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 3.1.1 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
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. 3.1.2 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 3.1.1 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. 3.1.3 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 3.1.4 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. 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 = 0. 3.1.5 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 = 1 3.1.6 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.
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 3.2.1 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 0. 3.2.2 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 2 1 1 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
3. ELEMENTARY THEORY OF L SPACES 37 5. 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 1 1 + 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 3.2.3 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 = 1. 3.2.4 REMARK. Point 5. of Proosition 3.2.2 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
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. 3.2.5 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 =1 3.3.1 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. 3.3.2 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 3.3.2 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
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=1 3.4.1 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
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 0 1 0 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.
3. ELEMENTARY THEORY OF L SPACES 41 Chater 3 Exercises 3.1. Prove Sublemma 3.2.3. 3.2 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 1 1 + 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 1 + 17e. 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.
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 ]. 3.8. 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 > 0. 2. 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). 3.4.2 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}.
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. 3.9. Prove Lemma 3.3.2. Hint.