Elementary theory of L p spaces

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1 CHAPTER 3 Elementary theory of L saces 3.1 Convexity. Jensen, Hölder, Minkowski inequality. We begin with two definitions. A set A R d is said to be convex if, for any x 0, x 1 2 A x = x 0 + (x 1 x 0 ) 2 A 8 2 [0, 1]. With this definition, the emty set and singletons are convex. If d = 1, A is convex if and only if A is an interval (ossibly unbounded). Thus all convex subsets of R are measurable. It is not hard to see that if A R d is convex, then all of its rojections onto the first d 1 coordinates A t = {x 0 (x 1,...,x d 1 ) : (x 0, t) 2 A}, t 2 R are also convex. By using inductively that A 2B(R d ) is Borel measurable if and only if A t 2B(R d 1 ) for all t 2 R, all convex sets are Borel sets. Let A R d be a convex set. A function f : A! R is said to be convex if f (x ) ale f (x 0 )+ ( f (x 1 ) f (x 0 )) 8 2 [0, 1]. Equivalently the set {(x, y) 2 A R : y f (x)} is convex in R d+1. Also equivalently, for all 2 (0, 1), x 0 6= x 1 f (x ) f (x 0 ) (3.1) ale f (x 1) f (x ) x x 0 x 1 x LEMMA. Let A R d be oen and convex set, and f : A! R be convex. Then f is a continuous function. Furthermore, for all v 2 R d the limits (3.2) D ± f (x + tv) f (x) f (x) := lim v t!0 ± t exist finite for all x 2 A and D v f (x) ale D + f (x). v PROOF. Let x 2 A. To show that f is continuous at x, it suffices to show that for all v 2 R d the function t 2 I x,v 7! g x,v (t)= f (x + tv), where I x,v = {t 2 R : x + tv 2 A} is an oen interval containing zero, is continuous at 0 2 R. We omit the subscrit in the notation from now on. It is easy to see that g is a convex function. Let a, s, t, b 2 I with a < s < 0 < t < c < d. From (3.1), we learn that g(0) g(a) g(s) ale g(a)+(s a), 0 a and similarly for the other air. This yields L(a) := g(0) g(a) ale L(s) := g(0) g(s) ale R(t) := g(t) g(0) ale R(b) := g(b) g(0) 0 a 0 s t 0 b 0 33

2 34 3. ELEMENTARY THEORY OF L SPACES Hence s! L(s) is increasing and bounded above, t! R(t) is increasing and bounded below, therefore the limits below exist finite and lim s"0 L(s) ale lim t#0 R(t), which establishes (3.2), and in articular, imlies that g is continuous at zero. A consequence of the above Lemma is that convex functions on R d are measurable. The next inequality is a very instructive and useful alication of convexity JENSEN S INEQUALITY. Let (X, F, µ) be a measure sace. Assume that µ(x ) < 1. Let f 2 L 1 (X, F, µ) be real-valued and : R! R be a convex function. Then 1 f dµ ale 1 f dµ. µ(x ) µ(x ) PROOF. Define t = 1 µ(x ) f dµ. Let = D (t). From the roof of Lemma one has It follows that (s) (t) (s t), 8s 2 R. ( f (x)) (t)+ ( f (x) t) and integrating both sides with resect to x in dµ yields the claimed inequality, since the integral of the second summand on the right hand side is zero REMARK. We do not know whether f 2 L 1 (X, F, µ). However, the roof shows that ( f ) 2 L 1 (X, F, µ). Thus either ( f ) + 2 L 1 (X, F, µ) or we may define the right hand side to be 1, and the inequality holds trivially EXAMPLE -YOUNG S INEQUALITY. Take for instance (t) =e t. Then, if µ(x )=1, the articular case of Jensen s inequality reads ex f dµ ale e f dµ Let us consider X = {x 1,...,x n } with F = P (X ) and µ defined via nx µ({x j })= j 2 [0, 1], j = 1,..., n, µ(x )= j = 1 Let y j be ositive numbers f : X! R be defined by f (x j )=log y j. The above inequality becomes the familiar relationshi between the arithmetic and the geometric mean, known as Young s inequality. ny nx (3.3) j=1 y j j ale j=1 j y j. j=1

3 3. ELEMENTARY THEORY OF L SPACES 35 A air (, 0 ) 2 [1, 1] 2 is called conjugate if there exists 0 ale ale 1 such that (3.4) = 1, 0 = 1 1 When we write = 1 and = 1, we mean that = HOLDER S INEQUALITYFORMEASURABLEFUNCTIONS. Let f, g : X! [0, 1] be measurable functions and (, 0 ) be a conjugate air with 1 < < 1. Then 1 1 fgdµ ale f 0 dµ g 0 dµ. PROOF. Call A and B resectively each factor in the right hand side. We may assume that A, B are both nonzero and finite, otherwise there is nothing to rove in each of the other cases. Writing F = f /A, G = g/b, note that F dµ = G 0 dµ = 1 Now, by Young s inequality alied with 1 =, 2 = 1, y 1 = F(x), y 2 = G(x) we obtain that FG ale F +(1 )G 0 so that 1 fgdµ = AB which is what we had to rove. FGdµ ale F dµ +(1 ) G 0 dµ = MINKOWSKI S INEQUALITY FOR MEASURABLE FUNCTIONS. Let f, g : X! [0, 1] be measurable and 1 ale < 1. Then ( f + g) dµ 1 ale f dµ 1 + g dµ PROOF. We already know the case = 1. So assume > 1. We can assume both summands on the right hand side are finite and nonzero. Otherwise the inequality holds trivially. In this case, since t 7! t is convex, ( f + g) ale 2 1 ( f + g ) and the left hand side is finite and nonzero as well. Write ( f + g) = f ( f + g) 1 + g( f + g) 1. Integrating both sides and alying Hölder inequality to both summands on the right hand side, we obtain 1 1! 1 ( f + g) dµ ale f dµ + g dµ ( f + g) dµ 0, and dividing by the second factor on the right hand side of the last dislay, which is finite and nonzero, we obtain the claimed inequality. 1.

4 36 3. ELEMENTARY THEORY OF L SPACES 3.2 L saces. Let (X, F, µ) be a measure sace which we assume fixed throughout unless otherwise mentioned, and let 0 < < 1. We define L (X, F, µ)= f : X! C measurable : f 2 L 1 (X, F, µ). If f 2 L (X, F, µ), the quantity (3.5) k f k := f dµ is finite, and moreover k f k = 0 () f = 0 1. We extend the definition of (3.6) to = 1 by (3.6) k f k 1 = inf M > 0:µ {x 2 X : f (x) > M} = 0. and define 1 L 1 (X, F, µ)={ f : X! C measurable : k f k 1 < 1}. A nontrivial remark is that, if k f k 1 < 1, it must be µ {x 2 X : f (x) > k f k 1 })=0 since {x 2 X : f (x) > k f k 1 } is the countable union of {x 2 X : f (x) > k f k 1 + n 1 }, each of which must have measure zero. Consequently, k f k 1 = M if and only if there exists g = f a.e.[µ] with su g(x) = M. x2x We record the fundamental Hölder s inequality, which is easily derived from (3.1.6). The missing case = 1 is certainly the easiest and is left as an exercise HÖLDER S INEQUALITY. Let (, 0 ) be a air of conjugate exonents with 1 ale ale1. Let f 2 L (X, F, µ), g 2 L 0 (X, F, µ). Then the roduct f g 2 L 1 (X, F, µ) and k fgk 1 alekf k kgk PROPOSITION. Let (X, F, µ) be a measure sace. 1. for all 0 < ale1,l (X, F, µ) is a linear sace, in the sense that f, g 2 L (X, F, µ), 2 C =) f + g 2 L (X, F, µ). 2. For 1 ale < 1, k k is a norm on L (X, F, µ), that is 2a. k f k = 0 =) f = 0; 2b. 2 C, f 2 L (X, µ, F )=)k f k = k f k 2c. f, g 2 L (X, µ, F )=)kf + gk alekf k + kgk 3. For 0 ale < 1, k k is a quasinorm on L (X, F, µ), that is roerties 2.a and 2.b above continue to hold and 2c. is relaced by (3.7) k f + gk ale k f k + kgk 4. For 0 < < 1, there holds Chebychev s inequality su "µ({x 2 X : f (x) > "}) 1 alekfk. ">0 1 Here, and everywhere else, we identify functions which are equal a.e.[µ] without exlicit mention

5 3. ELEMENTARY THEORY OF L SPACES The saces L (X, F, µ) are comlete, in the sense that if f n is a Cauchy sequence in L (X, F, µ), i.e. k f n f m k! 0, n, m!1, then there exists f 2 L (X, F, µ) such that f n L! f, i.e. k f n f k! 0 as n!1. 6. Simle functions with comact suort are dense in L (X, F, µ), 0 < < 1. PROOF. We leave the roof of oints 1. to 4. as an easy exercise. In articular, art 3. uses the inequalities (a + b) ale a + b Ä ; ä (a + b) 1 1 ale 2 1 a b. for a, b 0, 0 < ale 1, which are to be roved in Exercise 3.3. We turn to the roof of comleteness. There is nothing to rove in the case = 1. We turn to the case 0 < < 1. Let f n be a Cauchy sequence in L. Then, from Chebychev s inequality, f n is Cauchy in measure and furthermore k f n k is a bounded sequence. The main tool we will use is the following SUBLEMMA. Let f n be a sequence of measurable functions which is Cauchy in measure, in the sense that for all " > 0 µ({x 2 X : f n (x) f m (x) > "})! 0, m, n!1. Then there exists a subsequence f nj and f : X! C such that f nj! f ointwise. Moreover, f n! f in measure as well. We ostone the roof of the Sublemma to the exercises. We obtain f : X! C such that f n j! f ointwise. By Fatou s lemma so that f 2 L 1. At this oint, f nj f dµ ale lim inf j!1 f n j dµ<1 f ale 2 2 f for j large enough, thus k f n j f k! 0 by the dominated convergence theorem. It follows that f nj! f in L. But then k f n f k ale C k f n f n j k + C k f f n j k and both terms on the right hand side go to zero as n, n j!1, which imlies that f n! f in L, concluding the roof of comleteness. Density of simle functions with is roved in a totally analogous way to the case = REMARK. Point 5. of Proosition can be reformulated as comleteness of L (X, F, µ) with resect to the metric d ( f, g)=k f gk min{,1}, 0< ale1. The fact that the above is a metric is immediate when 1. In the exercises, the numerical inequality a b ale a + b 80 < ale 1

6 38 3. ELEMENTARY THEORY OF L SPACES is roved. This may be relied uon to infer that k f gk alekf hk + kh gk and conclude that d is a metric when 0 < < 1 as well. We conclude this introductory subsection with an aroximation theorem. The roof is identical to the case = 1 when 1 ale < 1, and very similar in the case 0 < < 1, where the quasi-triangle inequality relaces the triangle inequality THEOREM. Let 0 < < 1. and f 2 L (R d ) (R d with Lebesgue measure). Then there is a sequence f n 2C 0 (R d ) with f n! f in L. 3.3 Linear bounded functionals. Let (X, F, µ) be a measure sace and 1 ale ale1. A linear bounded, linear continuous or simly linear functional on L (X, F, µ) is a ma satisfying : L (X, F, µ)! C (linearity) f, g 2 L, 2 C =) ( f + g)=( f )+ (g) (boundedness) 9C 0 such that ( f ) ale Ck f k for all f 2 L. The least C > 0 such that the above holds for all f 2 L is denoted by kk. It is easy to see that ( f ) kk = su = su ( f ) f 2L :f 6=0 k f k k f k = REMARK (CONTINUITY). There is no difference between boundedness and (Lischitz) continuity for linear functionals. In other words a linear bounded functional is uniformly Lischitz continuous. This simly follows from ( f ) (g) = ( f g) alekkk f gk. A ivotal examle is as follows. Let g 2 L 0 (X, F, µ). Since f ḡ 2 L 1 whenever f 2 L we can define the ma (3.8) g : L (X, F, µ)! C, g ( f )= f ḡ dµ. It is easy to see that this is a linear ma, bounded by virtue of Hölder inequality which yields Therefore k g k alekgk 0. In fact, there holds g ( f ) alekfgk 1 alekgk 0k f k LEMMA. Let 1 ale ale1and define g as in (3.8). Then PROOF. Exercise 3.9. k g k = kgk 0 A natural question which arises from Lemma is whether all linear bounded functionals on L (X, F, µ) arise from g 2 L 0 as in (3.8). This is true if 1 ale < 1 and false for = 1. The first half of this statement goes under the name of Riesz reresentation theorem for the dual sace of L (X, F, µ). We will rove both statements in Chater 5 in a general

7 3. ELEMENTARY THEORY OF L SPACES 39 setting. In the next section, we devote ourselves to the secial case of this statement for ` saces. 3.4 The ` saces and their Riesz reresentation. In this aragrah, we consider the saces ` = L (N, P (N), ) where is the counting measure, namely, the sace of sequences ~f =(f 1,...,f j,...) such that ( Ä P 1 ä 1 1 > k f k` = f j=1 j 0 < < 1, su j2n f j = 1 The sequences b k =(b k 1,...,bk j,...), bk j = 0 j 6= k 1 j = k form an unconditional basis of `, in the sense that if f 2 `, the sequence nx f n = f k b k converges to f in `. k= THEOREM. Let 1 ale < 1 and : `! C be a linear bounded functional. Then there exists a unique g 2 `0 such that = g as in (3.8), namely and furthermore kk = kgk` 0. ( f )= g ( f )= 1X f j g j j=1 8 f 2 `, PROOF. We deal with 1 < < 1. The case = 1 follows with minor changes. The candidate g is obtained by setting g j = (b j ), j 2 N. Let f 2 `. By linearity, for all n Ç n å X ( f n )= f k b k = It follows that for n m nx j=m+1 k=1 nx f k b k = g ( f n ). k=1 f j g j = ( f n ) ( f m ) alekkk f n f m k`, which since f n is Cauchy, imlies that the series 1X g ( f )= f j g j j=1

8 40 3. ELEMENTARY THEORY OF L SPACES converges to ( f ). By Lemma 3.3.2, to finish the roof it suffices to rove that g 2 `0 and kgk 0 alekk. Define 0 g j = 0 h = {h j : j 2 N}, h j = g j 0 2 g j g j 6= 0 and let again h n = P n j=1 h j b j, g n = P n j=1 g j b j be the truncation to the first n comonents. Then h n 2 ` and kh n k = kg n k Furthermore, a comutation reveals that kg n k 0 0 = g (h n ) alekkkh n k = kkkg n k 0 1 0, and rearranging kg n k 0 alekk. The monotone convergence theorem finally yields kgk 0 ale kk as well.

9 3. ELEMENTARY THEORY OF L SPACES 41 Chater 3 Exercises 3.1. Prove Sublemma Converse of Jensen s inequality. In this roblem, X =[0, 1] with the Lebesgue measure. Suose : R! R is a function with the roerty that Ç å for all f 2 L 1 (X ). Prove that [0,1] f dµ ale [0,1] is a convex function. f dµ 3.3. Let 0 < ale 1, a, b 0 Prove the two numerical inequalities (a + b) ale a + b Ä ; ä (a + b) 1 1 ale 2 1 a b. 3.4 Log-convexity of L -norms and endoint saces. Let f 2 L \L q with 1 ale < q ale1. a. Prove that k f k r alekf k k f k1 q 8 2 [0, 1], 1 r = (1 ) + q. b. Prove that if f 2 L \ L 1 for large enough, lim!1 k f k = k f k 1. You should use a. for art of the statement. c. Assume in addition that µ(x )=1. Let f 2 L for all 0 < < 1 and such that k f k su 1ale<1 ale 17. Prove that e f 2 L 1 and ke f k 1 ale e. d. Assume again that µ(x )=1. Let f 2 L r for some 0 < r ale1. Prove that lim #0 k f k = ex with the convention that ex( 1)=0. log f dµ 3.5 Converses of Hölder s inequality. In this exercise, (X, F, µ) is a measure sace, µ(x )= 1, and 1 < q ale1is fixed. a. Let f : X! C be a measurable function with the following roerty: for all g 2 L q, the roduct fg2 L 1 and fg dµ alekgk q. Prove that f 2 L, with = q 0, and that k f k ale 1. b. Let f : X! C be a measurable function with the following roerty 2 : for all g 2 L q, the roduct fg2 L 1. Prove that f 2 L, with = q 0. 2 For this art, any solution using tools which do not fall into the scoe of the first three chaters will not be considered.

10 42 3. ELEMENTARY THEORY OF L SPACES 3.6. Here X = R with Lebesgue measure. Find a function F : X! [0, 1) such that {q 2 (0, 1) : F 2 L q } = {7}. Can there be a function G such that {q 2 (0, 1) : F 2 L q } = {e, }? 3.7. Let 0 < < 1, f 2 L (X, F, µ). Define for k 2, F k = {x 2 X : f (x) > 2 k } and Ç å 1 X [ f ] := 2 k µ(f k ). k2 Prove that there exist constants c < C such that Do the case with µ(x ) < 1 first. c [ f ] alekf k ale C [ f ] Let : [0, 1)! [0, 1) be a convex, strictly increasing function with (0) =0. This exercise is about finding a norm for the sace L = L (X, F, µ)= f : X! C measurable : x 7! ( f (x) ) 2 L 1 (X, F, µ). 1. Prove that if f : X! C I ( f ) := is a decreasing function of Ä ä f (x) dµ(x), > 0 and lim!0 + I ( f )=1 unless f = 0 a.e., in which case I ( f )=0 for all > Prove that if f 2 L (3.9) lim!1 I ( f )=0 and therefore k f k := inf{ > 0:I ( f ) ale 1} 2 [0, 1) is well defined. Furthermore rove that k f k = 0 if and only if f = 0 a.e. and that f 2 L, f 6= 0 =) I k f k ( f )=1. 3. Assume that f, g 2 L, 2 C. Prove that k f k = k f k, k f + gk alekf k + kgk and conclude that k k is a norm on L (in fact, arguing in the same way we used for L, L is comlete with resect to this norm) REMARK. In fact, using the above for (t) =t roves the triangle inequality for L saces without any aeal to Hölder s inequality. There is actually a way to derive Hölder s inequality in this framework, using the concet of conjugate function. Given : [0, 1)! [0, 1) convex, one define the conjugate : [0, 1)! [0, 1) by (s)=su {ts (t) : t 0}.

11 3. ELEMENTARY THEORY OF L SPACES 43 It turns out that is also convex and ( ) =. For instance, if (t)=t with 1 < < 1 then (s)=s 0. Using the definition of conjugate functions, it should be easy to rove the following: if f 2 L, g 2 L then fg2l 1 and k fgk 1 ale 2k f k kgk Prove Lemma Hint.