Decreasing Rearrangement and Lorentz L(p, q) Spaces

Transcription

1 Decreasing Rearrangement and Lorentz L(p, q) Spaces Erik Kristiansson December 22 Master Thesis Supervisor: Lech Maligranda Department of Mathematics

2 Abstract We consider the decreasing rearrangement of a measurable function and we prove some fundamental properties like the Hardy-Littlewood inequality. This notion leads us to the Lorentz L(p, q) spaces which are generalizations of the ordinary L p spaces. We will introduce a quasi-norm and also a norm on L(p, q) and examine their metrical and topological properties as well as boundedness of the maximal operator and the Hilbert transform.

3 Acknowledgment I want to take the opportunity to thank my supervisor Lech Maligranda for his excellent supervision, encouragement, and for introducing me to the world of functional analysis. I would also like to thank all the personal, and specially the Ph.D. student Johan Dasht, here at the Department of Mathematics of the Luleå University of Technology for all the help and support. 1

4 Contents 1 Introduction 3 2 Decreasing rearrangement Distribution function Decreasing rearrangement Examples of decreasing rearrangement Properties of decreasing rearrangement Decreasing rearrangement and L p -norms The Hardy-Littlewood inequality Operation ** Lorentz L(p, q) spaces Definition of the Lorentz L(p, q) spaces Normed Lorentz L(p, q) spaces Topological properties of the Lorentz L(p, q) spaces Applications PDEs and decreasing rearrangement The maximal operator The Hilbert transform

5 Chapter 1 Introduction The notion of rearrangement of a function has long been an important tool in classical analysis, playing a key role in many inequalities. Systematically introduced by Hardy and Littlewood it has been used by several authors in real and harmonic analysis, in investigations of singular integrals, function spaces and interpolation (see [1], [12] for instance). The contents of this report is as follows. The decreasing rearrangement f of a measurable function f is a decreasing right continuous function defined on (, ) which is equimeasurable with f. To make a formal definition of f the definition of the distribution function µ f is needed. These definitions can be found in Chapter 2 together with some fundamental properties like the uniqueness of f. In the last section of this chapter some examples of functions and their distribution function and decreasing rearrangement are given. In Chapter 3 some important theorems regarding the decreasing rearrangements and the famous Hardy-Littlewood inequality, f(x)g(x) dµ f (t)g (t) dt are proved. If we, in addition, assume that the measure space is nonatomic this inequality can also generalized to n functions. The last part of Chapter 3 contains some results regarding equality in the Hardy-Littlewood inequality and an explanation of the concept of resonant measure spaces. The decreasing rearrangement will lead us to the Lorentz L(p, q) space which 3

6 is the set of all classes of measurable functions such that the functional pq is finite, where ( ( t 1/p f (t) ) ) 1/q q dt if < p <, < q <, t f pq = sup t 1/p f (t) if < p, q =. t> The Lorentz L(p, q) spaces are investigated in Chapter 4. These spaces are generalizations of ordinary L p spaces and they coincides with L p when q = p. We show that the L(p, q) spaces are linear spaces (vector spaces) but to introduce a norm on L(p, q) is not a trivial task. It turns out that the functional pq will serve as a norm only for 1 q p < or p = q =. To deal with the case when q > p we need to introduce another functional pq which is defined as follows ( ( t 1/p f (t) ) ) 1/q q dt if < p <, < q <, f t pq = sup t 1/p f (t) if < p, q =. t> This functional is subadditive and equivalent to pq for all 1 < p < and 1 q which means that L(p, q) is a normed space with the norm pq. We will also show that in the remaining cases L(p, q) is not normable, that is, we can not on L(p, q) introduce an norm that is equivalent to pq. Next, we examine the topological properties of the Lorentz L(p, q) spaces, first its completeness and then the separability. Finally, the duality of L(p, q) spaces is proved in the last part of Chapter 4. In Chapter 5 we are giving application of rearrangement in PDE s. It is shown that from all the bodies in R 3 with density 1 and fixed volume, it is the ball that has the gravitational field of highest energy. Then, using our earlier investigations, we show that both classical Hardy-Littlewood maximal operator and the Hilbert transform are bounded in Lorentz L(p, q) spaces for 1 < p <. 4

7 Chapter 2 Decreasing rearrangement This chapter is devoted to the decreasing rearrangement. In the first section the distribution function of a measurable function is defined and some basic properties are proved. The next section contains the decreasing rearrangement and which is, in fact, a distribution function with respect to the Lebesgue measure. Then some fundamental properties of rearrangement are proved and we show by a counter-example that the operation * is neither subadditive nor submultiplicative. This fact will be very important when we discuss the Lorentz L(p, q) spaces in Chapter 4. We end this chapter with some examples, where the most important part are the distribution function and decreasing rearrangement of a simple function. We start by making a few assumptions which will remove much of the unnecessary troubles. Let (, Σ, µ) denote a σ-finite measure space. On we let f and g denote two Σ-measurable functions. We will keep these assumption till the end of Chapter 5 and sometimes we write instead of all (, Σ, µ). 2.1 Distribution function We begin with a definition of the distribution function. Definition 2.1. The function µ f : [, ) [, ] defined by µ f (λ) = µ({x : f(x) > λ}), λ (2.1) is called the distribution function of f. 5

8 The distribution function depends only on the absolute value of f and is clearly decreasing. Other useful properties are collected in the following theorem. Theorem 2.1. Let f n be Σ-measurable functions on, n = 1, 2,.... Then (i) µ f is decreasing and continuous from the right. (ii) If f(x) g(x) for x µ-a.e., then µ f (λ) µ g (λ) for any λ. (iii) µ f+g (λ 1 + λ 2 ) µ f (λ 1 ) + µ g (λ 2 ) for any λ 1, λ 2. (iv) µ fg (λ 1 λ 2 ) µ f (λ 1 ) + µ g (λ 2 ) for any λ 1, λ 2. (v) If f(x) lim inf n f n (x) for x µ-a.e., then µ f (λ) lim inf n µ fn (λ) for any λ. Proof. (i) Let λ 1 < λ 2 be arbitrary. Then {x : f(x) > λ 1 } {x : f(x) > λ 2 } so by the monotonicity of the measure we have that µ f (λ 1 ) µ f (λ 2 ), that is, µ f is decreasing. To prove that µ f is continuous from the right, let A λ = {x : f(x) > λ} and let λ be an arbitrary fixed number. The sets A λ increases as λ decreases, and A λ = A λ +1/n. n=1 Hence, by the continuity of the measure, µ f (λ + 1/n) = µ(a λ +1/n) µ(a λ ) = µ f (λ ) as n. Thus µ f is continuous from the right. (ii) If f(x) g(x) for x µ-a.e., then {x : g(x) > λ} {x : f(x) > λ}. Using the monotonicity of the measure, µ f (λ) µ g (λ) for all λ. (iii) Since {x : f(x) + g(x) > λ 1 + λ 2 } {x : f(x) > λ 1 } {x : g(x) > λ 2 } 6

9 it follows that µ f+g (λ 1 + λ 2 ) µ f (λ 1 ) + µ g (λ 2 ). (iv) As in (iii) we have {x : f(x)g(x) > λ 1 λ 2 } {x : f(x) > λ 1 } {x : g(x) > λ 2 } which implies that µ fg (λ 1 λ 2 ) µ f (λ 1 ) + µ g (λ 2 ). (v) Let A n = {x : f n (x) > λ}, n = 1, 2,.... By assumption and the definition of limes inferior, f(x) lim inf n f n(x) = sup m N inf f n(x). n>m This means, that for all x such that f(x) > λ there exists an integer m such that for all integers n > m, f n (x) > λ. Hence A λ m=1 n=m and for any m 1 ( ) µ A n inf µ (A n) sup n m m n=m A n, (2.2) inf µ(a n) = lim inf µ(a n) n m n Furthermore, by (2.2), the monotonicity of µ and since { n=ma n } m=1 is an increasing sequence of sets we finally get ( ) ( ) µ f (λ) = µ(a λ ) µ E n = lim µ E n m lim inf n µ f n (λ). m=1 n=m n=m 2.2 Decreasing rearrangement The next step is to introduce the decreasing rearrangement and its important properties. We start with a definition. 7

10 Definition 2.2. The decreasing rearrangement of f is the function f : [, ) [, ] defined by where we use the convention that inf =. f (t) = inf{λ : µ f (λ) t}, (2.3) If µ f is strictly decreasing, then f is clearly the inverse of µ f. Another very important relationship between the distribution function and the decreasing rearrangement is described in the following theorem. Theorem 2.2. The equation holds, where m is the Lebesgue measure. f (t) = m µf (t), t (2.4) Proof. Since µ f is a decreasing function by Theorem 2.1 (i) it follows that sup{λ : µ f (λ) > t} = m{λ : µ f (λ) > t}. Hence by Definition 2.1 we get f (t) = inf{λ : µ f (λ) t} = sup{λ : µ f (λ) > t} = m{λ : µ f (λ) > t} = m µf. Remark. The previous theorem shows that a decreasing rearrangement is in fact a distribution function with respect to the Lebesgue measure on the semi-axis [, ). Therefore the properties in Theorem 2.1 hold for all decreasing rearrangements. The next theorem establish some basic properties of the decreasing rearrangement. Theorem 2.3. The following properties holds. (i) f (t) > λ if and only if µ f (λ) > t. (ii) f and f are equimeasurable, that is, µ({x : f(x) > λ}) = m({t > : f (t) > λ}) (2.5) for all λ, where m is the Lebesgue measure. 8

11 (iii) If λ and µ f (λ) <, then f (µ f (λ)) λ and f (µ f (λ) + ε) λ for all < ε < f (t). If t and f (t) <, then µ f (f (t)) t and µ f (f (t) ε) t for all ε >. (iv) for < p < ( f p ) (t) = f (t) p (v) If A Σ, then (fχ A ) (t) f (t)χ [,µ(a)) (t) for all t. Proof. (i) First assume that µ f (λ) > t. Then, since µ f is a decreasing function, λ < inf{ν : µ f (ν) t}. Thus f (t) > λ. Now assume that f (t) > λ, that is, inf{ν : µ f (ν) t)} > λ. Thus, since µ f is a decreasing function, µ f (λ) > t. (ii) Let m be the Lebesgue measure on [, ). Then by (i) we get m f (λ) = m({t : f (t) > λ}) = m({t : µ f (λ) > t}) = m([, µ f (λ))) = µ f (λ) (iii) Assume that µ f (λ) <. Since µ f is a decreasing function we get Also, for any arbitrary ε >, f (µ f (λ)) = inf{s : µ f (s) µ f (λ)} λ. f (µ f (λ) + ε) = inf{s : µ f (s) µ f (λ) + ε} λ. Now assume that f (t) <. Then µ f (f (t)) = µ f (inf{λ : µ f (λ) t}) t by the right-continuity of µ f. Furthermore, for any ε >, we have by (ii) that µ f (f (t) ε) = µ({x : f(x) > f (t) ε}) (iv) Since λ < we get = m({s > : f (s) > f (t) ε}) t ( f p ) (t) = inf{λ : µ({x : f(x) p > λ}) t} = inf{ν p : µ({x : f(x) > ν}) t} = f (t) p 9

12 where ν = λ (1/p). (v) Since (fχ A )(x) f(x) for all x we have by Theorem 2.1(ii) and Theorem 2.2 that (fχ A ) (t) g (t), t. On the other hand, since µ({x : (fχ A )(x) > λ}) µ(a), we also have that (fχ A ) (t) =, t > µ(a). Combining these two estimates we can conclude that for all t. (fχ A ) (t) f (t)χ [,µ(a)) (t) Remark. Observe that property (ii) does not hold if we remove the strict inequalities, that is µ({x : f(x) λ) = m({t R : f (t) λ}) does not hold in the general case. This can easily be seen by taking Then f (t) 1 which means that f(x) = x 1 + x. µ({x : x/(1 + x) 1}) = = m({t R : 1 1}). An important question is if the decreasing rearrangement is unique and we will answer on this in the next theorem. Theorem 2.4 (Uniqueness). There exists only one right-continuous decreasing function f equimeasurable with f. Hence, the decreasing rearrangement is unique. Proof. Let f 1 (t) and f 2 (t) be two different right-continuous decreasing functions equimeasurable with f. Then there exist a t such that f 1 (t ) f 2 (t ) and we can assume without loss of generality that f 1 (t ) > f 2 (t ). Choose ε > such that f 1 (t ) > f 2 (t ) + ε. Then by the right continuity of f 1 (t) there exist a nonempty interval [t, t 1 ] such that f 1 (t) > f 2 (t ) + ε for all 1

13 t [t, t 1 ]. Using this estimate and the fact that both f 1 (t) and f 2 (t) are decreasing we get that and m({t : f 1 (t) > f 2 (t) + ε}) t 1 m({t : f 2 (t) > f 2 (t) + ε}) t 2 < t 1 which is a contradiction to the equimeasurability with f. Hence, f 1 (t) = f 2 (t), i.e. the decreasing rearrangement is unique. Remark. The operation * is neither subadditive, in the sense that (f + g) (t) f (t) + g (t) is not true in general, nor submultiplicative, that is, (fg) (t) f (t)g (t) does not hold for any t. The next example shows this. Example 2.1. Let A and B be measurable sets such that A B and < µ(a) < µ(b). Put f(x) = χ A (x) and g(x) = χ B (x). The decreasing rearrangements are f (t) = χ [,µ(a)) (t) and g (t) = χ [,µ(b)) (t) which means that 2 if t < µ(a), f (t) + g (t) = 1 if µ(a) t < µ(b), if t µ(b). Moreover, since (f + g)(x) = χ A (x) + χ B (x) it follows that 2 if t < µ(a B), (f + g) (t) = 1 if µ(a B) t < µ(a B), if t µ(a B). Hence, for any t such that µ(b) < t < µ(a B) we have (f + g) (t) = 1 > = f (t) + g (t). Similar calculations shows that * neither is submultiplicative. 11

14 Despite the previous remark and example we have the following theorem. Theorem 2.5. The inequalities (f + g) (t 1 + t 2 ) f (t 1 ) + g (t 2 ) and (fg) (t 1 + t 2 ) f (t 1 )g (t 2 ) hold for all t 1, t 2. In particular, (f + g) (t) f (t/2) + g (t/2) and for all t. (fg) (t) f (t/2)g (t/2) Proof. We start with the first inequality. Let f (t 1 ) + g (t 2 ) <, otherwise there is nothing to prove. Then for λ 1 = f (t 1 ) and λ 2 = g (t 2 ), by Theorem 2.3(iii), µ f (λ 1 ) t 1 and µ g (λ 2 ) t 2 and by Theorem 2.1(iii) µ f+g (λ 1 + λ 2 ) µ f (λ 1 ) + µ g (λ 2 ) t 1 + t 2. Using the definition of the decreasing rearrangement we obtain (f + g) (t 1 + t 2 ) λ 1 + λ 2 = f (t 1 ) + g (t 2 ). The proof of the second inequality is similar. We assume that f (t 1 )g (t 2 ) < and use Theorem 2.1(iv) to get that µ fg (λ 1 λ 2 ) µ f (λ 1 ) + µ g (λ 2 ). Again, using the definition of the decreasing rearrangement, we get (fg) (t 1 + t 2 ) λ 1 λ 2 = f (t 1 )g (t 2 ) which is our desired inequality. The rest of the theorem now follows by taking t 1 = t 2 = t/2. 12

15 2.3 Examples of decreasing rearrangement In this section we take a look on some examples of distribution function and decreasing rearrangement. The first example establish some important relations between a simple function, its distribution function and decreasing rearrangement. Example 2.2 (Decreasing rearrangement of a simple function). Let s be a simple function of the following form s(x) = α j χ Aj (x) where α 1 > α 2 > > α k >, A j = {x : s(x) = α j } and χ A is the characteristic function of the set A. Then µ s (λ) = µ({x : s(x) > λ}) = µ({x : = β j χ Bj (λ) α j χ Aj (x) > λ}) = where β j = j i=1 µ(a i), B j = [α j+1, α j ) for j = 1, 2,..., k and α k+1 = which shows that the distribution function of a simple function is a simple function itself (see Figure 2.1b). We can also find the decreasing rearrangement s (t) = inf{λ : µ s (λ) t} = inf{λ : = α j χ [βj 1,β j )(t) β j χ Bj (x) t} which is also a simple function (see Figure 2.1c). 13

16 s α 1 µ s s α 1 α 2 α 3 β 4 β 3 β 2 α 2 α 3 α 4 β 1 α 4 A 3 A 4 A 1 (a) A 2 x α 4 α 3 α 2 α 1 λ (b) β 1 β 2 β 3 β 4 (c) t Figure 2.1: Simple function, its distribution function and decreasing rearrangement Example 2.3. Let = [, ), Σ = all Lebesgue-measurable subsets of and µ = m, where m denotes the Lebesgue measure on. Define f : [, ) [, ) as { 1 (x 1) 2 if x 2, f(x) = if x > 2. The distribution function can be easily calculated { 2 1 λ if λ 1, m f (λ) = if λ > 1, and the decreasing rearrangement becomes { f 1 t 2 /4 if t 2, (t) = if t > 2. Figure 2.2 shows f, m f and f. Observe that the integral over f, m f and f are all the same, that is, f(x) dx = 2 1 (x 1) 2 dx = 1 2 (1 λ) dλ = 2 1 t 2 /4 dt = 4/3. 14

17 f 2 1 m f f λ 1 2 (a) (b) (c) x t Figure 2.2: The function f, its distribution function and decreasing rearrangement. Example 2.4. Let = [, ) and Σ and µ be as in the previous example. First we define an extended function f : [, ) [, ] as if x =, ln( 1 ) if < x < 1, 1 x f(x) = if 1 x 2, ln( 1 ) if 2 < x < 3, x 2 if x 3. Even if f is infinite over some interval the distribution function and the decreasing rearrangement are still defined and can be calculated and µ f (λ) = 1 + 2e λ, λ if t 1, f (t) = ln( 2 ) if 1 < t < 3, t 1 if t 3. Now, consider the function g : [, ) [, ) defined by g(x) = x for all x [, ). Then µ g (λ) = for all λ, which implies that f (t) = for 15

18 all t. Finally, let h : [, ) [, ) be defined as h(x) = x 1 + x, x. Observing Figure 2.3 we immediately get that { if λ < 1, µ h (λ) = if 1 λ. Thus, the unique decreasing right continuous function equimeasurable with h is the constant function h (t) 1. These examples shows that the distribution function and the decreasing rearrangement can be infinite, even if the original function is finite-valued. h 1 h Figure 2.3: The function h and its decreasing rearrangement h from Example 2.4. Example 2.5. Take = N, Σ = 2 N and µ({n}) = 1. Then any Σ- measurable function f : R is a sequence {a n }, n = 1, 2,... and the distribution function can be rewritten as µ f (λ) = µ({n N : a n > λ}) for λ. As always, the decreasing rearrangement is a function defined for all t but we can interpret it as a sequence if we define that is, a n = f (t) for n 1 t < n, a n = inf{λ : µ f (λ) k 1} = inf{λ : µ({n N : a n > λ}) n 1} Hence {a k } is the sequence { a n } permutated in a decreasing order. 16

19 Chapter 3 Properties of decreasing rearrangement The purpose of this third chapter is to establish some important and useful theorems about decreasing rearrangement which we will need later on. We start with a theorem stating that the L p norm of a function can be calculated from the distribution function or the decreasing rearrangement. This result can be found in the first section. In the next section we introduce and prove the famous Hardy-Littlewood inequality, first for two functions on an arbitrary σ-finite measure space and later for more functions when the measure µ is nonatomic. This inequality leads us to an operation **, which is an average of the decreasing rearrangement and this average turns out to be the largest possible. Section 3 is devoted to this averaging operation **, where we prove some basic properties and establish the very important subadditivity property. Indeed, its subadditivity is the key to the triangle inequality of the norm of Lorentz L(p, q) spaces which will be introduced in Chapter 4. We still keep the assumptions from Chapter 2, that is, (, Σ, µ) is a σ-finite measure space and f and g are Σ-measurable functions on. 3.1 Decreasing rearrangement and L p -norms The last equalities in Example 2.3 in the previous chapter comes from a more general theorem which states that the L p norm of a measurable function and 17

20 its decreasing rearrangement are the same. As we will see later on, this result is extremely important in evaluating the norm of elements in the Lorentz L(p, q) space. The case when p = 1 serves as a lemma and the general case follows directly after that. Lemma 3.1. The equalities f(x) dµ = and hold. sup λ> µ f (λ) dλ = f (t) dt. (3.1) λµ f (λ) = sup tf (t) (3.2) t> Proof. We first prove (3.1) for simple positive functions. Let s be a positive simple function on of the form s(x) = α j χ Aj (x) (3.3) where α 1 > α 2 > > α k > and all A j are pairwise disjoint sets. Then (see Example 2.2) µ s (λ) = β j χ Bj (λ) where β j = j i=1 µ(a i), B k = [α j+1, α j ),,2,...,k, and α k+1 =. Thus, by the definition of the integral of a simple function, Furthermore, µ s (λ) dλ = β j m([α j+1, α j )) = β j (α j α j+1 ). (3.4) β j (α j α j+1 ) = β 1 (α 1 α 2 ) + β 2 (α 2 α 3 ) + + β k α k = α 1 β 1 + α 2 (β 2 β 1 ) + + α k 1 (β k 1 β k 2 ) + α k β k = 18 α j µ(a j )

21 and using (3.4) and again the definition of the integral of a simple function, we get that µ s (λ) dλ = α j µ(a j ) = s(x) dµ. This proves the first part of (3.1) for simple functions. The next equality in (3.1) follows by observing that for a positive simple function s, so s (t) dt = s (t) = α j χ [βj 1,β j )(t) α j (β j β j 1 ) = α j µ(a j ) = s(x) dµ. The general case now follows from Theorem 2.1(v), Theorem 2.2 and the Monotone Convergence Theorem. Now we prove (3.2). If there exist a λ (, ) such that µ(λ ) = then sup λ> λµ f (λ) = and sup t> tf (t) sup tλ =. t> Conversely, if there exist a t (, ) such that f (t ) = then sup t> tf (t) = and sup λ> λµ f (λ) sup λt =. λ> This proves that (3.2) hold for the case when µ f (λ) or f (t) are infinite for a certain λ > or t >, respectively. We can therefore assume that µ f (λ) < for all λ and f (t) < for all t. To complete the proof of (3.2) we start by proving that sup λ> λµ f (λ) sup tf (t). t> Assume that sup t> tf (t) < otherwise there is nothing to prove. Since µ f (λ) < for all λ we get by Theorem 2.3(iii) that sup tf (t) µ f (λ)f (µ f (λ)) µ f (λ)λ t> 19

22 for all λ >. Taking supremum over all λ > on both sides we obtain that sup λ> λµ f (λ) sup tf (t). t> We now prove the other direction, that is, sup t> tf (t) sup λµ f (λ). λ> Assume that sup λ> λµ f (λ) < since otherwise there is nothing to prove. By hypothesis, f (t) < for all t > so by Theorem 2.3 we have for all < ε < f (t) that sup λµ f (λ) (f (t) ε)µ f (f (t) ε) (f (t) ε)t λ> Since this inequality holds for any ε > it follows that sup λµ f (λ) tf (t) λ> and taking the supremum over all t > on both sides we obtain that sup λ> Thus, the equality (3.2) is proved. λµ f (λ) sup tf (t). t> Theorem 3.2. Let < p <. Then f(x) p dµ = p λ p 1 µ f (λ) dλ = and sup λ> Moreover, for p = we have that f (t) p dt (3.5) λµ f (λ) 1/p = sup t 1/p f (t). (3.6) t> ess sup f(x) = inf{λ : µ f (λ) = } = f (). x 2

23 Proof. First, let < p <. Then, since f is a Σ-measurable function, f p is a also a Σ-measurable function for all < p < and it follows therefore by Lemma 3.1 and a change of variable that f(x) p dµ = µ f p(λ) dλ = µ f (λ 1/p ) dλ = µ f (ν) d(ν p ) = p ν p 1 µ f (ν) dν. This gives the first equality in (3.5). The second equality follows directly from the equimeasurability of f p and (f ) p. To prove (3.6) we use Theorem 2.3(iv) and Lemma 3.1 to get that ( sup t> t 1/p f (t) = sup t> = sup λ> (t [f (t)] p ) 1/p = sup (t( f p ) (t)) 1/p = t> ( λµf (λ 1/p ) ) 1/p = sup u> sup λµ f p(λ) λ> (u p µ f (u)) 1/p = sup u [µ f (u)] 1/p. u> ) 1/p Now for the case when p =. By the definition of essential supremum and since the distribution function is positive we have that ess sup f(x) = inf{λ : µ({x : f(x) > λ}) = } = inf{λ : µ f (λ) = } x = inf{λ : µ f (λ) } = f (). This completes the proof of the theorem. 3.2 The Hardy-Littlewood inequality In this section we prove the important Hardy-Littlewood inequality. This inequality says that the integral over the product of two measurable functions always is less than or equal to the integral over the product of the decreasing rearrangement of the functions. To prove this we first need the following lemma, which is the Hardy-Littlewood inequality for the case when one of the functions is the characteristic function. Lemma 3.3. The inequality f(x) dµ holds for all A Σ such that µ(a) a. A 21 a f (t) dt.

24 Proof. If a = the inequality is true by Lemma 3.1. Assume that a <. Then by Lemma 3.1 and Theorem 2.3(v) we obtain f(x) dµ = f(x)χ A (x) dµ = (fχ A ) (t) dt A f (t)χ [,µ(a)) (t) dt = µ(a) f (t) dt a f (t) dt. Using the previous lemma we can now prove the Hardy-Littlewood inequality for simple functions and using the Monotone Convergence Theorem for measurable functions on. Theorem 3.4 (Hardy-Littlewood). The inequality f(x)g(x) dµ f (t)g (t) dt. (3.7) holds. Proof. We prove this inequality for positive simple functions and the general result will follow by Theorem 2.1(v), Theorem 2.2 and the Monotone Convergence Theorem for measurable functions on. Let s be a simple function of the form s(x) = α j χ Aj where α 1 > α 2 > > α k > and all A j Σ are pairwise disjoint. Then we can rewrite s as s(x) = β j χ Bj where B j = j i=1 A i and β j = α j α j+1, α k+1 =. By Example 2.2 and 22

25 Lemma 3.3 we get that ( ) s(x)g(x) dµ = β j χ Bj (x) g(x) dµ = = = µ(bj ) β j g (t) dt = γj α j γ j 1 g (t) dt = s (t)g(t) dt where γ j = j i=1 µ(a j) and γ =. β j g(x) dµ B j (α j α j+1 ) µ(bj ) g (t) dt α j χ [γj 1,γ j )(t)g (t) dt Example 3.1. Let = N, Σ = 2 N and µ({n}) = 1. Then, as in Example 2.5, any Σ-measurable function on is a sequence. Thus, for two sequences {a n } n=1 and {b n } n=1 the Hardy-Littlewood inequality becomes a n b n a nb n where {a n} and {b n} are the sequences { a n } and { b n } permutated in decreasing order. The Hardy-Littlewood inequality arises many interesting questions. One of importance is how to characterize functions which give equality in the estimate 3.7. Another question is if there is any way to prove estimate 3.7 for n functions instead of only two. We will devote the rest of the chapter answering to these two questions and it turns out that both depends on the following result. Theorem 3.5. Let µ be a nonatomic measure and let a be any number satisfying a µ(). (i) Then the equality holds. sup f(x) dµ = µ(a) a A 23 a f (t) dt

26 (ii) If µ() <, then there is a set A Σ such that µ(a) = a and a f(x) dµ = f (t) dt. A (iii) If µ() = and f (a) > lim t f (t), then there is a set A Σ such that µ(a) = a and a f(x) dµ = f (t) dt. A Proof. First we prove (ii) and (iii). (ii) We need to consider two separate cases, according to whether the number a does or does not lie in the range of the distribution function µ f of f. Suppose first that there exist an α > for which µ f (α) = a. In that case, it follows that f (a) = inf{λ : µ f (λ) = a} α and so i.e., µ f (f (a)) = a. Therefore, let Then a = µ f (α) µ f (f (a)) a, A = {x : f(x) > f (a)}. µ fχa (λ) = µ f (max(λ, f (a)) and by the equimeasurability of f and f m f χ [,a) (λ) min(m f (λ), a) = min(µ f (λ), a) = min(µ f (λ), µ f (f (a))) = µ f (max(λ, f (a))) Let ε > be arbitrary and take t = min(µ f (λ), a) ε. Then by Theorem 2.3(iii) that is, f (t )χ [,a) (t ) = f (t ) f (µ f (λ) ε) λ ε > λ, m f χ [,a) (λ) = m { t > : f (t)χ [,a) (t) > λ } t. Since ε was arbitrary, we have m f χ [,a) (λ) = µ f (max(λ, f (a))) = µ fχa (λ) 24

27 for all λ. Hence fχ A and f χ [,a) are equimeasurable which implies that their integrals are the same a f(x) dµ = f (t) dt. A It is also clear that the A increases with a. Next, we consider the case where a is not in the range of µ f. Since µ() < and µ f is decreasing it follows that lim µ f(λ) = a. λ Let λ = f (a) and suppose first that λ >. Since a > (a = is in the range of µ f (λ) for λ µ()) it follows that λ <. Also, since a is not in the range of µ f, we see that µ f (λ ) < a < µ f (λ) for < λ < λ. Thus, if a 1 denotes the left-hand limit µ f (λ ) (which exists because µ f is decreasing), then a µ f (λ ) < a µ f (λ ) = a 1. (3.8) This shows in fact that We claim that f (t) = λ for a t < a 1. (3.9) a 1 = µ({x : f(x) λ }). (3.1) To see this, we need only to express the set on the right as the intersection of a decreasing sequence of sets B n = {x : f(x) > λ 1 }, n = 1, 2,..., n observe that µ() is finite, and then µ({x : f(x) λ }) = lim n µ(b n ) = lim n µ f It follows from (3.8) and (3.1) that the set G = {x : f(x) = λ } ( λ 1 ) = µ f (λ ). n 25

28 has measure a 1 a. Since µ is nonatomic we may select a subset B of G with measure a a. The set A defined as the disjoint union A = {x : f(x) > λ } B has measure µ f (λ ) + (a a ) = a, as desired. Moreover, f(x) dµ = f(x) dµ + f(x) dµ. A { f >λ } Now, by (3.8), the value a does lie in the range of µ f. The first part of the proof shows therefore that the first of the integrals has the value a f (t) dt. Also, since f = λ on B, the second integral has the value λ µ(b) = λ (a a ) = a B a f (t) dt, by (3.9). Combining these results we get that a f(x) dµ = f (t) dt. A In the remaining case λ =, we obtain in place of (3.8) the equality µ({x : f(x) > }) a < a. In that case, we may choose B disjoint from the support of f, with measure µ(b) = a a. Defining A again as A = {x : f(x) > } B and observing that f (t) vanishes for t a, we obtain as before a a f(x) dµ = f (t) dt = f (t) dt. A We observed in the first part of the proof that the sets A increases with a in the range of µ f. For the other values of a it is possible to show that the 26

29 sets B in the second part of the proof may be chosen so they increase with a. This completes the proof of (ii). (iii) The condition f (a) > lim t f (t) f ( ) implies that in the neighborhood of f (a) we have µ f (λ) <, and therefore If we denote and then µ({x : f(x) f (a)}) = m({t > : f (t) f (a)}) a. G = {x : f(x) > f (a)} H = {x : f(x) f (a)}, µ(g) a µ(h). Since the measure µ is nonatomic we can find a set A Σ such that G A H and µ(a) = a. The functions fχ A and f χ [,a) are equimeasurable and (fχ A ) (t) = f (t)χ [,a) (t). Indeed, by Theorem 2.3(v) we have (fχ A ) (t) f (t)χ [,a) (t). Assume that t < a. Then fχ A fχ G > f (a), that is, (fχ A ) (t) f (a). If x / A, then f(x) f (a) (fχ A ) (t) so Thus and B = {x : f(x) > (fχ A ) (t)} = {x A : f(x) > (fχ A ) (t)} = {x : f(x)χ A (x) > (fχ A ) (t)}. µ(b) = µ f ((fχ A ) (t)) = µ fχa ((fχ A ) (t)) t f (t) f (µ f ((fχ A ) (t))) (fχ A ) (t). Equimeasurability of fχ A and f χ [,a) implies the equality of the integrals. This completes the proof of (ii). (i) From Lemma 3.3 we immediately obtain the inequality sup µ(a)=a A f(x) dµ a f (t) dt. Therefore, in the case where f (a) > f ( ) from (iii) we have assertion. If f (a) = f ( ), then for G = {x : f(x) > f (a)} 27

30 and H = {x : f(x) > f (a) ε a } where ε > is an arbitrary small number we have µ(g ) a µ(h ) =. Therefore there exists a set A(ε) = A such that G A H and µ(a) = a. The functions (fχ G ) and f χ [,µ(g )) are equimeasurable. Thus G f(x) dµ = µ(g ) µ(g ) and we have f(x) dµ = f(x) dµ + A G µ(g ) f (t) dt + = µ(g ) f (t) dt + A\G f (t) dt, f(x) dµ A\G ( f ( ) ε ) a ( f ( ) ε a dt ) (a µ(g )) f (t) dt + f ( ) (a µ(g )) ε. On the interval (µ(g ), a) the function f is equal to f ( ) so a f(x) dµ f (t) dt ε. Hence (i) holds. A If we apply the Hardy-Littlewood inequality to the Σ-measurable functions f, g and h we get that f(x)g(x)h(x) dµ (fg) (t)h (t) dt. Thus, it does not immediately hold for more than two functions and if we try to use Theorem 2.5 we see that f(x)g(x)h(x) dµ f (t/2)g (t/2)h (t) dt 28

31 which leads to the following unsatisfactory result f (t/2)g (t/2)h (t) dt f (t)g (t)h (t) dt. Hence we need a stronger result than Theorem 2.5 and it turns out that the next theorem combined with Hardy s lemma (Lemma 3.7) will give us what we want. The only drawback is that we have to restrict considerations to nonatomic measure spaces. Theorem 3.6. Let µ be a nonatomic measure. Then a for any a >. In particular, (fg) (t) dt (fg) (t) dt a f (t)g (t) dt. f (t)g (t) dt. Proof. We begin with the case when a =. Then, by Lemma 3.1 and the Hardy-Littlewood inequality (Theorem 3.4) we immediately get that (fg) (t) dt = f(x)g(x) dµ f (t)g (t) dt. Now, if a is any positive number, we have by Theorem 3.5(i) that a (fg) (t) dt = f(x)g(x) dµ = f(x)g(x)χ A (x) dµ sup µ(a)=a A sup µ(a)=a Since we integrate f(x)g(x)χ A (x) over we can apply the Hardy-Littlewood inequality (Theorem 3.4) and Theorem 2.3(v) to get that f(x)g(x)χ A (x) dµ = sup f (t)(gχ A ) (t) dt sup µ(a)=a This completes the proof. = 29 µ(a)=a a sup µ(a)=a a f (t)g (t) dt f (t)g (t) dt.

32 Lemma 3.7 (Hardy s Lemma). Let f and g be positive Lebesgue measurable functions on (, ) and suppose that a f(t) dt a g(t) dt for all a >. Let h be any positive decreasing function on (, ). Then h(t)f(t) dt h(t)g(t) dt. Proof. We first prove this result for a simple functions. Let s be a simple positive decreasing function on (, ). Then s can be represented as s(t) = α j χ [,aj )(t) where α 1 > α 2 > α k > and a 1 > a 2 > > a k >. Therefore, s(t)f(t) dt = aj α j f(t) dt aj α j g(t) dt = s(t)g(t) dt and the general results follows by the Monotone Convergence Theorem. We can now combine Theorem 3.6 and Lemma 3.7 to generalize the Hardy- Littlewood inequality. Theorem 3.8. Let µ be nonatomic and let f, g and h be Σ-measurable functions on. Then f(x)g(x)h(x) dµ f (t)g (t)h (t) dt Proof. By Theorem 3.4, Theorem 3.6 and Lemma 3.7 with h(t) = f (t) we immediately get that f(x)g(x)h(x) dµ f (t)(gh) (t) dt f (t)g (t)h (t) dt 3

33 Corollary 3.9. Let µ be nonatomic and f 1, f 2,..., f n Σ-measurable functions on, n N. Then n f k (x) dµ k=1 n fk (t) dt. Proof. This follows from Theorem 3.8 and the principle of induction. We will now take a closer look on the question if there is a Σ-measurable function h equimeasurable with g such that f(x)h(x) dµ = f (t)g (t) dt The answer is in general no, and it depends if the measure space is nonatomic or completely atomic where all atoms are of equal measure and if the measure space is finite. To make the terminology less cumbersome we first make a definition which is taken from Bennett and Sharpley [1]. Definition 3.1. A measure space (, Σ, µ) or a measure µ is said to be resonant if it is σ-finite and either nonatomic or a countable union of atoms of equal measure. Theorem 3.1. If µ is resonant, then sup f(x) g(x) dµ = k=1 f (t)g (t) dt (3.11) where the supremum is taken over all Σ-measurable functions g equimeasurable with g, that is, µ g (λ) = µ g (λ), λ. Moreover, if also µ() <, then there is a Σ-measurable function g such that f(x) g(x) dµ = f (t)g (t) dt. (3.12) Proof. We can clearly assume that both f and g are positive. Let µ() <. If µ is completely atomic where all atoms are of equal measure, then g can be constructed by permutating the atoms. More specific, this permutation is the composition of the inverse of the permutation that takes f to f and the permutation that takes g to g. That is, g is g, but where all atoms are permutated so that the atom where g has its largest value is the same as the 31

34 atom where f has its largest value, the atom where g has its second largest value is the same as the atom where f has its second largest value, and so on. Clearly, g and g are equimeasurable which means that we have proved (3.12) for the case when µ is completely atomic where all atoms are of equal measure. Now let µ be nonatomic. Since g is positive we can find a sequence of simple positive functions {g n }, such that g n g. Then, for any arbitrary fixed integer n 1, we can represent g n as g n (x) = δ j χ Dj (x) where D 1 D 2 D k and δ j >, j = 1, 2,..., k. Since µ() < we can apply Theorem 3.5(ii) for each number µ(d j ) and get a sequence of sets E 1, E 2,..., E k such that µ(e j ) = µ(d j ), j = 1, 2,..., k and µ(dj ) f(x) dµ = f (t) dt (3.13) E j Define the simple function g n : [, ) by g n (x) = δ j χ Ek (x). Since the measures of E j and D j are equal for all j = 1, 2,..., k, the decreasing rearrangement of g n and g n are equal, that is gn(t) = g n(t) = δ j χ [,µ(ej )(t). Totally get a sequence of positive simple functions { g n }, such that g n and g n are equimeasurable, n = 1, 2,.... Using Theorem 2.1(v) it also follows that g = lim n g n is equimeasurable with g. Moreover, by (3.13), µ(dj ) f(x) g n (x) dµ = δ j f(x) dµ = δ j f E (t) dt j = f (t)δ j χ [,µ(dj ))(t) dt = 32 f (t)g n(t) dt.

35 Thus, we have proved (3.12) for g n, n = 1, 2,... and by Theorem 2.1(v), Theorem 2.2, the Monotone Convergence Theorem and the fact that g and g are equimeasurable, the general case follows. Now, let µ() =, µ be nonatomic or completely atomic where all atoms are of equal measure and α > be a real number such that α < f (t)g (t) dt. Since f and g are assumed to be positive it will suffice to show that there exists a positive function g equimeasurable with g such that α < f(x) g(x) dµ. Because µ is a σ-finite measure we can find a sequence of pairwise disjoint sets 1, 2,..., such that µ( n ) <, n = 1, 2,... and = n=1 n. Hence, we can therefore find two sequences of positive functions {f n } and {g n } such that f n f, g n g and both f n and g n has support in n, n = 1, 2,.... Since f n(t)g n(t) dt f (t)g (t) dt by Theorem 2.1(v), Theorem 2.2 and the Monotone Convergence Theorem it follows that there exist an integer N 1 such that α < f N(t)g N(t) dt. Then ( N, Σ, µ) is a finite measure space which is either nonatomic or completely atomic with all atoms of equal measure. We can therefore apply the first part of the proof and find a positive function ĥ on N equimeasurable with gχ N such that f(x)ĥ(x) dµ = (fχ N ) (t)(gχ N ) (t) dt. N Since f N fχ N and g N gχ N by the construction of {f n } and {g n } it follows that α < = N fn(t)g N(t) dt f(x)ĥ(x) dµ = 33 (fχ N ) (t)(gχ N ) (t) dt f(x)h(x) dµ

36 where Thus, if we take h(x) = {ĥ(x) if x N, if x / N. g(x) = h(x)χ N (x) + g(x)χ \N (x), x, then g is equimeasurable with g and h g. Hence α < f(x)h(x) dµ f(x) g(x) dµ which completes the proof. 3.3 Operation ** If we apply the Hardy-Littlewood inequality to g = χ A, where A is any measurable set with measure t we get that µ(a) t f(x)g(x) dµ = f(x) dµ f (s) ds = f (s) ds. If we divide both sides with µ(a) = t we get 1 f(x) dµ 1 µ(a) t A A t f (s) ds. The right hand side is a function which is defined for all < t. This function is of great importance for the Lorentz L(p, q) spaces and its application will be clear after Chapter 4. A formal definition is as follows. Definition 3.2. The function f : (, ) [, ] is defined as f (t) = 1 t t f (s) ds. Remark. f is sometimes called the maximal function of f since it is the largest of all average values over f. Even if the value of f (t) at t = is not included in the definition above, the 34

37 limit as t approaches zero from the right is defined for all rearrangements. In fact, lim f (t) = f () = ess sup f(x), t + where the last equality is from Theorem 3.2. Theorem Let f n be Σ-measurable functions on, n = 1, 2,.... Then (i) f is decreasing and continuous on (, ). (ii) f (t) f (t) for all t >. (iii) If f(x) g(x) for x µ-a.e., then f (t) g (t) for all t >. (iv) If {f n } is a sequence such that f n (x) f(x) for x µ-a.e., n = 1, 2,... and lim n f n (x) = f(x) for x µ-a.e., then fn (t) f (t) and lim n fn (t) = f (t) for all t >. Proof. (i) The continuity of f (t) follow directly from the continuity of the integral so the only thing left to prove is that f (t) is decreasing. Let < t 1 < t 2 be arbitrary. Then f (t 2 ) = 1 t2 f (s) dt = 1 t1 f (s) ds + 1 t2 f (s) ds t 2 t 2 t 2 x t 1 1 t1 f (s) ds + 1 f (t 1 )(t 2 t 1 ) t 2 t 2 = 1 t1 ( 1 f (s) ds + 1 ) t 1 f (t 1 ) t 2 t 1 t 2 1 t2 ( 1 f (s) ds + 1 ) t1 f (s) ds t 2 t 1 t 2 = 1 t1 f (s) ds = f (t 1 ). t 1 (ii) Since f is a decreasing function we get that f (t) = 1 t t f (s) ds 1 t t f (t) ds = f (t). 35

38 (iii) By Theorem 2.1(ii) and Theorem 2.2 we have that f (t) g (t) for all t. Moreover, since f and g are positive functions we get for t > that f (t) = 1 t t f (s) ds 1 t t g (s) ds = g (t). (iv) We know by Theorem 2.1(ii) that f n(t) f (t), n = 1, 2,... and by Theorem 2.1(v) that lim n f n(t) = f (t) for t. Since f (t) = 1 t t f (s) ds t > and since fn (t) f (t) by (iii) we get by the Monotone Convergence Theorem that fn (t) f (t) for all t >. Example 3.2. Let A be any measurable set and let f = χ A. Then, clearly, f (t) = χ [,µ(a)) so using the definition of f we get that f (t) = 1 t t ( f (s) ds = min 1, µ(a) ) t for t >. A graph of f can be found in Figure 3.1. f 1 µ(a) t Figure 3.1: Graph of f, where f(x) = χ A (x). Example 3.3. Let s be the simple function in the following form s(x) = α j χ Aj (x) 36

39 where α 1 > α 2 > > α k > and A j = {x : s(x) = α j }. Then, by Example 2.2, we have that s (t) = α j χ [βj 1,β j )(t), where β j = j i=1 µ(a j). Using the definition of s (t) we get that s (t) = 1 t t s (u) du = 1 t t α j χ [βj 1,β j )(t) for all t >. We consider two cases. First, if β m 1 < t β m for m = 1, 2,..., k, then s (t) = 1 t = 1 t t m 1 α j χ [βj 1,β j )(t) and the second one, if t > β k then α j µ(a j ) + α m t ( t m 1 µ(a j ) ) s (t) = 1 t α j µ(a j ), Hence, s (t) = 1 t 1 t [ m 1 α jµ(a j ) + α m (t ] m 1 µ(a j)) [ k ] α jµ(a j ) if β m 1 < t β m for m = 1, 2,..., k, if t > β k Figure 3.3 shows a graph of s when s is a simple function with k = 4. 37

40 α 1 α 2 α 3 α 4 t β 1 β 2 β 3 β 4 Figure 3.2: s and s where s is a simple function. Example 3.4. Let = [, ), Σ = all Lebesgue-measurable subsets of and m the Lebesgue measure. Define f : [, ) as { 1 (x 1) 2 if x 2, f(x) = if x > 2. From Example 2.3 we get for all t that { f 1 t 2 /4 if t 2, (t) = if t > 2, and we can obtain f by calculating the integral in Definition 3.2 { f 1 t 2 /12 if < t 2, (t) = 4/(3t) if t > 2, which can be seen in Figure

41 f Figure 3.3: The function f and f (dotted) from Example 3.4. t The last step before we introduce the Lorentz spaces is to derive the subadditivity of **. Theorem 3.12 (Subadditivity). If µ is a resonant measure, then (f + g) (t) f (t) + g (t) for all t >. Hence, operation ** is subadditive. Proof. If we apply Theorem 3.5 to the definition of f we see that f (t) = 1 t f (s) dt = 1 sup f(x) dµ. t t µ(a)=t A Hence, by the triangle inequality and the subadditivity of supremum (f + g) (t) = 1 sup f(x) + g(x) dµ t µ(a)=t A 1 ( ) sup f(x) dµ + g(x) dµ t µ(a)=t A A 1 f(x) dµ + 1 g(x) dµ t t sup µ(a)=t A = f (t) + g (t). sup µ(a)=t A 39

42 Chapter 4 Lorentz L(p, q) spaces In this chapter we introduce Lorentz L(p, q) spaces and prove their most important properties. The first section defines the general L(p, q) spaces and the Lorentz sequence space l(p, q). We show that these two definitions are almost equivalent. After that the linearity of L(p, q) is established we show that to introduce a norm on L(p, q) is not a trivial task. We end the first section with some relations between different Lorentz spaces. The second section deals with the problem of finding a suitable norm for L(p, q) and we show that a certain functional can be seen as a norm for some values of p and q. We also prove when L(p, q) is normable and not normable. The last section in this chapter is devoted to the topological properties of the Lorentz L(p, q) spaces. We prove completeness, density of simple functions and separability, and then we discuss the nature of the space of all bounded linear functionals on L(p, q), that is, the dual space of L(p, q). 4.1 Definition of the Lorentz L(p, q) spaces We start this section with a definition of the Lorentz L(p, q) space and the Lorentz sequence space l(p, q). Definition 4.1. Let (, Σ, µ) be a σ-finite measure space and < p, < q. Then the Lorentz L(p, q) space is the set of all classes of Σ- 4

43 measurable functions f such that the functional f pq <, where ( ( t 1/p f (t) ) ) 1/q q dt if < p <, < q <, t f pq = sup t 1/p f (t) if < p, q =. t> The Lorentz sequence space, denoted by l(p, q), is the set of all sequences a = {a n } c such that the functional a pq <, where ( ) ( ) 1/q n 1/p a q n n 1 if < p, < q <, a pq = n=1 sup n 1/p a n if < p, q =. n 1 and a = {a n} is the sequence { a n } permutated in a decreasing order. Remark. For the Lorentz L(p, q) space, the case when p = and < q < is not of any interest. The reason for this is that f q < implies f = µ-a.e on. This can be seen by using the following argument: Assume that L(, q) is a non-trivial space. Then there exists a nonzero function f L(, q) which means that there exists c > and a set A Σ of positive measure such that f(x) > c for all x A. Then f q q = f (t) q dt t (fχ A ) (t) dt t µ(a) c q dt t = Hence, we have a contradiction which implies that the only element in L(, q) is the zero function. The Lorentz sequence space l(p, q) and L(p, q), when = N, Σ = 2 N and µ({n}) = 1 are equivalent for < p <, < q <. In fact, if we let a n = f (t) for n 1 t < n as in Example 2.5 we get ( f pq = ( t 1/p f (t) ) ) ( 1/q q dt 1/q n = a q n t dt) q/p 1, t n=1 n 1 and since ( ) q/p 1 n n q/p 1 t q/p 1 dt 2n q/p 1 2 n 1 41

44 it follows that the two definitions in Definition 4.1 are equivalent and all results that hold for general L(p, q) will hold also for l(p, q) Observe that the space l(p, q) is not empty when p =. For example, all sequences which only have a finite number of nonzero elements are in l(, q) for all < q. This shows that there is a fundamental difference between L(, q) and l(, q). Lorentz L(p, q) spaces can be seen as generalizations of the ordinary L p spaces. The reason for this is that if we take q = p we obtain L(p, p) = L p for < p. In fact, for < p < we get by the definition of pq that ( p ( f pp = t 1/p f (t) ) ) 1/p p dt/t p ( ) 1/p = t (f (t)) p dt/t ( 1/p = f (t) dt) p. Using Theorem 3.2 we then obtain ( 1/p ( 1/p f pp = f (t) dt) p = f(x) dµ) p. For p = we get immediately, by the fact that f is decreasing and by Theorem 3.2, that f = sup t> f (t) = f () = ess sup f(x) = f x Hence, f pp = f p, which implies that L(p, p) = L p with equality of norms. Example 4.1. Let A be any Σ-measurable set of finite measure. Then ( ) 1/q p χ A pq = µ(a) 1/p q for < p < and < q <. If < p < and q = we get χ A p = µ(a) 1/p. 42

45 The next theorem consider the linearity of L(p, q) and the nature of the functional pq. It turns out that it can be seen as a quasi-norm. Theorem 4.1. The Lorentz L(p, q) space is a linear space and the functional pq is a quasi-norm 1. Proof. To prove that L(p, q) is a linear space we need to show that f + g L(p, q) and αf L(p, q) for any f, g L(p, q) and any scalar α. Let therefore f and g be two arbitrary elements in L(p, q). We start with the case when < p < and < q < and get by Theorem 2.5 that [ ( f + g pq = t 1/p (f + g) (t) ) ] 1/q q dt/t [ ( t 1/p [f (t/2) + g (t/2)] ) ] 1/q q dt/t [ ( = (2u) 1/p [f (u) + g (u)] ) ] 1/q q du/u [ ( [ 2 1/p 2 max u 1/p f (u), u 1/p g (u) ]) ] 1/q q du/u 2 1/p [ = 2 1/p+1 ( f q pq + g q pq ( [u 2 q 1/p f (u) ] q [ + u 1/p g (u) ] ) ] 1/q q du/u ) 1/q 2 1/p+1 [ 2 max( f q pq, g q pq) ] 1/q 2 1/p+1/q+1 ( f pq + g pq ). Moreover, for any scalar α we clearly have that αf pq = α f pq. Now, let < p and q =. Then, again by Theorem 2.5, we obtain 1 A quasi-norm is a functional that is like a norm except that it does only satisfy the triangle inequality with a constant C 1, that is, f + g C( f + g ) where C 1. 43

46 that f + g p = sup t> t 1/p (f + g) (t) sup t 1/p (f (t/2) + g (t/2)) t> = sup(2u) 1/p (f (u) + g (u)) u> 2 1/p (sup u> = 2 1/p ( f p + g p ). Clearly, we also have in this case that u 1/p f (u) + sup u 1/p g (u) u> ) αf pq = α f p Hence, L(p, q) is a linear space for all < p and < q <. Moreover, from our proof above follows that the functional pq is a quasi-norm. From now we will assume that the measure space is resonant. The reason for this is that resonant spaces are simpler than the general ones. The next theorem strengthen the previous result. It states that pq in some cases on p and q is even a norm. This result depends on monotonicity of the function t q/p 1. We prove that if t q/p 1 is a decreasing function then pq is a norm. Theorem 4.2. The functional pq is a norm if and only if either 1 q p or p = q =. Proof. To prove that pq is a norm we need only to show that the triangle inequality is satisfied. The rest follows from the last part of Theorem 4.1. First assume that 1 q p. Then we have q/p 1 which implies that the continuous function t q/p 1 is decreasing and hence (t q/p 1 ) = t q/p 1. We can therefore, since the measure space is resonant, use Theorem 3.1 for any 44

47 f and g in L(p, q) and obtain that [ f + g pq = (t 1/p (f + g) (t)) q dt/t [ = = [ sup [ = sup ] 1/q ] 1/q t q/p 1 (f + g) (t) q dt ] 1/q f(x) + g(x) q h(x) dµ ] f(x) h(x) 1/q + g(x) h(x) 1/q q 1/q dµ where the supremum is taken over all Σ-measurable functions h equimeasurable with t q/p 1. We can now use the Minkowski inequality and the subadditivity of the supremum to get [ ] f + g pq = sup f(x) h(x) 1/q + g(x) h(x) 1/q q 1/q dµ ( [ 1/q [ ] ) 1/q sup f(x) q h(x) dµ] + g(x) q h(x) dµ [ 1/q sup f(x) q h(x) dµ] + sup [ 1/q g(x) q h(x) dµ]. Finally, we use Theorem 3.1 again to get back to the old representation [ 1/q [ ] 1/q f + g pq sup f(x) q h(x) dµ] + sup g(x) q h(x) dµ [ 1/q [ ] 1/q = t q/p 1 f (t) dt] q + t q/p 1 g (t) q dt = f pq + g pq. To complete the proof we need to show that pq is not a norm for the remaining cases, that is, the triangle inequality doesn t hold. We start with the case when µ is nonatomic and we split it into the following parts. 1 1 p < q 2 < p < 1, < q 45

48 3 < q < 1 p < 1. We start with the case when 1 p < q <. Take and f(x) = (1 + ε)χ [,a+h) (x) + χ [a+h,a+2h) (x) g(x) = χ [,h) (x) + (1 + ε)χ [h,+a+2h) (x), where a, h, ε >. Clearly, f = g = f and since (f + g) (t) = (2 + 2ε)χ [,a) (t) + (2 + ε)χ [a,a+2h) (x) it follows that we can evaluate the norms of f, g and f + g by f q pq = g q pq = p [ (1 + ε) q (a + h) q/p + (a + 2h) q/p (a + h) q/p] q and f + g q pq = p q [ (2 + 2ε) q a q/p + (2 + ε) q ( (a + 2h) q/p a q/p)], respectively. Assume that the triangle inequality holds, that is, Then the inequality f + g pq f pq + g pq. (2 + 2ε) q a q/p + (2 + ε) q ( (a + 2h q/p a q/p) 2 q ( (1 + ε) q (a + h) q/p + (a + 2h) q/p (a + h) q/p) holds and it can be written as (a + 2h) q/p (a + h) q/p (1 + ε)q (1 + ε/2) q (1 + ε/2) q 1 Taking ε + we obtain that If we define a function f as ( (a + h) q/p a q/p). (a + 2h) q/p + a q/p 2(a + h) q/p. (4.1) f(x) = x 46 t q/p 1 dt,

49 then we can rewrite inequality (4.1) as f(a + 2h) + f(a) 2f(a + h) which implies, together with the fact that f is continuous, that it is a concave function. By the concavity of f, the derivative f (x) = x q/p 1 must be a decreasing function, that is, it must be that q p. This contradicts q > p, and we conclude that the triangle does not hold. For q =, take measurable sets A B such that 1 < and µ(b \ A) µ(a). If we let and then and Thus, ( ) 1/p µ(b) a µ(a) f(x) = aχ A (x) + χ B\A (x) g(x) = χ A + aχ B\A, f (t) = aχ [,µ(a)) + χ [µ(a),µ(b)) g (t) = aχ [,µ(b\a)) (t) + χ [µ(b\a),µ(b)) (t). f p, = max ( aµ(a) 1/p, µ(b) 1/p) = µ(b) 1/p, g p, = max ( aµ(b \ A) 1/p, µ(b) 1/p) = µ(b) 1/p and since f(x) + g(x) = (a + 1)χ B, Then f + g p = (a + 1)µ(B) 1/p. f + g p = (a + 1)µ(B) 1/p > 2µ(B) 1/p = f p + g p, which shows that the triangle inequality does not hold. Hence, the proof of the first case is complete. 2. Let < p, q < and A, B be two measurable sets such that A B =. Then ( ) 1/q p χ A + χ B pq = (µ(a) + µ(b)) 1/p q 47