2. Lpand `p
2.1. Duality of Lp-spaces: first bite. An important task in this Chapter will be that of characterizing the duals of
Lp-spaces for 1 ≤p < ∞. The theorem is easier to state than to prove: the dual of Lp“is” Lq, where p−1+q−1= 1
(and q=∞in the case p= 1). The Section B course included proofs of this duality for `p-spaces and for L2(which is a
Hilbert space).
Theorem 2.1. Let 1≤p≤ ∞, let qbe the conjugate index and let (Ω,F,µ)be a (σ-finite) measure space. There is a
linear isometric embedding Uq:Lq→L∗
pgiven by
(Uq(y))(x) = hx, yi=ZΩ
x(t)y(t)dµ(t).
Proof. By H¨older’s inequality the integral defining (Uq(y))(x) exists and we have
|(Uq(y))(x)|≤kykqkxkp.
Since linearity of Uq(y) is trivial we thus have Uq(y)∈L∗
pwith
kUq(y)k ≤ kykq, .
The operator Uq(which is also patently linear) is thus of norm at most 1; we need to show it is isometric. So let y∈Lq
be an element with kykq= 1. Initially assume that 1 <p<∞, set x=|y|q/psign yand observe that
kxkp= 1 ; (Uq(y))(x) = Zxy dµ=Z|y|q/p|y|dµ=Z|y|qdµ= 1.
We deduce that kUq(y)k= 1 as required.
If p=∞we simply set x= sign yand have
kxk∞= 1 ; (U1(y))(x) = Zxy dµ=Z|y|dµ=kyk1= 1.
Finally, in the case p= 1, need to be a little more careful, since σ-finiteness comes into the argument. Let y∈L∞be
an element with kyk∞= 1. For any r < 1 the set [|y| ≥ r] is non null and by σ-finiteness, we can choose a set A∈F
with A⊆[|y| ≥ r], and 0 < µ(A)<∞. We now set
x= (µ(A))−1sign y1A,
and have
x∈L1,kxk1= 1,(U∞(y))(x) = (µ(A))−1ZA
|y|dµ≥r.
The difficult part of our duality proofs will involve proving that Uqis surjective and we defer this to a later section.
Let us note that (except in the finite-dimensional case) U1is never a surjection from L1onto L∗
∞. We shall show, in due
course, that Uqis surjective in all other cases.
Notice, in passing, the use of the angle-brackets notation h·,·i for the duality between Lpand Lq. We shall use the
same notation in the context of the familiar duality between `pand `q, putting
hx, yi=
∞
X
n=1
x(n)y(n).
2.2. Disjointness in Lp.Another theme in this chapter is going to be the question: to what extent is Lp(0,1) “like” `p?
We know that when p= 2 these spaces are isometrically isomorphic. What happens when p6= 2? We start by observing
that there is at least some similarity between Lpand `p.
Definition 2.2. If xis a scalar-valued function on some set Γ we define the support of xto be
supp x={γ∈Γ : x(γ)6= 0}.
We say that two functions xand yare disjointly supported, or just disjoint, if supp x∩supp y=∅.When we are thinking
of elements of Lpwe say that xand yare disjoint if µ(supp x∩supp y) = 0, or equivalently xy = 0 a.e.
Proposition 2.3. Let 1≤p < ∞and let (xk)k∈Nbe a disjoint sequence of non-zero vectors in Lp(µ). Then
kX
k∈N
xkkp= X
k∈N
kxkkp
p!1/p
.
The closed linear span X=sph{xk:k∈N}i is isometrically isomorphic to `pand is complemented in Lp(µ)by a
projection Pof norm 1.
16
