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1 Lp Spaces and Banach Spaces IVbookroot July 6, 2011 Copyrighted Material 1 Lp Spaces and Banach Spaces In this work the assumption of quadratic integrability will be replaced by the integrability of jf(x)jp. The analysis of these function classes will shed a particu­ lar light on the real and apparent advantages of the exponent 2; one can also expect that it will provide essential material for an axiomatic study of function spaces. F. Riesz, 1910 At present I propose above all to gather results about linear operators defined in certain general spaces, no­ tably those that will here be called spaces of type (B)... S. Banach, 1932 Function spaces, in particular Lp spaces, play a central role in many questions in analysis. The special importance of Lp spaces may be said to derive from the fact that they offer a partial but useful generalization of the fundamental L2 space of square integrable functions. In order of logical simplicity, the space L1 comes first since it occurs already in the description of functions integrable in the Lebesgue sense. Connected to it via duality is the L1 space of bounded functions, whose supremum norm carries over from the more familiar space of continuous functions. Of independent interest is the L2 space, whose origins are tied up with basic issues in Fourier analysis. The intermediate Lp spaces are in this sense an artifice, although of a most inspired and fortuitous kind. That this is the case will be illustrated by results in the next and succeeding chapters. In this chapter we will concentrate on the basic structural facts about the Lp spaces. Here part of the theory, in particular the study of their linear functionals, is best formulated in the more general context of Ba­ nach spaces. An incidental benefit of this more abstract view-point is that it leads us to the surprising discovery of a finitely additive measure on all subsets, consistent with Lebesgue measure. IVbookroot July 6, 2011 Copyrighted Material 2 Chapter 1. LP SPACES AND BANACH SPACES 1 Lp spaces Throughout this chapter (X; F; ¹) denotes a σ-finite measure space: X denotes the underlying space, F the σ-algebra of measurable sets, and ¹ the measure. If 1 · p < 1, the space Lp(X; F; ¹) consists of all complex- valued measurable functions on X that satisfy Z p (1) jf (x)j d¹(x) < 1: X p p p To simplify the notation, we write L (X; ¹), or L (X), or simply L when the underlying measure space has been specified. Then, if f 2 p p L (X; F; ¹) we define the L norm of f by (Z ¶1=p p kfkLp(X;F;¹) = jf(x)j d¹(x) : X We also abbreviate this to kfkLp(X), kfkLp , or kfkp. 1 When p = 1 the space L (X; F; ¹) consists of all integrable functions 1 on X, and we have shown in Chapter 6 of Book III, that L together with k ¢ kL1 is a complete normed vector space. Also, the case p = 2 warrants special attention: it is a Hilbert space. We note here that we encounter the same technical point that we al­ ready discussed in Book III. The problem is that kfkLp = 0 does not imply that f = 0, but merely f = 0 almost everywhere (for the measure ¹). Therefore, the precise definition of Lp requires introducing the equiv­ alence relation, in which f and g are equivalent if f = g a.e. Then, Lp consists of all equivalence classes of functions which satisfy (1). However, in practice there is little risk of error by thinking of elements in Lp as functions rather than equivalence classes of functions. The following are some common examples of Lp spaces. (a) The case X = Rd and ¹ equals Lebesgue measure is often used in practice. There, we have (Z ¶1=p p kfkLp = jf(x)j dx : Rd (b) Also, one can take X = Z, and ¹ equal to the counting measure. p Then, we get the “discrete” version of the L spaces. Measurable functions are simply sequences f = fangn2Z of complex numbers, IVbookroot July 6, 2011 Copyrighted Material 1. Lp spaces 3 and à ! 1 1=p X p kf kLp = janj : n=¡1 2 When p = 2, we recover the familiar sequence space ` (Z). p The spaces L are examples of normed vector spaces. The basic prop­ erty satisfied by the norm is the triangle inequality, which we shall prove shortly. The range of p which is of interest in most applications is 1 · p < 1, and later also p = 1. There are at least two reasons why we restrict our attention to these values of p: when 0 < p < 1, the function k ¢ kLp does not satisfy the triangle inequality, and moreover, for such p, the space p 1 L has no non-trivial bounded linear functionals. (See Exercise 2.) 1 When p = 1 the norm k ¢ kL1 satisfies the triangle inequality, and L is a complete normed vector space. When p = 2, this result continues to hold, although one needs the Cauchy-Schwarz inequality to prove it. In the same way, for 1 · p < 1 the proof of the triangle inequality relies on a generalized version of the Cauchy-Schwarz inequality. This is H¨older’s p inequality, which is also the key in the duality of the L spaces, as we will see in Section 4. 1.1 The HÄolder and Minkowski inequalities If the two exponents p and q satisfy 1 · p; q · 1, and the relation 1 1 + = 1 p q holds, we say that p and q are conjugate or dual exponents. Here, 0 we use the convention 1=1 = 0. Later, we shall sometimes use p to denote the conjugate exponent of p. Note that p = 2 is self-dual, that is, p = q = 2; also p = 1; 1 corresponds to q = 1; 1 respectively. Theorem 1.1 (HÄolder) Suppose 1 < p < 1 and 1 < q < 1 are conju­ p q 1 gate exponents. If f 2 L and g 2 L , then fg 2 L and kfgkL1 · kfkLp kgkLq : 1 Note. Once we have defined L (see Section 2) the corresponding in­ equality for the exponents 1 and 1 will be seen to be essentially trivial. 1We will de¯ne what we mean by a bounded linear functional later in the chapter. IVbookroot July 6, 2011 Copyrighted Material 4 Chapter 1. LP SPACES AND BANACH SPACES The proof of the theorem relies on a simple generalized form of the arithmetic-geometric mean inequality: if A; B ¸ 0, and 0 · θ · 1, then θ 1¡θ (2) A B · θA + (1 ¡ θ)B: Note that when θ = 1=2, the inequality (2) states the familiar fact that the geometric mean of two numbers is majorized by their arithmetic mean. To establish (2), we observe first that we may assume B 6= 0, and θ replacing A by AB, we see that it suffices to prove that A · θA + (1 ¡ θ 0 θ¡1 θ). If we let f (x) = x ¡ θx ¡ (1 ¡ θ), then f (x) = θ(x ¡ 1). Thus f(x) increases when 0 · x · 1 and decreases when 1 · x, and we see that the continuous function f attains a maximum at x = 1, where f(1) = 0. Therefore f(A) · 0, as desired. To prove H¨older’s inequality we argue as follows. If either kfkLp = 0 or kfkLq = 0, then fg = 0 a.e. and the inequality is obviously verified. Therefore, we may assume that neither of these norms vanish, and after replacing f by f=kf kLp and g by g=kgkLq , we may further assume that kfkLp = kgkLq = 1. We now need to prove that kfgkL1 · 1. p q If we set A = jf(x)j , B = jg(x)j , and θ = 1=p so that 1 ¡ θ = 1=q, then (2) gives 1p 1 q jf(x)g(x)j · jf (x)j + jg(x)j : p q Integrating this inequality yields kfgkL1 · 1, and the proof of the H¨older inequality is complete. For the case when the equality kfgkL1 = kfkLp kgkLq holds, see Exer­ cise 3. p We are now ready to prove the triangle inequality for the L norm. p Theorem 1.2 (Minkowski) If 1 · p < 1 and f; g 2 L , then f + g 2 p L and kf + gkLp · kf kLp + kgkLp . Proof. The case p = 1 is obtained by integrating jf(x) + g(x)j · jf(x)j + jg(x)j. When p > 1, we may begin by verifying that f + g 2 Lp, when both f and g belong to Lp. Indeed, jf(x) + g(x)jp · 2p(jf(x)jp + jg(x)jp); as can be seen by considering separately the cases jf (x)j · jg(x)j and jg(x)j · jf(x)j. Next we note that jf(x) + g(x)jp · jf (x)j jf(x) + g(x)jp¡1 + jg(x)j jf (x) + g(x)jp¡1 : IVbookroot July 6, 2011 Copyrighted Material 1. Lp spaces 5 If q denotes the conjugate exponent of p, then (p ¡ 1)q = p, so we see p¡1 q that (f + g) belongs to L , and therefore H¨older’s inequality applied to the two terms on the right-hand side of the above inequality gives p p¡1 p¡1 (3) kf + gkLp · kfkLp k(f + g) kLq + kgkLp k(f + g) kLq : However, using once again (p ¡ 1)q = p, we get p¡1 p=q k(f + g) kLq = kf + gkLp : From (3), since p ¡ p=q = 1, and because we may suppose that kf + gkLp > 0, we find kf + gkLp · kfkLp + kgkLp ; so the proof is finished.
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