Definition and Elementary Properties of L p Spaces

Exercises for Chapter 3

4.2 Definition and Elementary Properties of L p Spaces

Theorem 4.4 (Tonelli).Let F (x, y) : ₁×₂ → R be a measurable function satisfying

(a)

₂|F (x, y)|dμ2<∞fora.e.x ∈1

and (b)

1

dμ₁

2

|F (x, y)|dμ₂<∞.

ThenF ∈L¹(₁×₂).

Theorem 4.5 (Fubini).Assume thatF ∈ L¹(1×2). Then for a.e. x ∈ 1, F (x, y) ∈ L¹_y(2)and

₂F (x, y)dμ2 ∈ L¹_x(1). Similarly, for a.e. y ∈ 2, F (x, y)∈L¹_x(₁)and

1F (x, y)dμ₁∈L¹_y(₂).

Moreover, one has

1

dμ1

2

F (x, y)dμ2=

2

dμ2

1

F (x, y)dμ1=

1×2

F (x, y)dμ1dμ2.

4.2 Definition and Elementary Properties of L

^p

Spaces

Definition. Letp∈Rwith 1< p <∞; we set L^p()=

f :→R;f is measurable and|f|^p∈L¹()

with

fL^p = fp= -

|f (x)|^pdμ .1/p

. We shall check later on that pis a norm.

Definition. We set L^∞()=

f :→R

f is measurable and there is a constant C such that|f (x)| ≤Ca.e. on

with

fL^∞ = f∞=inf{C; |f (x)| ≤Ca.e. on}. The following remark implies that _∞is a norm:

Remark1.Iff ∈L^∞then we have

|f (x)| ≤ f∞ a.e. on.

Indeed, there exists a sequenceC_nsuch thatC_n→ f∞and for eachn,|f (x)| ≤ C_n a.e. on. Therefore|f (x)| ≤ C_n for allx ∈ \E_n, with |E_n| = 0. We set

E= ∪^∞_n=1E_n, so that|E| =0 and

|f (x)| ≤C_n ∀n, ∀x∈\E; it follows that|f (x)| ≤ f_∞ ∀x∈\E.

Notation. Let 1≤p≤ ∞; we denote byptheconjugate exponent, 1

p + 1 p =1.

•Theorem 4.6 (Hölder’s inequality). Assume that f ∈ L^p and g ∈ L^p with 1≤p≤ ∞. Thenf g∈L¹and

(1)

|f g| ≤ fpgp.

Proof. The conclusion is obvious ifp = 1 orp = ∞; therefore we assume that 1< p <∞. We recallYoung’s inequality:¹

(2) ab≤ 1

pa^p+ 1

pb^p ∀a≥0, ∀b≥0.

Inequality (2) is a straightforward consequence of the concavity of the function log on(0,∞):

log 1

pa^p+ 1 pb^p

≥ 1

plog a^p+ 1

plog b^p =log ab.

We have

|f (x)g(x)| ≤ 1

p|f (x)|^p+ 1

p|g(x)|^p a.e.x ∈. It follows thatf g∈L¹and

(3)

|f g| ≤ 1 pf^p

p+ 1 pg^p

p. Replacingf byλf (λ >0)in (3), yields

(4)

|f g| ≤ λ^p−1 p f^p

p+ 1 λpg^p

p.

Choosing λ = f⁻_p¹g^pp^/p (so as to minimize the right-hand side in (4)), we obtain (1).

1It is sometimes convenient to use the formab≤εa^p+Cεb^pwithCε=ε^−1/(p−1).

4.2 Definition and Elementary Properties ofL^pSpaces 93 Remark2.It is useful to keep in mind the following extension of Hölder’s inequality:

Assume thatf₁, f₂, . . . , f_kare functions such that f_i ∈L^pi, 1≤i≤kwith 1

p = 1 p1+ 1

p2 + · · · + 1 pk ≤1.

Then the productf =f1f2· · ·fk belongs toL^pand fp≤ f1p₁f2p₂· · · fkp_k.

In particular, iff ∈L^p∩L^qwith 1≤p≤q≤ ∞, thenf ∈L^rfor allr,p≤r≤q, and the following “interpolation inequality” holds:

fr ≤ f^αpf^1−α

q , where1 r =α

p +1−α

q , 0≤α≤1; see Exercise 4.4.

Theorem 4.7.L^pis a vector space and pis a norm for anyp,1≤p≤ ∞.

Proof. The casesp=1 andp = ∞are clear. Therefore we assume 1< p < ∞ and letf, g∈L^p. We have

|f (x)+g(x)|^p≤(|f (x)| + |g(x)|)^p≤2^p(|f (x)|^p+ |g(x)|^p).

Consequently,f +g∈L^p. On the other hand, f +g^pp =

|f +g|^p−1|f +g| ≤

|f +g|^p−1|f| +

|f +g|^p−1|g|.

But|f+g|^p−1∈L^p, and by Hölder’s inequality we obtain f +g^pp≤ f +g^pp⁻¹(fp+ gp), i.e.,f +gp≤ fp+ gp.

•Theorem 4.8 (Fischer–Riesz).L^pis a Banach space for anyp,1≤p≤ ∞.

Proof. We distinguish the casesp= ∞and 1≤p <∞.

Case 1:p= ∞.Let(f_n)be a Cauchy sequence isL^∞. Given an integerk≥1 there is an integerN_k such thatf_m−f_n∞≤ ¹_k form, n ≥N_k. Hence there is a null setE_ksuch that

(5) |f_m(x)−f_n(x)| ≤ 1

k ∀x ∈\E_k, ∀m, n≥N_k. Then we letE=

kEk—so thatEis a null set—and we see that for allx ∈\E, the sequencef_n(x)is Cauchy (inR). Thusf_n(x)→f (x)for allx ∈\E. Passing to the limit in (5) asm→ ∞we obtain

|f (x)−fn(x)| ≤ 1

k for allx∈\E, ∀n≥Nk.

We conclude thatf ∈ L^∞andf −f_n∞ ≤ ¹_k ∀n ≥ N_k; thereforef_n → f inL^∞.

Case 2: 1≤p <∞.Let(f_n)be a Cauchy sequence inL^p. In order to conclude, it suffices to show that a subsequence converges inL^p.

We extract a subsequence(f_nk)such that f_nk+1−f_nkp≤ 1

2^k ∀k≥1.

[One proceeds as follows: choosen1such thatfm−fnp ≤ ¹₂ ∀m, n ≥ n1; then choosen₂≥n₁such thatf_m−f_np ≤ ₂¹2 ∀m, n≥n₂etc.] We claim that f_nk converges inL^p. In order to simplify the notation we writef_kinstead off_nk, so that we have

(6) fk+1−fkp ≤ 1

2^k ∀k≥1.

Let

g_n(x)= n k=1

|f_k+1(x)−f_k(x)|, so that

g_np≤1.

As a consequence of the monotone convergence theorem,g_n(x)tends to a finite limit, sayg(x), a.e. on, withg∈L^p. On the other hand, form≥n≥2 we have

|f_m(x)−f_n(x)| ≤ |f_m(x)−f_m−1(x)|+· · ·+|f_n+1(x)−f_n(x)| ≤g(x)−g_n−1(x).

It follows that a.e. on, f_n(x)is Cauchy and converges to a finite limit, sayf (x).

We have a.e. on,

(7) |f (x)−f_n(x)| ≤g(x) forn≥2,

and in particular f ∈ L^p. Finally, we conclude by dominated convergence that fn−fp→0, since|fn(x)−f (x)|^p →0 a.e. and also|fn−f|^p ≤g^p ∈L¹. Theorem 4.9.Let(fn)be a sequence inL^pand letf ∈L^pbe such thatfn−fp

→0.

Then, there exist a subsequence(fn_k)and a functionh∈L^psuch that (a)f_nk(x)→f (x)a.e.on,

(b)|f_nk(x)| ≤h(x) ∀k,a.e.on.

Proof. The conclusion is obvious whenp= ∞. Thus we assume 1≤p <∞. Since (f_n)is a Cauchy sequence we may go back to the proof of Theorem 4.8 and consider

4.3 Reflexivity. Separability. Dual ofL^p 95 a subsequence(f_nk)—denoted by(f_k)—satisfying (6), such thatf_k(x)tends a.e. to a limit²f(x)withf ∈L^p. Moreover, by (7), we have|f(x)−f_k(x)| ≤ g(x)

∀k, a.e. onwithg ∈L^p. By dominated convergence we know thatf_k → f in L^p and thusf =fa.e. In addition, we also have|f_k(x)| ≤ |f(x)| +g(x), and the conclusion follows.

4.3 Reflexivity. Separability. Dual of L

^p

We shall consider separately the following three cases:

(A) 1< p <∞, (B) p=1, (C) p= ∞.

A. Study ofL^p()for 1< p <∞.

This case is the most “favorable”:L^pis reflexive, separable, and the dual ofL^p isL^p.

•Theorem 4.10.L^pis reflexive for anyp, 1< p <∞.

The proof consists of three steps:

Step1 (Clarkson’s first inequality). Let 2≤p <∞.We claim that

(8)

f +g 2

^p

p

+ f −g

2 ^p

p

≤ 1

2(f^pp+ g^pp) ∀f, g∈L^p. Proof of(8).Clearly, it suffices to show that

a+b 2

^p+ a−b

2

^p≤ 1

2(|a|^p+ |b|^p) ∀a, b∈R. First we note that

α^p+β^p ≤(α²+β²)^p/2 ∀α, β ≥0 (by homogeneity, assumeβ =1 and observe that the function

(x²+1)^p/2−x^p−1

increases on[0,∞)). Choosingα= |^a+₂^b|andβ = |^a−₂^b|, we obtain a+b

2 ^p+

a−b 2

^p≤

a+b 2

²+ a−b

2 ²

p/2

= a²

2 +b² 2

p/2

≤1

2(|a|^p+|b|^p)

2A priori one should distinguishfandf: by assumptionfn→finL^p, and on the other hand, fn_k(x)→f(x)a.e.

(the last inequality follows from the convexity of the function x → |x|^p/2 since p≥2).

Step2:L^p is uniformly convex, and thus reflexive for2 ≤ p < ∞. Indeed, let ε >0 and letf, g∈L^pwithfp≤1,gp≤1, andf−gp > ε. We deduce from (8) that

f +g 2

^p

p

<1−/ε 2

0p

and thus^f+₂^gp < 1−δ withδ = 1− [1−(^ε₂)^p]^1/p > 0. Therefore, L^p is uniformly convex and thus reflexive by Theorem 3.31.

Step3:L^pis reflexive for1< p≤2.

Proof. Let 1< p <∞. Consider the operatorT :L^p→(L^p)defined as follows:

Letu∈L^pbe fixed; the mappingf ∈L^p →

uf is a continuous linear functional onL^p and thus it defines an element, sayT u, in(L^p)such that

T u, f =

u f ∀f ∈L^p. We claim that

(9) T u_(L^p₎ = up ∀u∈L^p.

Indeed, by Hölder’s inequality, we have

|T u, f| ≤ up fp ∀f ∈L^p and thereforeT u_(L^p₎ ≤ up.

On the other hand, set

f0(x)= |u(x)|^p−2u(x) (f0(x)=0 ifu(x)=0).

Clearly we have

f₀∈L^p, f₀p =u^p−1

p and T u, f₀ = u^pp; thus

(10) T u_(L^p₎ ≥ T u, f0 f₀p

= up.

Hence, we have shown thatT is an isometry fromL^pinto(L^p), which implies that T (L^p)is a closed subspace of(L^p)(becauseL^pis a Banach space).

Assume now 1< p≤2. SinceL^pis reflexive (by Step 2), it follows that(L^p)

4.3 Reflexivity. Separability. Dual ofL^p 97 is also reflexive (Corollary 3.21). We conclude, by Proposition 3.20, thatT (L^p)is reflexive, and as a consequence,L^pis also reflexive.

Remark3.In fact,L^pis also uniformly convex for1< p≤2. This is a consequence ofClarkson’s second inequality, which holds for 1< p≤2:

f +g 2

^p

p

+ f −g

2 ^p

p

≤ 1

2f^p

p+1 2g^p

p

1/(p−1)

∀f, g∈L^p.

This inequality is trickier to prove than Clarkson’s first inequality (see, e.g., Prob- lem 20 or E. Hewitt–K. Stromberg [1]). Clearly, it implies thatL^pis uniformly convex when 1< p≤2; for another approach, see also C. Morawetz [1] (Exercise 4.12) or J. Diestel [1].

•Theorem 4.11 (Riesz representation theorem).Let1 < p < ∞and letφ ∈ (L^p). Then there exists a unique functionu∈L^p such that

φ, f =

uf ∀f ∈L^p. Moreover,

up =φ

(L^p).

Remark4.Theorem 4.11 is very important. It says that every continuous linear func- tional onL^pwith 1< p <∞can be represented “concretely” as an integral. The mapping φ → u, which is a linear surjective isometry, allows us to identify the

“abstract” space(L^p)withL^p.

In what follows, we shall systematically make the identification (L^p)=L^p.

Proof. We consider the operatorT : L^p → (L^p) defined byT u, f = uf

∀u ∈ L^p,∀f ∈ L^p. The argument used in the proof of Theorem 4.10 (Step 3) shows that

T u(L^p) = up ∀u∈L^p.

We claim thatT is surjective. Indeed, letE=T (L^p). SinceEis a closed subspace, it suffices to prove thatE is dense in(L^p). Let h ∈ (L^p)satisfyh, T u = 0

∀u∈L^p. SinceL^pis reflexive,h∈L^p, and satisfies

uh=0∀u∈L^p. Choosing u= |h|^p−2h, we see thath=0.

Theorem 4.12.The spaceC_c(R^N)is dense inL^p(R^N)for anyp,1≤p <∞. Before proving Theorem 4.12, we introduce some notation.

Notation. Thetruncation operationT_n:R→Ris defined by

T_nr=

⎧⎨

⎩

r if|r| ≤n, nr

|r| if|r|> n.

Given a setE⊂, we define thecharacteristic function³χ_E to be χ_E(x)=

1 ifx∈E, 0 ifx∈\E.

Proof. First, we claim that given f ∈ L^p(R^N)andε > 0 there exist a function g∈L^∞(R^N)and a compact setKinR^Nsuch thatg=0 outsideKand

(11) f−gp < ε.

Indeed, letχ_n be the characteristic function ofB(0, n)and letf_n = χ_nT_nf. By dominated convergence we see that f_n −fp → 0 and thus we may choose g = f_n withn large enough. Next, givenδ > 0 there exists (by Theorem 4.3) a functiong1∈C_c(R^N)such that

g−g11< δ.

We may always assume thatg1∞≤ g∞; otherwise, we replaceg1byTng1with n= g∞. Finally, we have

g−g1p≤ g−g1^1/p₁ g−g1¹_∞^−(1/p)≤δ^1/p(2g_∞)^1−(1/p). We conclude by choosingδ >0 small enough that

δ^1/p(2g∞)^1−(1/p)< ε.

Definition. The measure spaceis calledseparableif there is a countable family (E_n)of members ofMsuch that theσ-algebra generated by(E_n)coincides with M(i.e.,Mis the smallestσ-algebra containing all theE_n’s).

Example.The measure space=R^Nis separable. Indeed, we may choose for(E_n) any countable family of open sets such that every open set inR^Ncan be written as a union ofE_n’s. More generally, ifis aseparable metric spaceandMconsists of the Borel sets (i.e.,Mis theσ-algebra generated by the open sets in), thenis a separable measure space.

Theorem 4.13.Assume thatis a separable measure space. ThenL^p()is sepa- rable for anyp,1≤p <∞.

We shall consider only the case = R^N, since the general case is somewhat tricky. Note that as a consequence,L^p() is also separable for any measurable set⊂R^N. Indeed, there is a canonical isometry fromL^p()intoL^p(R^N)(the

3Not to be confused with theindicator functionIEintroduced in Chapter 1.

4.3 Reflexivity. Separability. Dual ofL^p 99 extension by 0 outside); thereforeL^p()may be identified with a subspace of L^p(R^N)and henceL^p()is separable (by Proposition 3.25).

Proof of Theorem4.13when= R^N.LetRdenote the countable family of sets inR^N of the form R = 1_N

k=1(a_k, b_k)witha_k, b_k ∈ Q. LetE denote the vector space overQgenerated by the functions(χ_R)_R∈R, that is,Econsists of finite linear combinations with rational coefficients of functionsχ_R, so thatEis countable.

We claim that E is dense inL^p(R^N). Indeed, givenf ∈ L^p(R^N)andε > 0, there exists somef₁∈C_c(R^N)such thatf −f₁p < ε. LetR ∈Rbe any cube containing suppf₁(the support off₁). Givenδ >0 it is easy to construct a function f2 ∈ E such thatf1−f2_∞ < δ andf2 vanishes outsideR: it suffices to split R into small cubes ofRwhere the oscillation (i.e., sup−inf) off1is less thanδ.

Therefore we havef1−f2p ≤ f1−f2∞|R|^1/p < δ|R|^1/p. We conclude that f −f2p<2ε, providedδ >0 is chosen so thatδ|R|^1/p< ε.

B. Study ofL¹().

We start with a description of the dual space ofL¹().

•Theorem 4.14 (Riesz representation theorem).Letφ∈(L¹). Then there exists a unique functionu∈L^∞such that

φ, f =

uf ∀f ∈L¹. Moreover,

u_∞= φ(L¹).

•Remark5.Theorem 4.14 asserts that every continuous linear functional onL¹can be represented “concretely” as an integral. The mappingφ →u, which is a linear surjective isometry, allows us to identify the “abstract” space(L¹)withL^∞.In what follows, we shall systematically make the identification

(L¹)=L^∞.

Proof. Let(_n)be a sequence of measurable sets insuch that= ∪^∞_n=1nand

|_n|<∞ ∀n. Setχ_n=χ_n.

Theuniquenessofuis obvious. Indeed, supposeu∈L^∞satisfies

uf =0 ∀f ∈L¹.

Choosingf =χnsignu(throughout this book, we use the convention that sign 0= 0), we see thatu=0 a.e. onnand thusu=0 a.e. on.

We now prove the existence of u. First, we construct a function θ ∈ L²() such that

θ (x)≥ε_n >0 ∀x ∈_n.

It is clear that such a functionθ exists. Indeed, we defineθto beα₁on₁, α₂ on₂\₁, . . . , α_non_n\_n−1, etc., and we adjust the constantsα_n>0 in such a way thatθ∈L².

The mappingf ∈L²()→ φ, θfis a continuous linear functional onL²().

By Theorem 4.11 (applied withp=2) there exists a functionv∈L²()such that

(12) φ, θf =

vf ∀f ∈L²().

Setu(x)=v(x)/θ (x). Clearly,uis well defined sinceθ >0 on; moreover,uis measurable anduχ_n ∈L²(). We claim thatuhas all the required properties. We have

(13) φ, χ_ng =

uχ_ng ∀g∈L^∞() ∀n.

Indeed, it suffices to choosef = χ_ng/θ in (12) (note thatf ∈ L²()sincef is bounded on_nandf =0 outside_n).

Next, we claim thatu∈L^∞()and that

(14) u_∞≤ φ(L¹).

Fix any constantC >φ(L¹) and set

A= {x∈; |u(x)|> C}.

Let us verify that A is a null set. Indeed, by choosingg=χ_Asignuin (13) we obtain

A∩n

|u| ≤ φ(L¹)|A∩_n| and therefore

C|A∩n| ≤ φ(L¹)|A∩n|.

It follows that|A∩n| =0 ∀n, and thus A is a null set. This concludes the proof of (14).

Finally, we claim that

(15) φ, h =

uh ∀h∈L¹().

Indeed, it suffices to chooseg =T_nh(truncation ofh) in (13) and to observe that χnTnh→hinL¹().

In order to complete the proof of Theorem 4.14 it remains only to check that u∞= φ(L¹). We have, by (15),

|φ, h| ≤ u∞h1 ∀h∈L¹(),

4.3 Reflexivity. Separability. Dual ofL^p 101

and thereforeφ(L¹) ≤ u_∞. We conclude with the help of (14).

•Remark6.The spaceL¹()isnever reflexiveexcept in the trivial case where consists of a finite number of atoms—and thenL¹()is finite-dimensional. Indeed suppose, by contradiction, thatL¹()is reflexive and consider two cases:

(i)∀ε >0∃ω⊂measurable with 0< μ(ω) < ε.

(ii)∃ε >0 such thatμ(ω)≥εfor every measurable setω⊂withμ(ω) >0.

In Case (i) there is a decreasing sequence(ω_n)of measurable sets such that μ(ωn) > 0 ∀n andμ(ωn) → 0 [choose first any sequence(ω_k)such that 0 <

μ(ω_k) <1/2^kand then setω_n=_∞

k=nω_k].

Let χ_n = χ_ωn and define u_n = χ_n/χ_n1. Since u_n1 = 1 there is a subsequence—still denoted by u_n—and someu ∈ L¹ such that u_n u in the weak topologyσ (L¹, L^∞)(by Theorem 3.18), i.e.,

(16)

u_nφ→

uφ ∀φ∈L^∞. On the other hand, for fixedj, and n > j we have

u_nχ_j = 1. At the limit, as n→ ∞, we obtain

uχ_j =1∀j. Finally, we note (by dominated convergence) that uχ_j →0 asj → ∞—a contradiction.

In Case (ii) the spaceis purely atomic and consists of a countable union of distinct atoms (a_n)(unless there is only a finite number of atoms!). In that case L¹()is isomorphic to¹and it suffices to prove that¹is not reflexive. Consider the canonical basis:

e_n=(0,0, . . . , 1

(n),0,0. . . ).

Assuming¹is reflexive, there exist a subsequence(e_nk)and somex∈¹such that e_nk xin the weak topologyσ (¹, ^∞), i.e.,

ϕ, e_nk −→

k→∞ϕ, x ∀ϕ ∈^∞. Choosing

ϕ=ϕj =(0,0, . . . , 1

(j ),1,1, . . . )

we find thatϕj, x = 1 ∀j. On the other handϕj, x → 0 asj → ∞ (since x ∈¹)—a contradiction.

C. Study ofL^∞.

We already know (Theorem 4.14) that L^∞ = (L¹). Being a dual space,L^∞ enjoys some nice properties. In particular, we have the following:

(i) The closed unit ball BL^∞ is compact in the weak topology σ (L^∞, L¹) (by Theorem 3.16).

(ii) Ifis a measurable subset inR^Nand(f_n)is a bounded sequence inL^∞(), there exists a subsequence(f_nk)and somef ∈ L^∞()such thatf_nk f in

the weak topology σ (L^∞, L¹)(this is a consequence of Corollary 3.30 and Theorem 4.13).

HoweverL^∞()isnot reflexive, except in the trivial case whereconsists of a finite number of atoms; otherwiseL¹()would be reflexive (by Corollary 3.21) and we know thatL¹is not reflexive (Remark 6). As a consequence, it follows that thedual space(L^∞)ofL^∞containsL¹(sinceL^∞=(L¹))and(L^∞)isstrictly biggerthanL¹. In other words, there are continuous linear functionalsφ onL^∞ whichcannotbe represented as

φ, f =

uf ∀f ∈L^∞and someu∈L¹.

In fact, let us describe a “concrete” example of such a functional. Letφ₀:C_c(R^N)→ Rbe defined by

φ₀(f )=f (0)forf ∈C_c(R^N).

Clearlyφ0is a continuous linear functional onC_c(R^N)for the _∞norm. By Hahn–

Banach, we may extendφ0 into a continuous linear functionalφonL^∞(R^N)and we have

(17) φ, f =f (0) ∀f ∈Cc(R^N).

Let us verify that there existsnofunctionu∈L¹(R^N)such that

(18) φ, f =

uf ∀f ∈L^∞(R^N).

Assume, by contradiction, that such a functionuexists. We deduce from (17) and

(18) that

uf =0 ∀f ∈C_c(R^N)andf (0)=0.

Applying Corollary 4.24 (with=R^N\{0})we see thatu=0 a.e. onR^N\{0}and thusu=0 a.e. onR^N. We conclude (by (18)) that

φ, f =0 ∀f ∈L^∞(R^N), which contradicts (17).

Remark7.The dual space ofL^∞does not coincide withL¹but we may still ask the question: what does(L^∞) look like? For this purpose it is convenient to view L^∞(;C) as a commutative C-algebra (see, e.g., W. Rudin [1]). By Gelfand’s theoremL^∞(;C)is isomorphic and isometric to the spaceC(K;C)of continuous complex-valued functions on some compact topological spaceK(Kis the spectrum of the algebraL^∞;Kis not metrizable except whenconsists of a finite number of atoms). Therefore(L^∞(;C)) may be identified with the space of complex- valuedRadon measuresonKandL^∞(;R)may be identified with the space of

4.3 Reflexivity. Separability. Dual ofL^p 103 real-valued Radon measures onK; for more details, see Comment 3 at the end of this chapter, W. Rudin [1] and K. Yosida [1] (p. 118).

Remark8.The spaceL^∞()isnot separableexcept when consists of a finite number of atoms. In order to prove this fact it is convenient to use the following.

Lemma 4.2.LetEbe a Banach space. Assume that there exists a family(O_i)_i∈I such that

(i)for eachi∈I,O_i is a nonempty open subset ofE, (ii)O_i∩O_j = ∅ifi =j ,

(iii)I isuncountable.

ThenEisnotseparable.

Proof of Lemma4.2.Suppose, by contradiction, thatEis separable. Let (u_n)_n∈N denote a dense countable set inE. For eachi∈I, the setO_i∩(u_n)_n∈N = ∅and we may choosen(i)such thatu_n(i) ∈Oi. The mappingi→n(i)is injective; indeed, if n(i) = n(j ), thenun(i) = un(j ) ∈ Oi ∩Oj and thus i = j. Therefore,I is countable—a contradiction.

We now establish that L^∞()is not separable. We claim that there is an un- countable family(ωi)i∈I of measurable sets inwhich are all distinct, that is, the symmetric differenceωi ωj has positive measure fori =j. We then conclude by applying Lemma 4.2 to the family(Oi)i∈I defined by

O_i = {f ∈L^∞(); f −χ_ωi∞<1/2}

(note thatχ_ω−χ_ω∞=1 ifωandωare distinct). The existence of an uncountable family(ω_i)is clear whenis an open set inR^Nsince we may consider all the balls B(x₀, r)withx₀∈andr >0 small enough.

Whenis a general measure space we splitinto its atomic part_a and its nonatomic (=diffuse) part_d; then we distinguish two cases:

(i) _dis not a null set.

(ii) _dis a null set.

In Case (i), then for each real numbert, 0< t < μ(d), there is a measurable setωwithμ(ω)=t; see, e.g., P. Halmos [1], A. J. Weir [1], or J. Neveu [1]. In this way, we obtain an uncountable family of distinct measurable sets.

In Case (ii) consists of a countable union of distinct atoms(a_n)(unless consists of a finite number of atoms). For any collection of integers,A⊂N, we define ω_A=

n∈Aa_n. Clearly,(ω_A)is an uncountable family of distinct measurable sets.

The following table summarizes the main properties of the spaceL^p()when is a measurable subset ofR^N:

Reflexive Separable Dual space L^pwith 1< p <∞ YES YES L^p

L¹ NO YES L^∞

L^∞ NO NO Strictly bigger thanL¹