1.4 Lorentz Spaces
1.4.3 Duals of Lorentz Spaces
Raise (1.4.10) to the powerq, multiply bytq/p, integrate with respect todt/tover (0,∞), and apply Fatou’s lemma to obtain
f−fnk
0
q
Lp,q≤lim inf
k→∞
fnk−fnk
0
q
Lp,q. (1.4.11)
Now letk0→∞in (1.4.11) and use the fact that{fn}is Cauchy to conclude that fnk converges to f inLp,q. It is a general fact that if a Cauchy sequence has a convergent subsequence in a quasinormed space, then the sequence is convergent to the same
limit. It follows that fnconverges to f inLp,q.
Remark 1.4.12.It can be shown that the spacesLp,q are normable when p,qare bigger than 1; see Exercise 1.4.3. Therefore, these spaces can be normed to become Banach spaces.
It is natural to ask whether simple functions are dense inLp,q. This is in fact the case whenq6=∞.
Theorem 1.4.13.Simple functions are dense in Lp,q(X,µ)when0<q<∞.
Proof. Let f ∈Lp,q(X,µ). Assume without loss of generality that f ≥0. Given n=1,2,3, . . ., we find a simple function fn≥0 such that
fn(x) =0 when f(x)≤1/n, and
f(x)−1
n ≤fn(x)≤f(x)
when f(x)>1/n, except on a set of measure less than 1/n. It follows that µ({x∈X: |f(x)−fn(x)|>1/n})<1/n;
hence(f−fn)∗(t)≤1/nfort≥1/n. Thus
(f−fn)∗(t)→0 asn→∞and fn∗(t)≤f∗(t) for allt>0.
Since(f−fn)∗(t)≤2f∗(t/2), an application of the Lebesgue dominated conver- gence theorem gives that
fn−f
Lp,q→0 asn→∞.
Remark 1.4.14.One may wonder whether simple functions are dense inLp,∞. This turns out to be false for all 0<p≤∞. However, ifX isσ-finite, countable linear combinations of characteristic functions of sets with finite measure are dense in Lp,∞(X,µ). We call such functionscountably simple. See Exercise 1.4.4 for details.
T
Z∗= sup
kxkZ=1
|T(x)|.
Observe that the dual of a quasi-Banach space is always a Banach space.
We are now considering the following question: What are the dual spaces(Lp,q)∗ ofLp,q? The answer to this question presents some technical difficulties for general measure spaces. In this exposition we restrict our attention toσ-finite nonatomic measure spaces, where the situation is simpler.
Definition 1.4.15.A subsetAof a measure space(X,µ)is called anatomifµ(A)>
0 and every subset Bof A has measure either equal to zero or equal to µ(A). A measure space(X,µ)is callednonatomicif it contains no atoms. In other words,X is nonatomic if and only if for anyA⊆Xwithµ(A)>0, there exists a proper subset B$Awithµ(B)>0 andµ(A\B)>0.
For instance,R with Lebesgue measure is nonatomic, but any measure space with counting measure is atomic. Nonatomic spaces have the property that every measurable subset of them with strictly positive measure contains subsets of any given measure smaller than the measure of the original subset. See Exercise 1.4.5.
Definition 1.4.16.A measure space is calledσ-finiteif there is a sequence of mea- surable setsKNwithµ(KN)<∞such that
∞ [
N=1
KN=X.
For instance,Rnequipped with Lebesgue measure is aσ-finite measure space. So isZnwith the usual counting measure.
Theorem 1.4.17.Suppose that(X,µ)is a nonatomicσ-finite measure space. Then (i) (Lp,q)∗={0}, when0<p<1,0<q≤∞, (ii) (Lp,q)∗=L∞, when p=1,0<q≤1, (iii) (Lp,q)∗={0}, when p=1,1<q<∞, (iv) (Lp,q)∗6={0}, when p=1, q=∞,
(v) (Lp,q)∗=Lp0,∞, when1<p<∞,0<q≤1, (vi) (Lp,q)∗=Lp0,q0, when1<p<∞,1<q<∞, (vii) (Lp,q)∗6={0}, when1<p<∞, q=∞, (viii) (Lp,q)∗6={0}, when p=q=∞.
Proof. Since X is σ-finite, we have X =S∞N=1KN, where KN is an increasing sequence of sets with µ(KN)<∞. Given T ∈(Lp,q)∗, where 0< p<∞ and 0 <q ≤∞, consider the measure σ(E) =T(χE). Since σ satisfies |σ(E)| ≤ (p/q)1/q
T
µ(E)1/p when q<∞ and|σ(E)| ≤ kT
µ(E)1/p, it follows that σ is absolutely continuous with respect to the measure µ. By the Radon–Nikodym
theorem, there exists a complex-valued measurable function g (which satisfies R
KN|g|dµ<∞for allN) such that
σ(E) =T(χE) = Z
X
gχEdµ. (1.4.12)
Linearity implies that (1.4.12) holds for any simple function onX. The continuity ofT and the density of the simple functions onLp,q(whenq<∞) gives
T(f) = Z
X
g f dµ (1.4.13)
for every f∈Lp,q. We now examine each case (i)–(viii) separately.
(i) We first consider the case 0<p<1. Let f =∑nanχEn be a simple function onX (take f to be countably simple whenq=∞). IfX is nonatomic, we can split eachEnas a union ofNdisjoint setsEjneach having measureN−1µ(En). Let fj=
∑nanχEjn. We see that fj
Lp,q=N−1/p f
Lp,q. Now ifT∈(Lp,q)∗, it follows that
|T(f)| ≤
N j=1
∑
|T(fj)| ≤ T
N
∑
j=1fj
Lp,q≤ T
N1−1/p f
Lp,q.
LetN→∞and use that p<1 to obtain thatT =0.
(ii) We now consider the casep=1 and 0<q≤1. Clearly, everyg∈L∞gives a bounded linear functional onL1,q, since
Z
X
f g dµ
≤ g
L∞
f
L1≤Cq g
L∞
f
L1,q.
Conversely, suppose thatT ∈(L1,q)∗whereq≤1. The functionggiven in (1.4.12) satisfies
Z
E
g dµ
≤ T
µ(E) for allE⊆KN, and hence|g| ≤
T
µ-a.e. on everyKN. See Rudin [229, p. 31]
(Theorem 1.40) for a proof of this fact. It follows that g
L∞ ≤ T
and hence (L1,q)∗=L∞.
(iii) Let us now takep=1, 1<q<∞, and suppose thatT ∈(L1,q)∗. Then
Z
X
f g dµ
≤ T
f
L1,q, (1.4.14)
whereg is the function in (1.4.13). We show thatg=0 a.e. Suppose that|g| ≥δ on some setE0withµ(E0)>0. Let f =g|g|−1χE0h, whereh≥0. Then (1.4.14) implies that
h
L1(E0)≤ T
δ−1 h
L1,q(E0)
for allh≥0. SinceX is nonatomic, this can’t happen unlessT =0. See Exercise 1.4.8.
(iv) In the casep=1,q=∞something interesting happens. Since every contin- uous linear functional onL1,∞extends to a continuous linear functional onL1,qfor 1<q<∞, it must necessarily vanish on all simple functions by part (iii). However, (L1,∞)∗contains nontrivial linear functionals. For details we refer to the articles of Cwikel and Fefferman [63], [64].
(v) We now take up the case p>1 and 0<q≤1. Using Exercise 1.4.1(b) and Proposition 1.4.10, we see that ifg∈Lp0,∞, then
Z
X
f g dµ
≤ Z ∞
0
t1pf∗(t)t
1 p0
g∗(t)dt t
≤ f
Lp,1
g Lp0,∞
≤Cp,q f
Lp,q
g Lp0,∞.
Conversely, suppose thatT∈(Lp,q)∗when 1<p<∞and 0<q≤1. Letgsatisfy (1.4.13). Taking f =g|g|−1χ|g|>α and using that
Z
X
f g dµ
≤ T
f
Lp,q,
we obtain that
α µ({|g|>α})≤(p/q)1/q T
µ({|g|>α})1p.
It follows that g
Lp0,∞≈ T
.
(vi) Using Exercise 1.4.1(b) and H¨older’s inequality, we obtain
Z
X
f g dµ
≤ Z ∞
0
t1pf∗(t)t
1 p0
g∗(t)dt t ≤
f Lp,q
g Lp0,q0;
thus everyg∈Lp0,q0 gives a bounded linear functional onLp,q. Conversely, letT be in(Lp,q)∗. By (1.4.13),T is given by integration against a locally integrable function g. It remains to prove thatg∈Lp0,q0. For all f inLp,q(X)we have
Z ∞
0
f∗(t)g∗(t)dt= sup
h:dh=df
Z
X
hg dµ
≤ T
f
Lp,q, (1.4.15) where the equality is a consequence of the fact thatX is nonatomic (see Exercise 1.4.5). Pick a function f onXsuch that
f∗(t) = Z ∞
t/2
s
q0 p0−1
g∗(s)q0−1ds
s . (1.4.16)
This can be achieved again by Exercise 1.4.5. The fact that the integral in (1.4.16) converges is a consequence of the observation that the function f∗ defined in
(1.4.16) lies in the spaceLq(0,∞)with respect to the measuretq/p−1dt. This fol- lows from the inequality
f
Lp,q = Z ∞
0
t
q p
Z ∞ t/2
s
q0 p0−1
g∗(s)q0−1ds s
q
dt t
1q
≤C1(p,q) Z ∞
0
(t
1 p0
g∗(t))q0dt t
1q
=C1(p,q) g
q0/q Lp0,q0 <∞,
which is a consequence of Hardy’s second inequality in Exercise 1.2.8 withb=q/p.
Using (1.4.15), we conclude that Z ∞
0
f∗(t)g∗(t)dt≤ T
f
Lp,q ≤C1(p,q) T
g
q0/q
Lp0,q0. (1.4.17) On the other hand, we have
Z ∞
0
f∗(t)g∗(t)dt ≥ Z ∞
0 Z t
t/2
s
q0 p0−1
g∗(s)q0−1ds s g∗(t)dt
≥ Z ∞
0
g∗(t)q0 Z t
t/2
s
q0 p0−1ds
s dt
=C2(p,q) g
q0 Lp0,q0.
(1.4.18)
Combining (1.4.17) and (1.4.18), we obtain g
Lp0,q0≤C(p,q) T
. This estimate is valid only when we have a priori knowledge that
g
Lp0,q0 <∞. Suitably modifying the preceding proof and using that
g Lp0,q0
(KN)<∞, we obtain that g
Lp0,q0
(KN)≤ C(p,q)
T
for allN=1,2, . . .. LettingN→∞, we obtain the required conclusion.
(vii) For a complete characterization of this space, we refer to the article of Cwikel [62].
(viii) The dual ofL∞can be identified with the set of all bounded finitely additive
set functions. See Dunford and Schwartz [77].
Remark 1.4.18.Some parts of Theorem 1.4.17 are false ifXis atomic. For instance, the dual of`p(Z)containsl∞when 0<p<1 and thus it is not{0}.