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section5b - 246 C Functional Analysis and Operator Theory...

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Unformatted text preview: 246 C Functional Analysis and Operator Theory o.5.2 The Dual of LP(E) The fact that the dual space of Hilbert space L2(E) is (antilinearly) 1somor— phic to L2(E) has a generalization to L'P(E) for 1 < p < 00. By Holder 3 Inequality, if g 6 LP (B) then( (f, 119): fEf (m) g(m) )dzt defines a bounded lin- ear functional on LP(E), and the following exercise shows that the operator norm of My equals the L1” norm of 9. Exercise C. 41. Let E be a Lebesgue measurable subset of R and fix 1 < p < 00. For each g E L” (E), define ,ug: LP(E) —> C by my) = [Eran—(Eda, feLp(E)- (0.10) Show that #9 e LP(E)* and ”#in = ”9le- Detailed Solutions 519 0.41 Holder’s Inequality implies that. ”pg” 5 “gflpr. (a) Fix 1 < p < 00. Choose 9 E DUKE). Ifg = 0 a.e., then pg 2 0 and ”pg“ = “ngr. Therefore, we can assume that g is not the zero vector. Let law)! = 1 satisfy cu($)g($) = |g(s;)], and define re) : a($)lg(x)lp L. Hgllir I Then since we have (p’—1)p=p’, so urn: = Mflfll—‘Y‘fi -= new” if : 1_' H9! E Halli: Also, _ z :t:— a($)i9(x)lpf—I :3 CL" um» — [Eflxm )d — fE——-”gngcl g< )d ugngi ”gig—1 This shows that ”My“ 2 ”9”?" Hence we have that ling“ = l|g||pr. ”slip!- (b) Now consider the case p = 1. Fix any nonzero g E L°°(lR). Choose 5 > 0. Then there exists a set A Q E with 0 < {AI < 00 such that [9(3)] 3 llglloo. — s for a.e. as e A. Let f : W‘A- Then ||f|f1 = 1, and 1 l (m) = 1741/ sends: 2 171? [(llgllw-Eldfli : ”gum—e- Hence “pg“ 2 “g”00 — e, and since E is arbitrary we conclude that ”pg“ 2 llglloo. (c) Finally, consideii= 00. Fix any nonzero g E LlaR). As before, let |a(q:)| = 1 satisfy 05(33):; :12) = |g(m)[, and define f(f£) = a(a:) for all :12. Then llflloo = 1: and my) = [Eammdm = L Igendw =llglh. This shows that “n9“ 2 [MIL so we have that may” 2 “9H1. 0.5 Some Dual Spaces 247 kl‘bhoug-h-mdw-i-H-net-fiwe-it, the next theorem states that if 1 S p < 00 then every bounded linear functional on LITE) has the form My for some 9 E Lpt(e). Consequently, LP(E)* and LIKE) are (antilinearly) isomorphic. However, for p = 1, although we obtain an embedding of L1(E) into L°° (E)*, this mapping is not surjective. The proof of Theorem 0.42 relies on the Radon— Nikodym Theorem (see Theorem 13.59). Theorem (3.42 (Dual Space of Lp(E)). Let E: be a Lebesgue measurable subset of R, and fix 1 S p < 00. For each 9 E L"(E), define #9 as in equa- tion (0.10). Then the mapping T: LPI(E) —> Lp(E')* defined by T(g) 2 ,ug is an antilinear isometric isomorphism of LPYE) onto LP(E)*. There is an analogous result for the £1” spaces, and generalizations to L? (X) for arbitrary measure spaces X , see [F0199]. REAL ANALYSIS LECTURE NOTES: THE DUAL OF LP CHRISTOPHER HEIL Notation: Integrals with unspecified measures are taken to be Lebesgue integrals. That is, 'we write f f = firearm. We begin with a kind of converse to the fact that each element of U'(E) defines a bounded linear functional on EU?) Theorem 1. Let E be a Lebesgue measurable subset of R, and fix 1 S p S 00. Let S be the set of simple functions on E that vanish outside a set of finite measure, i.e., S = {gin E —> (C : gt is simple and |{¢> 79 OH < co}. . Suppose that: (a) g: E —> C is Lebesgue measurable, (b) dig E L1(E) for each qt 6 S, and (c) M, = sup{l]¢g‘=¢es, Il¢||p=1} < oo. - Then 9 E LP'(E) and ”9H,; =IMQ. Proof. By Holder’s Inequality, we have ' (fagl s Haunt, so we automatically have My 3 l|g||pn Let a(:c) be the complex scalar of unit modulus such that |9(CC)| = 0420982), er. Step 1. Note that if q!) E S, then gt E PH?) and by definition of Mg we have l/asg We will extend this formula to bounded measurable functions f on E that vanish outside a set of‘finite measure. . ' ‘ < My ||¢|lp- 1 2 THE DUAL OF LP Suppose that f is bounded and measurable, and that f is nunzero only on a set A of finite measure. Then XA E 3, so XA g E L1(E). Since f is bounded and zero outside of A, we therefore have f g E L1(E). By our standard approximation theorems, we can find simple functions 913;, such that Q51, —> f pointwise a.e. and [de g I f | Then each qbk belongs to S, and we have 45;, g —> f g pointwise ac. and lei], 9| 3 | f g| E L1 (E) Hence, by the Lebesgue Dominated Convergence Theorem, W = tinsel/w Step 2. Suppose now that 1 < p < 00, so we also have 1 < p’ < 00. Let qbg, be simple functions on E such that 0 S 925;, /‘ |glp'. If necessary, by replacing qbk with qbk - X Efll-kfi] we may assume that each qbk vanishes outside a set of finite measure. Since each gt], is nonnegative, we can define Sgymmtsawt 0) 9!: = CH ' 9511/?- Motivation: 1 1 1"? r 1—-1 r_ _ Im= f=ameW)s=mpa and therefore lgkgl e W- . . 1 More premsely, Since ngl = qbk/p, we have na:(fwflw=va r351: = if” if" S aimlgl = aft/Fag = m- Further, each 9;, is bounded and vanishes outside a set of finite measure, so by equation (1) we have HMWWW=wa=fagfiwsaMt=amW. If g = 0 then there is nothing to prove. Otherwise, we will have qty, 75 0 for all large enough It, and therefore we can divide through in the equation above to obtain wmtsa- By Fatou’s Lemma, we therefore have “ME/Mt=flmmmsnmflmhwgymmgwi k—roo k—mo 1/19 - 1/ = ll¢k||1 p and Thus 9 e L?” (E) and “9“,, g My. Step 5’. SuppOse that p = 1. Fix a > 0, and define A = {M Z Mg+e}. THE DUAL OF LP 3 If [A] > 0, choose B Q A with 0 < IBI < oo. Define f = CK'XB. Then f is bounded and vanishes outside a set of finite measure, so by equation (1) we have ‘fmlswwun=lm. fn=1gg=ém:ca+ma (My +5) IBI g My IBI: which is a contradiction; Therefore we must have |A| = 0. Hence |g| 5 Mg + 5 a.e. Since this is true for every 5 > 0, we conclude that “gr”00 S Mg < oo. However, Therefore Step .4. Suppose p 2 be, so we have p’ = 1. Set 91: = 04 XEn[—Ic,k]~ Then gk is bounded an vanishes outside a set of finite measure, so we have f M=fW1<Ammn=%- EanJc] Since this is true for every k, we have Hm=LMS%<w Hence g E L1(E). #3 Remark 2. This theorem extends to arbitrary positive measure spaces (X , 2, pi) if we assume either that the set 39 = {g gé 0} is (Ir—finite, or that the measure it is semifinite. See Theorem 6.14 in Folland for details. Now we can characterize the dual of LP (E) for finite p. Theorem 3 (Dual Space of LP(E )). Let E be a Lebesgue measurable subset of R, and fix 1<p<oo. ForeacthB’UE‘) ,definengby <fMQ= [rm 7da feme. ' (h Then the mapping T. LP(E )-—> *defined by T(g ) 2 n9 is an antiiinear isometric isomorphism of LP(E) onto LP(E)£JP 4 7 THE DUAL OF L” Proof. We already know that T is an antilinear isometric map of EDIE) into .U’(E)*. There— fore T is injective, and hence we only need to prove that T is surjective. Case 1: [El < 00. Suppose that ,u E LP(E)*, i.e., u: LP(E) —> (C is bounded and linear. Define 1/(A) = (X A, ,u), measurable A Q E. Our immediate'goal is to show that V is a complex measure on (E, £13), where EE is the o—algebra of Lebesgue measurable subsets of E. First, since E has finite measure, we have X A E LP(E) for each measurable AC E. Therefore 1/(A) is a well—defined complex scalar for each A Q E, and also 12(0): (0, ,u.)— «— 0. Second, suppose that A1, A2, . .. are disjoint measurable subsets of E, and let A: UAj . Then we have pointwise that = ZXAj(:E), :6 E E1 j=1 since for each :1: the series on the right has at most one nonzero term. Exercise: Show that the series X A = 2:1 X A}. converges in IP-norm (note that we use the fact that [El < 00 here). Consequently, using both the linearity and the continuity of ,u., we have V(A) = (99.1,“ )=<:XA, a): if XAJJJ.) 2;?!” Therefore 1/ is countably additive, and hence is a complex measure on E. Moreover, if A Q E satisfies [AI = 0, then XA = 0 a.e., so we have VlA) = (XAnu) = (0: M) = 0- Hence 1/ << dm. The RadoniNikodym Theorem therefore implies that there exists a g E L1(E) such that .dr/ = 9 dm. We will show that g E LP’(E) and that ,u = ,ug = T(g), which will show that T is surjective. Now, the symbols dy = @033: mean that 1/(A) = ] mobs, measurable A Q E. A Consequently, for each measurable A Q E we have (95.4%) = ”(Al = fXAdV = fXAE- By linearity, given any simple function 45 on E, we therefore have >=/¢§. THE DUAL OF L" 5 Since |E| < 00, any simple function belongs to IKE). Since it is bounded on LP(E), we therefore have that fM=mm1nwwp It therefore follows from Theorem 1 that g E Li’(E ). Hence #9 is a bounded linear functional on D” (E), i.e., as, E LP(E)*. For any simple function we have (We —/¢5§—( - (as) Thus ,u and pig agree on the simple functions. Since these operators are continuouus and since the simple functions are dense in LP(E), we conclude that ,u = My. This finishes the proof for this case. Case 2: |E| = 00. Set E], = E H [—k, k] (the important fact here being that the real line under Lebesgue measure is a—finite). Let me be the restriction of ,u to LP(E;,) Then pig, 6 IP(E;,)*, so it follows from Case 1 that there exists a function g, E LPIEk) such that pk = ugh. By taking functions to be zero outside of Egg, we can consider LP(Ek) to be a subspace of L10 (E) Therefore, we have (full!) : (firu'k) ”T ffia fELp(Ek)- Ek Furthermore, “gkllp’ = HMH S Hell, where in, is the operator norm of ,u restricted to LP(Ek), and “n“ is the operator norm of ,u, on LP(E). " Now, if f is any function in LP(E;,), then we have f E LP(Ek+1) as well, and since f is zero outside of EC, we have f9? = (fuu’) = f9k+1 = f9k+r E1: EH1 Es: Consequently, we must have 9;, = 9k+1 for almost every x 6 BE. Hence we can define a function g on E by setting g(x) = 9143) if a: 6 Big. Then 9 is measurable, and if 1 < p < 00 then we have by Fatou’s Lemma that ' muswgumsmm<m If p = 1 then p’ = 00 and we have llglloo = Slip llgklloo 3 Hell < 00- In any case, we conclude that g E LPTE). Finally, if f E LP(E) then fXE,c E H’UE';c , so ) kaw= £3ka- .<fXEkHu‘) = 5 THE DUAL OF LP However, fXEk —> f in LP—norm and ,u, is continuous, so On the other hand, fg E L1(E) by Holder’s Inequality, so by the Lebesgue Dominated Convergence Theorem, we have fa a» f M. Ek E Hence (m = [Era = (my) This is true for all f 6 LP (E), so we conclude that ,u = My. III Remark 4. This theorem extends to arbitrary positive measure spaces (X , 23, pt). For 1 < p < 00 we can even prove this Without any restrictions on a, i.e., even if a is not cr—finite. However, for p = 1 we must assume that p, is a—finite. See Theorem 6.15 in Folland for details. 0.5 Some Dual Spaces 247 0.5.3 The Relation between LP'(E) and LP(E)* We have chosen to consider the relation between LP’(E) and LP(E)* in a way that most directly generalizes the inner product on a Hilbert space and the characterization of the dual space of a Hilbert space given by the Riesz Representation Theorem. Under our choice, we write the action of ,u E L?’(E)* on f E LNE) as (f, ,u), and regard this as a sesquilinear form, linear in f but antilinear in ,u. With this notation, the following statements hold. ,(a) LNE), EDIE), and LP(E)* are linear spaces. (b) LP(E)* is the space of bounded linear functionals on L'P(E). (c) T: LP'(E) —> LP(E)* given by T(g) : rig is an isometric isomorphism, but is antilinear. To illustrate one advantage of this approach, consider the special case p 2 2. Since L2(E) is both a Hilbert space and a particular Lp space, we have intro— duced two different uses of the notation (-, -) with regard to L2(E). On the one hand, ( f, g) denotes the inner product of f, g 6 L2(E), while, on the other hand, (f, ,u) denotes the action of ,u E L2(E)* on f E L2(E). Fortunately, (f, g) = ( f, pg), so our linear functional notation is not in conflict with our inner product notation. This notationally simplifies certain calculations. For example, if A: L2(E) —> L2(E) is unitary then we have for f, g E L2(E) that (f, g) = (A 1", Ag), where (-, -) denotes the inner product on H, and also (f, pg) 2 (A f, ,u Ag). However, we do have to accept that our identification of g with its. is antilinear rather than linear. There are several alternative approaches, each with their own advantages and disadvantages. Let us discuss two of these. A second choice is to base our notation on the usual convention that if 1/ is a linear functional, then the notation 1/( f ) is linear in both f and 1/. If we , follow this convention, then we will associate a function g 6 L10 {E} with the functional 1/9: LPGR) —> C defined by vg(f) = ff(:t)g(:r:)das. With this notation, we have the following facts. (3.) LP(E), HIE), and LP(E)* are linear spaces. (b) LP(E)* is the space of bounded linear functionals on LP(E). (c) U: LPIE) —) LP(E)* given by U (9) = 1/9 is an isometric isomorphism, and is linear. This is a natural choice except for the fact that the notation VU') is not an extension of the inner product on L2(E). Specifically, although we identify 5] E LZOR) with 1/9 E L2(E)*, the inner product (f, 9) does not coincide with V9 (f). Hence for p = 2, if A: L2 (E) —> L2 (E) is unitary, then although we have (f, g) = (Af,Ag), we do not have 1290“) 2 12,49 (Af). Another consequence is 248 C Functional Analysis and Operator Theory that if L: L2(E) —i L2(E) is linear, then the adjoint L* of L defined by the requirement that (Lf, g) = (f, L*g) is different than the adjoint defined by the requirement that 129(Lf) = VL*g(f) (adjoints are considered in Section 0.6). Essentially, we end up with different notions for concepts on L2(E) depend- ing on whether we regard L2(E) as a Hilbert space under the inner product or a member of the class of Banach spaces LP(E) with the identification be- tween Lp (E)* and L? (E) given by U. The isomorphism U between L2 (E) and L2(E)* is different than the one given by the Riesz Representation Theorem (Exercise C. 38). A third possibility is to let the functionals on L1" (E) be antilinear func- tionals instead of linear. For example, we can associate g E LPI(E) with the functional pg: LPUR) u» C given by [f.pgl= /f(:c)g(zv) Then the the dual space is a space of antilinear functionals, i.e., the dual space is LP(E)"' = {,0: UTE) —-> (C : p is bounded and antilinear}. In this case, we have the following facts. (a) LIKE), EDIE), and LP(E)" are linear spaces. (b) LIKE)" is the linear space whose elements are the bounded antilinear functionals on LP(E). (c) V: L”,(E) —» UTE)" given by V(g) : pg is an isometric isomorphism, and is linear. While V is linear, we again have a disagreement between our functional no— tation and the inner product on L2(E). Despite the fact that Our discussion of notation has been quite lengthy, , in the end the difference between these choices comes dOWn to nothing more than convenience — each choice makes certain formulas “pretty” and others “unpleasan .” As our main concern is the use of these notations in harmonic analysis, our choice is motivated by the formulas of harmonic analysis, and in particular the Parsevai formula for the Fourier transform. We choose a notation that directly generalizes the inner product, and consequently obtain the simplest notational representation for generalizing the Fourier transform to distributions and measures (see Chapters 4 and 5). ...
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