lp space p d ∗ L ( ) := f : kfk p d < 1 : weak R Lweak(R ) p −p Equivalently, f 2 Lweak if and only if jfx : jf(x)j > λgj . λ . Warning. The quantity in (1) does not define a norm. This is the reason we append the asterisk to the usual norm notation. To make a side-by-side comparison with the usual Lp norm, we note that ZZ 1=p p−1 kfkLp = pλ dλ dx 0≤λ<jf(x)j Z 1 dλ1=p = jfjfj > λgj pλp 0 λ 1=p 1=p = p λjfjfj > λgj p dλ L ((0;1); λ ) and, with the convention that p1=1 = 1, ∗ 1=1 1=p kfkLp = p λjfjfj > λgj 1 dλ : weak L ((0;1); λ ) This suggests the following definition. Definition 2. For 1 ≤ p < 1 and 1 ≤ q ≤ 1 we define the Lorentz space Lp;q(Rd) as the space of measurable functions f for which ∗ 1=q 1=p (2) kfkLp;q := p λjfjfj > λgj q dλ < 1: L ( λ ) p;p p p;1 p From the discussion above, we see that L = L and L = Lweak. Again ∗ k · kLp;q is not a norm in general. Nevertheless, it is positively homogeneous: for all a 2 C, ∗ −1 1=p ∗ (3) kafkLp;q = λ fjfj > jaj λg Lq (dλ/λ) = jaj · kfkLp;q (strictly the case a = 0 should receive separate treatment). In lieu of the triangle inequality, we have the following: ∗ 1=p kf + gkLp;q = λ fjf + gj > λg Lq (dλ/λ) 1 1 1=p ≤ λ fjfj > 2 λg + fjgj > 2 λg Lq (dλ/λ) 1 1=p 1 1=p ≤ λ fjfj > 2 λg Lq (dλ/λ) + λ fjgj > 2 λg Lq (dλ/λ) by the subadditivity of fractional powers and the triangle inequality in Lq(dλ/λ). Thus ∗ ∗ ∗ (4) kf + gkLp;q ≤ 2kfkLp;q + 2kgkLp;q : Combining (3), (4), and the fact that kfkLp;q = 0 implies f ≡ 0 almost every- ∗ where, we see that k · kLp;q obeys the axioms of a quasi-norm. When p > 1, this quasi-norm is equivalent to an actual norm (see below). When p = 1 and q 6= 1, there cannot be a norm that is equivalent to our quasi-norm. However there is a metric that generates the same topology. In either case, we obtain a complete metric space. ∗ ∗ Notice that (i) if jfj ≥ jgj then kfkLp;q ≥ kgkLp;q and (ii) The quasi-norms are ∗ ∗ rearrangement invariant, which is to say that kfkLp;q = kf ◦φkLp;q for any measure preserving bijection φ : Rd ! Rd. 1 2 p;q P Proposition 3. Given f 2 L , we write f = fm where fm(x) := f(x)χfx:2m≤|f(x)j<2m+1g: Then ∗ kfk p;q ≈p;q kfmk p d L Lx(R ) q `m(Z) p;q1 p;q2 In particular, L ⊆ L whenever q1 ≤ q2. P m Proof. It suffices to consider f of the form f = 2 χEm with disjoint sets Em (cf. m m+1 Em = f2 ≤ jfj < 2 g). Now Z 1 ∗ q q q=p dλ kfkLp;q = p λ jfjfj > λgj 0 λ Z 2m q=p X q X dλ = p λ jEnj m−1 λ m 2 n≥m 1=p q X m X ≈ 2 jEnj : m n≥m To obtain a lower bound, we keep only the summand n = m; for an upper bound, we use the subadditivity of fractional powers. This yields m 1=p ∗ X m 1=p (5) 2 jE j q kfk 2 jE j : m ` . Lp;q . n m q m≤n `m m m 1=p p As k2 χEm kL = 2 jEmj , we have our desired lower bound. To obtain the upper bound, we use the triangle inequality in `q(Z): 1 1 X −k m+k X −k m RHS(5) = 2 k2 χE kLp ≤ 2 k2 χE kLp m+k m q q `m k=0 `m k=0 This completes the proof of the upper bound. Lemma 4. Given 1 ≤ q < 1 and a finite set A ⊂ 2Z, X X q q X Aq ≤ A ≤ 2 max A ≤ 2q Aq A2A where all sums are over A 2 A. More generally, for any subset A of a geometric series and any 0 < q < 1, q X q X A ≈ A where the implicit constants depend on q and the step size of the geometric series. Proposition 5. For 1 < p < 1 and 1 ≤ q ≤ 1, R ∗ ∗ (6) sup j fgj : kgkLp0;q0 ≤ 1 ≈ kfkLp;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 0 0 q 6= 1, the dual Banach space is Lp ;q , under the natural pairing. R Remark. When p = 1 (and q 6= 1), the LHS(6) is typically infinite; indeed, E jfj 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;1(R) can be deduced by computing N ∗ N X −1 X −1 ∗ jx − nj ≈ N log(N) and jx − nj L1;1 ≈ N: 1;1 n=0 L n=0 3 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 P n P m that f = 2 χFn and g = 2 χEm . By the normalization just mentioned, 0 X n 1=pq X m 1=p0 q (7) 2 jFnj ≈ 1 ≈ 2 jEmj n m Combining the above with Lemma 4, we obtain q X X n 1=p X X n 1=pq (8) 2 A ≈ 2 jFnj ≈ 1: A22Z n:jFn|≈A A22Z n:jFn|≈A and similarly for g. Now we compute: Z X n m jfgj dx = 2 2 jFn \ Emj n;m X X n X m ≤ 2 · min(A; B) · 2 A;B22Z n:jFn|∼A m:jEm|∼B 1 1 0 X X n 1=p A 0 B p X m 1=p ≤ 2 A · min p ; · 2 B : B A A;B22Z n:jFn|∼A m:jEm|∼B Notice that this has the structure of a bilinear form: two vectors (indexed over 2Z) with a matrix sitting between them. Moreover, by Schur's test, the matrix is a bounded operator on `q(2Z). Thus, Z X n 1=p X m 1=p0 jfgj dx . 2 A · 2 B ≈ 1 0 `q (A22Z) `q (B22Z) n:jFn|∼A m:jEm|∼B by (8) and the corresponding statement for g. This completes proof of the . part P n p;q of (6). We turn now to the opposite inequality. Given f = 2 χFn 2 L , we choose q−1 1 − 1 q−p X n p p0 X n(q−1) p g = 2 jFnj jFnj χFn = 2 jFnj χFn : n n Then Z q−1 q q 1 1− 1 1 X n p n p0 X n p ∗ fg = 2 jFnj 2 jFnj = 2 jFnj ≈ kfkLp;q ≈ 1: n n ∗ It remains to show that kgk 0 0 1. By Proposition 3, Lp ;q . q0=p0 q0 0 q−p ∗ X q X n(q−1) p kgk p0;q0 ≈ A jFnj where n 2 N(A) , 2 jFnj ≈ A: L A22Z n2N(A) Notice that for each A, the sum in n is over part of a geometric series; indeed, p −n p(q−1) n 2 N(A) () jFnj ≈ A q−p 2 q−p : Thus Lemma 4 applies and yields 0 ∗ q X q0 X q0=p0 X nq q=p kgkLp0;q0 ≈ A jFnj ≈ 2 jFnj ≈ 1: A22Z n2N(A) n 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 Lp;q is complete in this norm differs little from the usual Riesz{Fischer argument. p;q 1=p Let ` be a continuous linear functional on L . By definition, j`(χE)j . jEj and so the measure E 7! `(χE) is absolutely continuous with respect to Lebesgue measure and so is represented by some locally L1 function g. This is the Radon{ Nikodym Theorem. By linearity this representation of the functional extends to 4 simple functions. Boundedness when tested against simple functions suffices to 0 0 show that g 2 Lp ;q . When q 6= 1, the simple functions are dense in Lp;q and so our linear functional admits the desired representation. When q = 1 the simple functions are not dense. For example, one cannot approximate jxj−d=p 2 Lp;1(Rd) by simple functions. Indeed, inspired by the Banach limit linear functionals on `1(Z) we can construct a non-trivial linear functional on Lp;1 that vanishes on simple functions. Let L denote the vector space of f 2 Lp;1 such that `(f) := lim jxjd=pf(x) exists. x!0 Notice that L contains the simple functions and that ` vanishes on these.
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