3. the Lp Spaces (1 ≤ P < ∞)
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3. The Lp spaces (1 ≤ p < ∞) In this section we discuss an important construction, which is extremely useful in virtually all branches of Analysis. In Section 1, we have already introduced the space L1. The first construction deals with a generalization of this space. Definitions. Let (X, A, µ) be a measure space, and let K be one of the fields R or C. A. For a number p ∈ (1, ∞), we define the space Z Lp (X, A, µ) = f : X → : f measurable, and |f|p ∈ dµ < ∞ . K K X R Here we use the convention introduced in Section 1, which defines X h dµ = ∞, for those measurable functions h : X → [0, ∞], that are not integrable. Of course, in this definition we can allow also the value p = 1, and in this case we get the familiar definition of L1 (X, A, µ). K B. For p ∈ [1, ∞), we define the map Q : L1 (X, A, µ) → [0, ∞) by p K Z Q (f) = |f|p dµ, ∀ f ∈ L1 (X, A, µ). p K X Remark 3.1. The space L1 (X, A, µ) was studied earlier (see Section 1). It K has the following features: (i) L1 (X, A, µ) is a -vector space. K K (ii) The map Q : L1 (X, A, µ) → [0, ∞) is a seminorm, i.e. 1 K (a) Q (f + g) ≤ Q (f) + Q (g), ∀ f, g ∈ L1 (X, A, µ); 1 1 1 K (b) Q (αf) = |α| · Q (f), ∀ f ∈ L1 (X, A, µ), α ∈ . 1 1 K K (iii) R f dµ ≤ Q (f), ∀ f ∈ L1 (X, A, µ). X 1 K Property (b) is clear. Property (a) immediately follows from the inequality |f +g| ≤ |f| + |g|, which after integration gives Z Z Z Z |f + g| dµ ≤ |f| + |g| dµ = |f| dµ + |g| dµ. X X X X In what follows, we aim at proving similar features for the spaces Lp (X, A, µ) K and Qp, 1 < p < ∞. The following will help us prove that Lp is a vector space. Exercise 1 ♦. Let p ∈ (1, ∞). Then one has the inequality (s + t)p ≤ 2p−1(sp + tp), ∀ s, t ∈ [0, ∞). Hint: The inequality is trivial, when s = t = 0. If s + t > 0, reduce the problem to the case t + s = 1, and prove, using elementary calculus techniques that min tp + (1 − t)p = 21−p. t∈[0,1] Proposition 3.1. Let (X, A, µ) be a measure space, let K be one of the fields R or C, and let p ∈ (1, ∞). When equipped with pointwise addition and scalar multiplication, Lp (X, A, µ) is a -vector space. K K 300 §3. Lp spaces (1 ≤ p < ∞) 301 Proof. It f, g ∈ Lp (X, A, µ), then by Exercise 1 we have K Z Z p Z Z |f + g|p dµ ≤ |f| + |g| dµ ≤ 2p−1 |f|p dµ + |g|p dµ < ∞, X X X X so f + g indeed belongs to Lp (X, A, µ). K It f ∈ Lp (X, A, µ), and α ∈ , then the equalities K K Z Z Z |αf|p dµ = |α|p · |f|p dµ = |α|p · |f|p dµ X X X clearly prove that αf also belongs to Lp (X, A, µ). K Our next task will be to prove that Qp is a seminorm, for all p > 1. In this direction, the following is a key result. (The above mentioned convention will be used throughout this entire section.) Theorem 3.1 (H¨older’s Inequality for integrals). Let (X, A, µ) be a measure space, let f, g : X → [0, ∞] be measurable functions, and let p, q ∈ (1, ∞) be such 1 1 1 that p + q = 1. Then one has the inequality Z Z 1/p Z 1/q (1) fg dµ ≤ f p dµ · gq dµ . X X X R p R p Proof. If either X f dµ = ∞, or X g dµ = ∞, then the inequality (1) is trivial, because in this case, the right hand side is ∞. For the remainder of the R p R q proof we will assume that X f dµ < ∞ and X g dµ < ∞. ∞ ∞ 1 Use Corollary 2.1 to find two sequences (ϕn) , (ψn) ⊂ L (X, A, µ), n=1 n=1 R,elem such that • 0 ≤ ϕ1 ≤ ϕ2 ≤ ... and 0 ≤ ψ1 ≤ ψ2 ≤ ... ; p q • limn→∞ ϕn(x) = f(x) and limn→∞ ψn(x) = g(x) , ∀ x ∈ X. By the Lebesgue Dominated Convergence Theorem, we will also get the equalities Z Z Z Z p q (2) f dµ = lim ϕn dµ and g dµ = lim ψn dµ. X n→∞ X X n→∞ X 1/p 1/q Remark that the functions fn = ϕn , gnψn , n ≥ 1 are also elementary (because they obviously have finite range). It is obvious that we have • 0 ≤ f1 ≤ f2 ≤ ... , and 0 ≤ g1 ≤ g2 ≤ ... ; • limn→∞ fn(x) = f(x), and limn→∞ gn(x)] = g(x), ∀ x ∈ X. With these notations, the equalities (2) read Z Z Z Z p p q q (3) f dµ = lim (fn) dµ and g dµ = lim (gn) dµ. X n→∞ X X n→∞ X Of course, the products fngn, n ≥ 1 are again elementary, and satisfy • 0 ≤ f1g1 ≤ f2g2 ≤ ... ; • limn→∞[fn(x)gn(x)] = f(x)g(x), ∀ x ∈ X. Using the General Lebesgue Monotone Convergence Theorem, we then get Z Z fg dµ = lim fngn dµ. X n→∞ X 1 Here we use the convention ∞1/p = ∞1/q = ∞. 302 CHAPTER IV: INTEGRATION THEORY Using (3) we now see that, in order to prove (1), it suffices to prove the inequalities Z Z 1/p Z 1/q p q fngn dµ ≤ (fn) dµ · (gn) dµ , ∀ n ≥ 1. X X X In other words, it suffices to prove (1), under the extra assumption that both f and g are elementary integrable. Suppose f and g are elementary integrable. Then (see III.1) there exist pair- m wise disjoint sets (Dj)j=1 ⊂ A, with µ(Dj) < ∞, ∀ j = 1, . , m, and numbers α1, β1, . , αm, βm ∈ [0, ∞), such that f = α + ··· + α 1κ D1 mκ Dm g = β + ··· + β 1κ D1 mκ Dm Notice that we have fg = α β + ··· + α β , 1 1κ D1 m mκ Dm so the left hand side of (1) is the given by m Z X fg dµ = αjβjµ(Dj). X j=1 1/p 1/q Define the numbers xj = αjµ(Dj) , yj = βjµ(Dj) , j = 1, . , m. Using these 1 1 numbers, combined with p + q = 1, we clearly have m Z X (4) fg dµ = (xjyj). X j=1 At this point we are going to use the classical H¨olderinequality (see Appendix D), which gives m m 1/p m 1/q X X p X q (xjyj) ≤ (xj) · (yj) , j=1 j=1 j=1 so the equality (4) continues with Z m 1/p m 1/q X p X q fg dµ ≤ (xj) · (yj) = X j=1 j=1 m 1/p m 1/q X p X q = (αj) µ(Dj) · (βj) µ(Dj) = j=1 j=1 Z 1/p Z 1/q p q = f dµ · g dµ . X X Corollary 3.1. Let (X, A, µ) be a measure space, let K be one of the fields 1 1 R or C, and let p, q ∈ (1, ∞) be such that p + q = 1. For any two functions f ∈ Lp (X, A, µ) and g ∈ Lq (X, A, µ), the product fg belongs to L1 (X, A, µ) and K K K one has the inequality Z fg dµ ≤ Qp(f) · Qq(g). X §3. Lp spaces (1 ≤ p < ∞) 303 Proof. By H¨older’sinequality, applied to |f| and |g|, we get Z |fg| dµ ≤ Qp(f) · Qq(g) < ∞, X 1 1 so |fg| belongs to L+(X, A, µ), i.e. fg belongs to L (X, A, µ). The desired inequal- R RK ity then follows from the inequality X fg dµ ≤ X |fg| dµ. Notation. Suppose (X, A, µ) is a measure space, K is one of the fields R or C, and p, q ∈ (1, ∞) are such that 1 + 1 = 1. For any pair of functions f ∈ Lp (X, A, µ), p q K g ∈ Lq (X, A, µ), we shall denote the number R fg dµ ∈ simply by hf, gi. With K X K this notation, Corollary 3.1 reads: p q hf, gi ≤ Qp(f) · Qq(g), ∀ f ∈ L (X, A, µ), g ∈ L (X, A, µ). K K The following result gives an alternative description of the maps Qp, p ∈ (1, ∞). Proposition 3.2. Let (X, A, µ) be a measure space, let K be one of the fields or , let p, q ∈ (1, ∞) be such that 1 + 1 = 1. and let f ∈ Lp (X, A, µ). Then R C p q K one has the equality q (5) Qp(f) = sup hf, gi : g ∈ L (X, A, µ),Qq(g) ≤ 1 . K Proof. Let us denote the right hand side of (5) simply by P (f). By Corollary 3.1, we clearly have the inequality P (f) ≤ Qp(f). To prove the other inequality, let us first observe that in the case when Qp(f) = 0, there is nothing to prove, because the above inequality already forces P (f) = 0. Assume then Qp(f) > 0, and define the function h : x → K by |f(x)|p if f(x) 6= 0 f(x) h(x) = 0 if f(x) = 0 It is obvious that h is measurable. Moreover, one has the equality |h| = |f|p−1, which using the equality qp = p + q gives |h|q = |f|qp−q = |f|p.