CHAPTER 8
Introduction to Banach Spaces and Lp Space
1. Uniform and Absolute Convergence As a preparation we begin by reviewing some familiar properties of Cauchy sequences and uniform limits in the setting of metric spaces. Definition 1.1. A metric space is a pair (X, ⇢), where X is a set and ⇢ is a real-valued function on X X which satisfies that, for any x, y, z X, ⇥ 2 (a) ⇢(x, y) 0 and ⇢(x, y) = 0 if and only if x = y, (b) ⇢(x, y)= ⇢(y, x), (c) ⇢(x, z) ⇢(x, y)+⇢(y, z). (Triangle inequality) The function ⇢ is called the metric on X. Any metric space has a natural topology induced from its metric. A subset U of X is said to be open if for any x U there exists some r>0 such that Br(x) U. Here Br(x)= y X : ⇢(x, y)
The next two theorems highlight some important features of Cauchy sequences and uniform convergence. Theorem 1.1. (Cauchy Sequences) Consider sequences in a metric space (X, ⇢). (a) Any convergent sequence is a Cauchy sequence. (b) Any Cauchy sequence is bounded. (c) If a subsequence of a Cauchy sequence converges, then the Cauchy sequence converges to the same limit.
Proof. (a) Suppose xn converges to x.Given" > 0, there is an N N such that ⇢(xn,x) < "/2 for any n N.Thesequence{ } x is Cauchy because 2 { n} ⇢(x ,x ) < ⇢(x ,x)+⇢(x, x ) < " for any n, m N. n m n m (b) Let xn be a Cauchy sequence. Choose N N such that ⇢(xn,xm) < 1 for all n, m N. Then for any{ x } X, 2 2 ⇢(x ,x) ⇢(x ,x )+⇢(x ,x) n n N N < max ⇢(x1,xN ), ⇢(x2,xN ), , ⇢(xN 1,xN ), 1 + ⇢(xN ,x), { ··· } where the last equation is a finite bound independent of n. (c) Let x be a Cauchy sequence with a subsequence x converging to x.Given" > 0, { n} { nk } choose K, N N such that 2 " ⇢(x ,x) < for any k K, nk 2 " ⇢(x ,x ) < for any n, m N. n m 2 Taking n such that k K and n N,then k k ⇢(x ,x) ⇢(x ,x )+⇢(x ,x) < " for any n N. n n nk nk This shows that x converges to x. { n} ⇤ Theorem 1.2. (Uniform Convergence) Given a sequence of functions fn from a nonempty set S to a metric space (X, ⇢).Suppose f converges uniformly to a function{ }f : S X. { n} ! (a) If each fn is bounded, then so is f. (b) Assume S is a topological space, E S. If each f is continuous on E, then so is f. ⇢ n Proof. (a) Choose N N such that ⇢(f(s),fn(s)) < 1 for any n N and s S.Givenx X, choose M>0 such that ⇢(2f (s),x)
(a) x 0 and x = 0 if and only if x = 0. (b) k↵k x = ↵ xk . k (c) kx +ky | |kxk+ y . (Triangle inequality) k kk k k k A norm of V defines a metric ⇢ on V via ⇢(x, y)= x y . All concepts from metric and topological spacesk · k are applicable to normed spaces. k k There are multiple ways of choosing norms once a norm is selected. A trivial one is to multiply the original norm by a positive constant. Concepts like neighborhood, convergence, and complete- ness are independent of the choice of these two norms, and so we shall consider them equivalent norms. A more precise characterization of equivalent norms is as follows.
Definition 1.6. Let V be a vector space with two norms , 0. We say these two norms are equivalent if there exists some constant c>0 such that k · k k · k 1 x 0 x c x 0 for any x V. c k k kk k k 2 Example 1.1. The Euclidean space Fd, F = R or C, with the standard norm defined by k · k 2 2 1 x =(x + + x ) 2 k k | 1| ··· | d| is a normed space. Now consider two other norms of Fd defined by
x = max x1 , , xd , called the sup norm; k k1 {| | ··· | |} x = x + + x , called the 1-norm. k k1 | 1| ··· | d| Verifications for axioms of norms are completely straightforward. In the case of sup norm, “balls” in Rd are actually cubes in Rd with faces parallel to coordinate axes. In the case of 1-norm, “balls” in Rd are cubes in Rd with vertices on coordinate axes. These norms are equivalent since x x x 1 d x . k k1 k kk k k k1 Definition 1.7. A complete normed vector space is called a Banach space. Example 1.2. Consider the Euclidean space Fd, F = R or C, with the standard norm .The k · k normed space (Rd, ) is complete since every Cauchy sequence is bounded and every bounded k · k sequence has a convergent subsequence with limit in Rd (the Bolzano-Weierstrass theorem). The d d spaces (R , 1) and (R , ) are also Banach spaces since these norms are equivalent. k · k k · k1 Example 1.3. Given a nonempty set X and a normed space (Y, )overfieldF. The space k · k of functions from X to Y form a vector space over F, where addition and scalar multiplication are defined in a trivial manner: Given two functions f, g, and two scalars ↵, F,define↵f + g by 2 (↵f + g)(x)=↵f(x)+ g(x),xX. 2 Let b(X, Y ) be the subspace consisting of bounded functions from X to Y . Define a real-valued function on b(X, Y )by k · k1 f =sup f(x) . k k1 x X k k 2 It is clearly a norm on b(X, Y ), also called the sup norm. Convergence with respect to the sup norm is clearly the same as uniform convergence. If (Y, ) is a Banach space, then any Cauchy sequence fn in b(X, Y ) converges pointwise to some functionk · k f : X Y ,since f (x) is a Cauchy sequence{ } in Y for any fixed x X.In ! { n } 2 118 8. INTRODUCTION TO BANACH SPACES AND Lp SPACE fact, the convergence fn f is uniform. To see this, let " > 0 be arbitrary. Choose N N such ! 2 that fn fm < "/2 whenever n, m N. For any x X, there exists some mx N such that f (kx) f(xk)1< "/2. Then for any n N, 2 k mx k f (x) f(x) f (x) f (x) + f (x) f(x) < ". k n kk n mx k k mx k This proves that the convergence fn f is uniform. By Theorem 1.2(a), f b(X, Y ), and so the space b(X, Y ) with the sup norm is a! Banach space. 2 Example 1.4. Let (X, ) be a topological space and let (Y, ) be a Banach space over T k · k F = R or C. Denote by C(X, Y ) the space of continuous functions from X to Y . Let Cb(X, Y )= C(X, Y ) b(X, Y ), the space of bounded continuous functions from X to Y . Given any sequence f in C\ (X, Y ) which converges uniformly to f b(X, Y ). By Theorem 1.2(b), f C (X, Y ). { n} b 2 2 b This shows that Cb(X, Y ) is a closed subspace of b(X, Y ), and is therefore a Banach space (see Exercise 1.1).
Definition 1.8. A series k1=1 ak in a normed space X is said to be convergent (or summable) n if its partial sum ak converges to some s X as n .Wesay 1 ak is absolutely k=1 P 2 !1 k=1 convergent (or absolutely summable)if 1 ak < . P k=1 k k 1 P In the following we prove some usefulP criteria for completeness and uniform convergence of series. Theorem 1.3. AnormedspaceX is complete if and only if every absolutely convergent series is convergent.
Proof. Suppose X is complete, k1=1 ak is absolutely convergent. We need to show the conver- n gence of sn = ak.Given" > 0, choose N N such that 1 ak < ",then sn sm < " k=1 P 2 k=N k k k k whenever n, m N.Thus sn n1=1 is a Cauchy sequence, and so it converges. Conversely,P suppose every{ absolutely} convergent series in XPconverges. Let s be a Cauchy { n}n1=1 sequence in X. By Theorem 1.1 it su ces to show that sn n1=1 has a convergent subsequence s . Now choose n such that { } { nk }k1=1 k 1 nk
Given > 1, a, b ⌃ ,let 2 2 1 a b ⇢ (a, b)= | k k|. k Xk=1 Show that (⌃2, ⇢ ) is a complete metric space. 1.3. Consider the space of real sequences s. Let
1 ak bk ⇢(a, b)= k | | . 2 (1 + ak bk ) Xk=1 | | Show that (s, ⇢) is a complete metric space.
1.4. Show that all norms in Fd, F = R or C, are equivalent. Consequently, Fd with any norm is complete.
1.5. Consider the space C0[a, b] Ck(a, b), k N, and let \ 2 (k) f Ck = f + f 0 + f 00 + + f . k k k k1 k k1 k k1 ··· k k1 0 k Show that (C [a, b] C (a, b), k ) is a Banach space. \ k · kC 1.6. Consider the space BV [a, b] of functions on [a, b] with bounded variations. Let f = f(a) + V b(f). k kBV | | a Show that (BV [a, b], ) is a Banach space. k · kBV 2. The `p Space Definition 2.1. Let F = R or C.Given0 The space `1 consists of bounded sequences in F. Addition and multiplication of sequences are defined componentwise: (a ,a , )+(b ,b , )=(a + b ,a + b , ) 1 2 ··· 1 2 ··· 1 1 2 2 ··· (a ,a , ) (b ,b , )=(a b ,a b , ). 1 2 ··· · 1 2 ··· 1 1 2 2 ··· p Clearly ` with any 0 (a) (Young’s inequality) If u, v 0, 1 Fix any M N.Given" > 0, there exists some N N such that 2 2 1 1 M p p (n) (m) p 1 (n) (m) p ak ak ak ak < " n, m N. ! ! 8 k=1 k=1 X X Let m ,thenlet M ,wefind !1 !1 1 p p (n) 1 (n) a a p = ak ak " for any n N. k k ! Xk=1 (n) (n) (n) p Thus a a p 0 as n , and so a p a a p + a p < , a ` .Thisverifies k k p! !1 k k k k k k 1 2 completeness of ` . ⇤ Theorem 2.3. The space `p is separable if 1 p< ,andthespace` is not separable. 1 1 Proof. The space `1 is not separable because it has an uncountable subset s = a =(a ,a ,...) `1 : a = 0 or 1 n { 1 2 2 n 8 } with a b = 1 for any a = b s. Considerk k1 1 p< . Let6 be2 the set of finite sequences with rational coordinates. Clearly 1 p D p p D is countable. Given a ` and any " > 0, we can choose N N such that 1 ak < " /2. 2 2 k=N+1 | | Now choose b , ,b such that N a b p < "p/2. Let b =(b , ,b , 0, 0, ) . 1 N Q k=1 k k 1 P N Then ··· 2 | | ··· ··· 2 D P N 1 1 a b p = a b p = a b p + a p < "p. k kp | k k| | k k| | k| Xk=1 Xk=1 k=XN+1 Thus a b < ". This shows that is dense since " > 0 is arbitrary. k kp D ⇤ Exercises. p p 2.1. Consider the ` space with 0 2.3. Consider sequences of real numbers. Show that the space c0 of sequences converging to zero with sup norm is a Banach space, and for any 1 p 3. The Lp Space In this section we consider a space Lp(E)whichresembles`p on many aspects. After general concepts of measure and integral were introduced, we will see that these two spaces can be viewed as special cases of a more general Lp space. Definition 3.1. Given a measurable set E Rd. For 0 The space L1(E) and the real-valued function on L1(E) are given by k · k1 L1(E)= f : f is measurable on E and ess sup f < , f =esssup f . { E | | 1} k k1 | | Functions in L1(E) are said to be essentially bounded. The measurable function f in the definition of Lp(E) for 0 m f>ess sup f = m 0 f ↵ 1 =0. E { } ✓⇢ ◆ ess supE f<↵ ↵[ B 2Q C @ A This tells us that f ess supE f a.e. on E. Therefore, ess supE f can be characterized as the smallest number to which f is less than or equal almost everywhere. Being “essentially bounded” is di↵erent from being “bounded” in that the former is regardless of measure zero sets. For instance, the function x3 x Q f(x)= 2 sin xx R Q ⇢ 2 \ is unbounded but essential bounded with ess sup f = 1. p For any 0 p When 1 p , the function p is a norm on L (E). This follows from the theorem below, the proof for which1 is similar to thatk · ofk `p. Theorem 3.1. Given 1 p, q with 1 + 1 =1. Let f,g be measurable functions on E Rd. 1 p q ⇢ (a) (Holder’s¨ Inequality for Lp) If f Lp(E), g Lq(E), then fg L1(E) and 2 2 2 fg f g . k k1 k kpk kq (b) (Minkowski’s Inequality for Lp) If f,g Lp(E), then f + g Lp(E) and 2 2 f + g f + g . k kp k kp k kp Proof. (a) The cases p = 1, q = and p = , q = 1 are obvious. Consider 1 q 1 1 Proof. Let r = q p ,then q/p + r = 1. Therefore, p 1 p q q r q q p f p ( f p) p 1r = f q m(E) q . | | | | | | ZE ✓ZE ◆ ✓ZE ◆ ✓ZE ◆ Then the corollary follows from 1 p q p 1 1 p f = f f m(E) qp = f m(E) p q . k kp | | k kq k kq ✓ZE ◆ ⇤ Example 3.1. Consider f(x)=xr, r = 0, defined on [0, ). 6 1 When r<0, f Lp[1, ) if and only if p> 1/r, 2 1 f Lp(0, 1) if and only if 0 gp lim inf gp M p. n n ZE !1 ZE 3. THE Lp SPACE 125 p Therefore g is finite almost everywhere and g L (E). When g(x)isfinite, k1=1 fk(x) is absolutely convergent. Let s(x) be its value, and set 2s(x) = 0 elsewhere. Then the function s is defined everywhere, measurable on E, and P n f = s s almost everywhere on E. k n ! Xk=1 Since s (x) g(x) for all n,wehave s(x) g(x), where hence s Lp(E) and | n | | | 2 s (x) s(x) 2g(x) Lp(E). | n | 2 By the Lebesgue dominated convergence theorem, s s p 0 as n . | n | ! !1 ZE p p This proves that 1 f converges to s L (E), and thus proves completeness of L (E). k=1 k 2 ⇤ d Definition 3.2P. Consider cubes in R of the form [k1,k1 + 1] ... [kd,kd + 1], k1 ...,kd Z. ⇥ ⇥ 2 Bisect each of these cubes into 2d congrument subcubes, and repeat this process. The collection of all these cubes are called dyadic cubes. Theorem 3.4. If 1 p< , then the set of finite linear combinations of characteristic 1 D functions on dyadic cubes is dense in Lp(Rd). In particular, the set of step functions is dense in Lp(Rd). p d Proof. All we need to prove is: given any f L (R ), there exists a sequence fk such 2 2 D that fk f p 0 as k . Itk is su kcient! to consider!1 the case f 0 since + + + f = f f , f f f f + f f (by Minkowski’s inequality). k k kp k k kp k k kp In fact, it is su cient to consider the case f 0 with compact support since p p fk f =lim fk f , m Rd | | [ m,m]d | | Z !1 Z according to the monotone convergence theorem. Let gk be an increasing sequence of nonnegative simple functions such that gk f, f 0 has compact{ } support. Then, by the monotone convergence theorem, % p d p gk L (R ), gk f 0 as k . 2 d | | ! !1 ZR Therefore, it is su cient to consider the case f 0, f is a simple function with compact support. N For a compactly supported simple function f = ak Ek , k=1 N P p p p d f g = ak g for any g L (R ). d R | | Ek | | 2 Z Xk=1 Z Therefore, it is su cient to consider the case when f is the characteristic function of some bounded measurable set F . In this case we can choose a G set G containing F with m(G F ) = 0, so we \ only need to consider the case F being a G set. 126 8. INTRODUCTION TO BANACH SPACES AND Lp SPACE 1 Let F = k, 1 2 be a nested sequence of bounded open sets. Then, by the k=1 O O O ··· monotone convergenceT theorem, p k F 0 as k . d | O | ! !1 ZR Therefore, it is su cient to consider f = ,where is a bounded open set. But in this case, O O 1 f = Qk for some dyadic cubes Qk. By monotone convergence theorem again, the finite sum k=1 n P p d p d Qk converges to f in L (R ) as n . It follows that is dense in L (R ). ⇤ k=1 !1 D P Theorem 3.5. If 1 p< and E Rd is measurable, then Lp(E) is separable. 1 ⇢ Proof. The set in Theorem 3.4 has a countable dense subset 0 which consists of linear combinations of characteristicD functions with rational coe cients. D d p ˜ Given measurable set E R . Let E0 = g E : g 0 . For any f L (E), let f = f on E, ˜ d ˜ ⇢p d D { · 2 D } 2 f = 0 on R E.Thenf L (R ). Choose fk 0 such that \ 2 { } ⇢ D p fk f˜ 0 as k . d | | ! !1 ZR Then p p p fk E f = fk E f˜ fk f˜ 0 as k . | · | d | · | d | | ! !1 ZE ZR ZR Therefore is a dense subset of Lp(E). DE0 ⇤ d Given h R . Let ⌧hf(x)=f(x + h) be the translation operator. 2 Theorem 3.6. (Continuity of Variable Translations) If 1 p< and f Lp(Rd), then 1 2 lim ⌧hf f p =0. h k k !1 Proof. Let be the collection of functions in Lp(Rd) satisfying this property. It follows easily C from the Minkowski inequality that it is a subspace of Lp(Rd). Apparently any characteristic function on a bounded cube belongs to . Therefore, the linear C subspace given in Theorem 3.4 is contained in . Then by Theorem 3.4, is dense in Lp(Rd), so all we haveD to show is that is closed. C C p d C Given f L (R ) and sequence fk in such that fk f p 0 as k .Then 2 { } C k k ! !1 ⌧ f f ⌧ f ⌧ f + ⌧ f f + f f k h kp kh h kkp k h k kkp k k kp ⌧ f f +2 f f kh k kkp k kkp Let h 0, then let k ,wefindlimsuph 0 ⌧hf f p = 0. Therefore is closed. ⇤ ! !1 ! k k C Exercises. 3.1. Use the generalized Young’s inequality in Exercise 2.2 to formulate a generalization of H¨older’s inequality for Lp(E). 3.2. Suppose m(E) < . Show that f =limp f p. How about if m(E)= ? 1 k k1 !1 k k 1 4. DENSE SUBSPACES OF Lp 127 3.3. Let f be a real-valued measurable function on E, m(E) > 0. Define the essential infimum on E by ess inf f =sup ↵ ( , ]:m( f<↵ )=0 . E { 2 1 1 { } } Show that, if f 0, then ess inf f =1/ ess sup (1/f). E E p p p 3.4. Consider L (E)with0 3.8. Show that L1(E) is not separable whenever m(E) > 0. 4. Dense Subspaces of Lp In the proof of Theorem 3.4 we constructed a countable collection of step functions which is dense in Lp(E). This implies that the space of simple functions is also dense in Lp(Rd). In this section we prove that the space of smooth functions with compact supports, and the space of functions with rapidly decreasing derivatives are also dense in Lp(Rd). These results have important applications to partial di↵erential equations. Let us begin with some applications of H¨older’s inequality. We will frequently encounter func- tions in Lp(Rd) and Lq(Rd)with1 p, q and 1 + 1 = 1, in which case we call p, q conjugate 1 p q exponent of each other. Lemma 4.1. Given 1 p . 1 (a) If f Lp(Rd), g L1(Rd), then f g Lp(Rd) and 2 2 ⇤ 2 f g f g . k ⇤ kp k kpk k1 (b) (Minkowski’s integral inequality) If 1 Here ↵ is called a multi-index. Its length is ↵ = ↵1 + + ↵d. Its factorial is ↵!=↵1! ↵d!. We can also use multi-index to express monomials:| | let x =(···x , ,x ), then ··· 1 ··· d x↵ = x↵1 x↵d . 1 ··· d p d m d m d Lemma 4.2. Given 1 p .Iff L (R ), g C0 (R ), 0 m , then f g C0 (R ) and D↵(f g)=f D↵g for any1 multi-index2 ↵ with 2↵ m. 1 ⇤ 2 ⇤ ⇤ | | p d Proof. Let q = p 1 be the conjugate exponent of p. Consider m = 0. Given h R , 2 (f g)(x + h) (f g)(x) = f(x + h y)g(y)dy f(x y)g(y)dy | ⇤ ⇤ | d d ZR ZR = f(x y)g(y + h)dy f(x y)g(y)dy d d ZR ZR = f(x y)[g(y + h) g(y)]dy d ZR f p ⌧hg g q (by H¨older’s inequality) . k k k k By Theorem 3.6, the last term converges to zero as h 0. Note that when p = 1, the term ! 0 d ⌧hg g q converges to zero as h 0 since g is uniformly continuous. This proves f g C (R ). k Consider k m = 1. Given t>0.! By the mean-value theorem, there exist s [0,t] such⇤ 2 that 2 1 1 ((f g)(x + tei) (f g)(x)) = f(y) [g(x + tei y) g(x y)] dy t ⇤ ⇤ d t ZR @ = f(y) g(x + sei y)dy. d @x ZR i By uniform continuity of @ g, and the fact that it has compact support, the last integral converges @xi to f @ g(x) as t 0. Since @ g C0(Rd), f @ g C0(Rd), we see that @ (f g) exists and ⇤ @xi ! @xi 2 0 ⇤ @xi 2 @xi ⇤ equals f @ g.Thisimpliesf g C1(Rd)sincei is arbitrary. The proof for general m follows by ⇤ @xi ⇤ 2 induction. ⇤ It is not hard to see that (L1(Rd), ) is an algebra without identity; that is, there is no L1(Rd) ⇤ 2 such that f = f for every f L1(Rd) (see Exercise 4.3). However, there are “well-behaved” ⇤ 1 d 2 functions K" in L (R ) which “approximate the identity” in certain sense: { } p d Definition 4.1. We say K" ">0 is an approximation of the identity in L (R )if { } f Lp, f K f 0 as " 0+. 8 2 k ⇤ " kp ! ! Here is an explicit way of constructing approximation of the identity. Given K L1(Rd) and " > 0, define 2 1 x 1 x x K (x)= K = K 1 ,..., n . " "d " "d " " ⇣ ⌘ ⇣ ⌘ Lemma 4.3. If K L1(Rd), then for any " > 0, 2 K" = K and lim K"(x) dx =0 r>0. " 0+ Rd Rd x >r | | 8 Z Z ! Z{| | } 4. DENSE SUBSPACES OF Lp 129 x 1 Proof. Let T"(x)= " .ThenT" is linear, invertible, and det(T")= "d .Thus 1 1 1 K" = d K T" = d K = K. d " d " d det(T ) ZR ZR ✓ZR ◆ | " | Z Moreover, for any r>0, 1 x K"(x) dx = d K dx = K(x) dx 0 as " 0+ x >r | | " x >r | " | x > r | | ! ! Z{| | } Z{| | } ⇣ ⌘ Z{| | " } since K L1(Rd). 2 ⇤ 1 d Theorem 4.4. Given 1 p< , K L (R ) and d K =1. Then K" is an approximation 1 2 R { } of the identity in Lp( d). R R Proof. Observe that f K"(x) f(x) f(x y) f(x) K"(y) dy. | ⇤ | d | || | ZR By Minkowski’s integral inequality, 1 p p f K" f p f(x y) f(x) K"(y) dy dx k ⇤ k d d | || | R R Z Z 1 p p p f(x y) f(x) K"(y) dx dy d d | | | | R R Z Z 1 p p = f(x y) f(x) dx K"(y) dy d d | | | | ZR ZR = ⌧ yf f p K"(y) dy. d k k | | ZR Given "0 > 0, choose ⌘ > 0 such that ⌧ yf f p < "0 whenever y < ⌘.Then k k | | f K" f p "0 K"(y) dy + 2 f p K"(y) dy. k ⇤ k y <⌘ | | y ⌘ k k | | Z| | Z| | The theorem follows by letting " 0 and then " 0. ! 0 ! ⇤ The support supp(f) of a measurable function f is defined as the intersection of set in d c f = K R : K is closed and f = 0 a.e. on K . K { ⇢ } Lemma 4.5. Given measurable functions f and g. We have f g =0on (supp(f)+supp(g))c. In particular, supp(f g) supp(f)+supp(g). ⇤ ⇤ ⇢ Proof. Observe that (f g)(x)= f(x y)g(y)dy = f(x y)g(y)dy =0 ⇤ d ZR Zsupp(g) if x y/supp(f) for every y supp(g). That is, f g = 0 on (supp(f)+supp(g))c. 2 2 ⇤ ⇤ We are now ready to prove main theorems in this section. d p d Theorem 4.6. C1(R ) is dense in L (R ), 1 p< . 0 1 130 8. INTRODUCTION TO BANACH SPACES AND Lp SPACE Proof. Given f Lp(Rd). For any N N,let 2 2 f(x)iff(x) , x Theorem 4.7. For any 1 p< , the Schwartz class (Rd) is a dense subspace of Lp(Rd). 1 S d d d Proof. It follows easily from Theorem 4.6, C01(R ) (R ), and the observation that (R ) p d ⇢ S S is a subspace of L (R ). ⇤ Exercises. 4.1. Prove the Lemma 4.1(a). More generally, given 1 p, q, r such that 1 + 1 =1+1 , 1 p q r and given f Lp(Rd), g Lq(Rd), prove the following Young’s convolution inequality: 2 2 f g f g . k ⇤ kr k kpk kq 1 1 1 Hint: Find suitable p0, q0 such that + + = 1, then apply generalized H¨older’s inequality. p0 q0 r 4.2. Use H¨older’s inequality to prove Minkowski’s integral inequality. 4.3. Suppose p is the conjugate exponent of q [1, ]. Show that if f Lp(Rd), g Lq(Rd), 2 1 2 2 then f g C0(Rd). Use it to show that the commutative algebra (L1(Rd), ) has no identity. Hint: Consider⇤ 2 the convolution with characteristic function of a ball. ⇤ 1 2 2 x 4.4. Show that f(x)=e x (0, ) belongs to (R), and g(x)=f(x a)f(b x) C01(R), where a 4.5. Given bounded open sets G1 G2 such that G1 G2. Construct a function f C01 such c ⇢ ⇢ 2 that f = 1 on G1 and f = 0 on G2. 4.6. Given f L ( d), K L1( d) and K = 1. Show that f K (x) f(x) as " 0 1 R R Rd " provided f is continuous2 at x. 2 ⇤ ! ! R 5. LINEAR TRANSFORMATIONS AND DUAL SPACE 131 5. Linear Transformations and Dual Space This section is brief introduction to the concepts of bounded linear operators and dual spaces. Those who unacquainted with undefined concepts herein are referred to any standard textbook for linear algebra. Those who interested in further discussions and examples are referred to any standard textbook for functional analysis. For concepts intimately related to our discussions for Lp spaces, we prove them here. In this section we consider only normed spaces, even though many concepts can be extended to more general topological vector spaces. For convenience, we shall use the same notation for norms of various spaces, and we shall simply write X for (X, ), for instance. It is oftenk evident· k which norm we are referring to. When it is necessary to avoidk · k confusion, we denote its norm by . k · kX Definition 5.1. Given two normed spaces X and Y , a linear transformation (operator) T : X Y is said to be bounded if there exists some M>0 such that ! Tx M x for any x X. k k k k 2 Denote the space of bounded linear operators from X to Y by B(X, Y ). Theorem 5.1. A linear operator T : X Y is bounded if and only if it is continuous. ! Proof. Clearly any bounded linear operator is Lipschitz continuous. If T is continuous, then T 1(B (0)) B (0) for some > 0. Then, whenever x 1, we have 1 k k 2 2 Tx = T x . k k 2 ✓ ◆ x 2 For general x,wehave Tx = T x x x .Thus T is bounded. ⇤ k k k k k kk k k k It is a simple exercise to show that Tx sup Tx =supk k =supTx =infM : Tx M x x X . x =1 k k x=0 x x 1 k k { k k k k8 2 } k k 6 k k k k Their common value is denoted by T . This notation is justified in the following theorem. k k Theorem 5.2. B(X, Y ) with the function T T is a normed space. If Y is a Banach space, then so is B(X, Y ). 7! k k Proof. We only verify the triangle inequality for , for the other two axioms of norm are obvious here. Given S, T B(X, Y ), and x V . k · k 2 2 (S + T )x = Sx + Tx Sx + Tx S x + T x . k k k kk k k kk kk k k kk k This implies the triangle inequality S + T S + T . Given a Cauchy sequence T kin B(X,kk Y ). Fork anyk k x V ,thesequence T x is a Cauchy { n} 2 { n } sequence in Y , and so it converges. Let T : X Y be defined by Tx =limn Tnx. Clearly T is linear. Given " > 0, choose N such that T !T < " for all n, m N.Fix!1n N and x X, k n mk 2 Tnx Tx =lim Tnx Tmx " x . m k k !1 k k k k Therefore, Tx Tx T x + T x (" + T ) x . k kk N k k N k k N k k k 132 8. INTRODUCTION TO BANACH SPACES AND Lp SPACE This shows that T is bounded. Furthermore, Tn T =sup Tnx Tx " for any n N. k k x =1 k k k k This implies that T T 0 as n ,since" > 0 is arbitrary. k n k! !1 ⇤ Observe that any composition of bounded linear operators is a bounded linear operator. Indeed, if S B(X, Y ), T B(Y,Z), then TS B(X, Z) and TS T S since 2 2 2 k kk kk k TSx T Sx T S x for any x X. k kk kk kk kk kk k 2 In particular, the space B(X, X) of bounded linear operators on X with composition is an algebra with identity. Example 5.1. Given 1 q and let p be its conjugate exponent. Given g Lq(Rd). By 1 2 H¨older’s inequality, the linear map G : Lp( d) defined by G(f)= fg is bounded and R R Rd G g . ! k kk kq R Example 5.2. Given g L1(Rd), 1 p . Define the convolution operator G : Lp(Rd) p d 2 1 ! L (R )byG(f)=f g. Lemma 4.1 tells us that G is a bounded linear operator and G g 1. 1 ⇤ 1 1 q d k kk k More generally, if p + q =1+r , g L (R ), then Young’s convolution inequality (exercise 4.1) p d r d 2 tells us that G : L (R ) L (R ) is a bounded linear operator and G g q. ! k kk k Definition 5.2. A linear functional f on a normed space X over F (= R or C) is a linear transformation from X to F. The space B(X, F) of bounded linear functionals on X is called the dual space of X. It is usually denoted by X⇤. Here is an immediate corollary of Theorem 5.2: Corollary 5.3. The dual space X⇤ of any normed space X over R or C is a Banach space. Definition 5.3. Given T B(X, Y ). The adjoint of T , denoted by T , is the linear operator 2 ⇤ from Y ⇤ to X⇤ defined by (T ⇤y⇤)(x)=y⇤(Tx), where y Y , x X. In other words, T y = y T . ⇤ 2 ⇤ 2 ⇤ ⇤ ⇤ It is straightforward to verify that T ⇤ is linear. T ⇤y⇤ is indeed a bounded linear functional on X since it is simply composition of y and T . Moreover, T B(Y ,X )since ⇤ ⇤ 2 ⇤ ⇤ T ⇤y⇤ = y⇤ T =sup y⇤(Tx) sup y⇤ Tx T y⇤ for any y⇤ Y ⇤. k k k k x =1 k k x =1 k kk kk kk k 2 k k k k This observation tells us that T ⇤ T . We leave it as an exercise to show that T ⇤ = T . A commonly used notationk forkkx (x)isk x ,x ,wherex X and x X. k k k k ⇤ h ⇤ i ⇤ 2 ⇤ 2 Remark 5.1. If S B(X, Y ), T B(Y,Z), then (TS)⇤ = S⇤T ⇤. This follows trivially from the definition of adjoint: 2 2 (TS)⇤z⇤ = z⇤ (TS)=(z⇤ T ) S = S⇤(T ⇤z⇤)=(S⇤T ⇤)z⇤ for any z⇤ Z⇤. 2 Definition 5.4. Two normed spaces X, Y are said to be isomorphic if there exists a bijection 1 T B(X, Y ) with bounded inverse T . In this case we say T is invertible. Such an operator T is called2 an isomorphism.WesayX, Y are isometrically isomorphic if there exists an isomorphism T : X Y which is also an isometry. ! 5. LINEAR TRANSFORMATIONS AND DUAL SPACE 133 Two normed spaces are considered equivalent if they are isometrically isomorphic. This clearly defines an equivalence relation on the class of Banach spaces. 1 1 Remark 5.2. We remark here that T is invertible implies T ⇤ is invertible, and (T ⇤) =(T )⇤. Indeed, given any x X , ⇤ 2 ⇤ 1 1 1 T ⇤(T )⇤(x⇤)=T ⇤(x⇤ T )=x⇤(T T )=x⇤. 1 Thus (T )⇤ is a right inverse of T ⇤. The proof for being a left inverse is similar. Now we state some facts about isometric isomorphisms without proof. They can be served as incentives to consider isometric isomorphisms and separable Banach spaces. Example 5.3. Given any normed space X, there exists a canonical isometric isomorphism ⌧ from X to a subspace of X⇤⇤.Itisdefinedby⌧(x)(x⇤)=x⇤(x). The mapping ⌧ is often called the canonical embedding from X to X⇤⇤. For convenience, the term ⌧(x) is often written x, and thus we regard elements in X as elements in X⇤⇤. Definition 5.5. We say a normed space X is reflexive if the canonical embedding ⌧ : X X⇤⇤ is an isomorphism. ! Note that, by Corollary 5.3, any reflexive normed space is a Banach space. Example 5.4. Every separable Banach is isometrically isomorphic to a closed subspace of C0[0, 1]. This is known as the Banach-Mazur theorem. Note that the space Q[t] of polynomi- als with rational coe cients is a countable dense subset of C0[0, 1] (by Weierstrass approximation theorem, or Stone-Weierstrass theorem). The Banach-Mazur theorem tells us that C0[0, 1] is the “largest” separable Banach space. Exercises. 5.1. Verify identities above Theorem 5.2. 5.2. Determine the operator norm G in Examples 5.1 (without using any theorem in the next section), and Example 5.2 for the casekq =k 1. 5.3. Given 1 (a) For any x X, x =sup x⇤(x) : x⇤ X⇤, x⇤ 1 . (Hint: Consider2 k a functionk {| which| sends2↵x to k↵ xk,where} ↵ is a scalar.) (b) For any T B(X, Y ), T =sup y (Tx) : y k Yk , y 1,x X, x 1 . 2 k k {| ⇤ | ⇤ 2 ⇤ k ⇤k 2 k k } 1It is a special case of a more general statement known as the Hahn-Banach theorem,whichisaconsequenceof the axiom of choice. 134 8. INTRODUCTION TO BANACH SPACES AND Lp SPACE (c) T = T for any T B(X, Y ). k k k ⇤k 2 (d) The canonical embedding ⌧ : X X⇤⇤ in Example 5.3 is an isometric isomorphism from X to ⌧(X). ! 6. Dual Space of Lp In this section we characterize the dual space of Lp(I), where I is an interval. A more general theorem will be proved after general measures and integrals were introduced. The key ingredients of the proof are the following two lemmas. The first one says that, roughly speaking, any bounded linear functional on Lp(I) induces a canonical indefinite integral. Gener- alization of this lemma to higher dimensional spaces requires further knowledge about indefinite integrals. p Lemma 6.1. Let I be a bounded interval, I¯ =[a, b], 1 p< .ForanyG L (I)⇤, there exists g L1(I) such that G( )= g for any measurable set A 1I. 2 2 A A ⇢ Proof. Without loss of generality,R assume I =[a, b]. First note that G( A)=G( B)ifA B has measure zero. Let (s)=G( ), s I. Then for any subinterval [s, t] I, [a,s] 2 ⇢ (t) (s)=G( ) G( )=G( )=G( ). [a,t] [a,s] [a,t] [a,s] [s,t] Given nonoverlapping subintervals [a ,b ]: i =1, ,n of I with total length . { i i ··· } n n (b ) (a ) = ( (b ) (a )) sgn ( (b ) (a )) | i i | i i i i i=1 i=1 X Xn = G( ) sgn ( (b ) (a )) [ai,bi] i i i=1 X n = G sgn ( (b ) (a )) i i [ai,bi] i=1 ! Xn 1/p G sgn ( (bi) (ai)) [ai,bi] G . kk kk i=1 p X This implies that AC[a, b]. By the fundamental theorem of calculus, g = L1(I) and 2 0 2 t G( )= (t) (s)= g(u) du for any subinterval [s, t] I. [s,t] ⇢ Zs It follows that G( A)= A g for any measurable set A in I (exercise 6.1). ⇤ Lemma 6.2. Let 1 Rp< , 1 + 1 =1. Let E R be measurable. Suppose g L1(E) and for 1 p q ⇢ 2 some M>0, 'g M ' k kp ZE q for any simple function '. Then g L ( E) and g q M. 2 k k Proof. Assume 1 Then, 1 p ' g M ' = M , n n k nkp n ZE ZE ✓ZE ◆ q q q implying that E n M . By the monotone convergence theorem, E g M . The case p =1 is left to the reader (exercise 6.2). | | R R ⇤ Theorem 6.3. (Riesz Representation Theorem for Lp(I), I R) 1 1 ⇢p Suppose 1 p< , + =1, I R is an interval. For any G L (I)⇤, there exists unique 1 p q ⇢ 2 g Lq(I) such that 2 G(f)= fg for any f Lp(I). 2 ZE Moreover, G = g .ThemapG g is an isometric isomorphism from Lp(I) to Lq(I). k k k kq 7! ⇤ Proof. Denote the -algebra of measurable sets in I by . Consider the case I¯ =[a, b]. By 1 B Lemma 6.1, there exists g L (I) such that G( A)= A g for any A . Given any simple n 2 2 B function ' = a ,where E are disjoint. i=1 i Ei { i} ⇢ B R P n n n G(')= aiG( Ei )= ai g = ai Ei g = 'g. Ei E I Xi=1 Xi=1 Z Xi=1 Z Z Therefore, 'g = G(') G ' . | | k kk kp ZI q p By Lemma 6.2, g L (I) and g q G . Consider the linear functional G˜ on L (I)definedby ˜ 2 k k k k p G(f)= I fg. It is continuous, by H¨older’s inequality. Since simple functions are dense in L (I), we see that G = G˜. H¨older’s inequality also tells us that G g , and so g = G .The R q q function g Lq(I) is unique, for if G(f)= fg˜ for all f kLpkk(I), wek would havek k k k 2 E 2 R f(g g˜) = 0 for all f Lp(I). 2 ZI Then g g˜ gives the zero functional on Lp(I), so that g g˜ = 0. That is, g =˜g in Lq(I). k kq Now we consider unbounded interval I. Let In be an increasing sequence of bounded intervals { } q c such that 1 I = I. Then for any n,thereexistsuniqueg L (I) such that g = 0 on I and n=1 n n 2 n n S G(f)= fg for any f Lp(I)withf = 0 on Ic . n 2 n ZI Moreover, g G .Byuniqueness,g = g a.e. on I for each n. We may assume without k nkq k k n+1 n n loss of generality that gn+1 = gn on In. Then the function g(x):=g (x),xI n 2 n is well-defined on I. Furthermore, gn(x) g(x) as n . By the monotone convergence theorem, | | % | | !1 q q q g =lim gn G . | | n | | k k ZI !1 ZI 136 8. INTRODUCTION TO BANACH SPACES AND Lp SPACE q p In particular, g L (I), g q G . For any f L (I), let fn = f In .Thenfn f , fn f pointwise on I as2n .k Itk followsk k that g g 2 0, f f 0 (see exercise| | 3.5), | and| hence! !1 k n kq ! k n kp ! f g fg (f f)g + f(g g) k n n k1 kn nk1 k n k1 f f g + f g g 0 as n . kn kp k nkq k kp k n kq ! !1 Then, by continuity of G on Lp(I), fg =lim fngn =limG(fn)=G(f). n n ZI !1 ZI !1 Moreover, G g , and so g = G . Uniqueness of such g follows as in the case of bounded k kk kq k |q k k I. This completes the proof. ⇤ Corollary 6.4. If 1 6.3. Let I R be an interval. Suppose g L1(I). Show that for any " > 0 there exists f L1(I), f ⇢= 0, such that 2 2 k k1 6 fg f 1( g "). k k k k1 ZI 6.4. Let I =[0, 1]. Consider the linear functional G on C(I)definedbyG(f)=f(1). Use this linear functional and the assumption in Exercise 5.4 to show that L1(I) is not isometrically isomorphic to L1(I)⇤. 6.5. Determine a representation for (`p) ,1 p< . ⇤ 1