Banach Lattices and the Dual of C(X)

Ma 110a B. Simon
Fall 2006
GENERAL LINEAR FUNCTIONALS ON C(X),
HAHN AND JORDAN DECOMPOSITIONS
c○ 2006 by Barry Simon
1. Banach Lattices
A lattice is a partially ordered set, L, (i.e., a ≤ b and b ≤ c ⇒ a ≤ c
and a ≤ b plus b ≤ a ⇔ a = b) so that every a, b have a greatest lower
bound (i.e., c ≤ a, c ≤ b so that d ≤ a plus d ≤ b ⇒ d ≤ c) denoted
a ∧ b, and a least upper bound (analog of glb with ≤ replaced by ≥),
denoted a ∨ b. a ∧ b is also called min(a, b) and a ∨ b is max(a, b).
A vector lattice or Riesz space, V, is a real vector space with a partial
order in which it is a lattice and so that for any x ∈ V, a ≤ b ⇒ x+a ≤
x + b and any λ ∈ (0, ∞), a ≤ b ⇒ λa ≤ λb.
Clearly, x+(a∧b) = (x+a)∧(x+b) and x+(a∨b) = (x+a)∨(x+b).
Also, adding −y − x to y ≤ x shows
Thus, multiplication by −1 reverses order, so
x ≤ y ⇔ −y ≤ −x (1.1)
x ∧ y = −[(−x) ∨ (−y)]
Since (x + y) + [(−x) ∨ (−y)] = y ∨ x, this implies
If V is a vector lattice, define
x ∧ y + y ∨ x = x + y (1.2)
V+ = {x | x ≥ 0} V− = {x | −x ≥ 0}
V+ is called the positive elements. Clearly, by (1.1), V− = {x | x ≤ 0},
so the partial ordering property shows
Given a vector lattice, V, define
V+ ∩ V− = {0}
x+ = x ∨ 0 x− = (−x) ∨ 0
Each is clearly a positive element. Moreover,
x+ − x = −x + (x ∨ 0) = 0 ∨ (−x) = x−
1

2 NOTES FOR MA 110A
so x = x+ − x−. We define |x| by
|x| = x+ + x− x = x+ − x− (1.3)
A Banach lattice is a vector lattice, X, which is a Banach space and
which obeys
|x| ≤ |y| ⇒ x ≤ y (1.4)
for any x ∈ X. Notice since | |x| | = |x|, (1.4) implies
|x| = x (1.5)
Examples. If X is a compact Hausdorff space, C(X) is a Banach
lattice with pointwise order (i.e., f ≤ g ⇔ f(x) ≤ g(x) for all x) and
· ∞. Similarly, L p (X, dµ) is a Banach lattice.
We need the following:
Lemma. Let V be a vector lattice.
(i) If x = a+ − a− with a± ≥ 0, then x± ≤ a±.
(ii) |x| = x+ ∨ x−.
Proof. (i) Since x + a− = a+, we have a+ ≥ x, and clearly, a+ ≥ 0, so
a+ ≥ x ∨ 0 = x+. Similarly, a− ≥ (−x) ∨ 0 = x−.
(ii) By (1.2), this is equivalent to x+ ∧x− = 0. Since 0 ≤ x+, 0 ≤ x−,
y = x+ ∧ x− ≥ 0. We have x = (x+ − y) − (x− − y) is a difference of
positive elements since x± − y ≥ 0, so by (i), y ≤ 0. Thus, y = 0.
Proposition. We have that in any vector lattice,
−g ≤ f ≤ g ⇒ |f| ≤ g (1.6)
Proof. First, −g ≤ g ⇒ 2g ≥ 0 ⇒ g ≥ 0. Next, f ≤ g and g ≥ 0
implies f+ ≤ g. Similarly, f− ≤ g. By the lemma,
|f| = f+ ∨ f− ≤ g
2. Decomposition of BLFs on Banach Lattices
The main result here is
Theorem. Let X be a Banach lattice. Any bounded linear function
ℓ: X → R can be written
ℓ = ℓ+ − ℓ−
(2.1)
where ℓ± ≥ 0 (i.e., f ≥ 0 ⇒ ℓ±(f) ≥ 0). Moreover, the decomposition
can be done so that
(i) If ℓ = k+ − k− with k± ≥ 0, then ℓ± ≤ k±.

(ii) For any f ≥ 0, we have
NOTES FOR MA 110A 3
ℓ+(f) + ℓ−(f) ≤ ℓ f (2.2)
Remark. It is an interesting exercise to show this implies X ∗ with
the order ℓ ≥ k if and only if ℓ(f) ≥ k(f) for f ≥ 0 is a lattice and
ℓ± = 0 ∨ (±ℓ).
The interest for us will be what it says about C(X) ∗ .
Lemma. In a vector lattice, if 0 ≤ f ≤ g + h with g ≥ 0, h ≥ 0, then
we can find g1, h1 so 0 ≤ g1 ≤ g, 0 ≤ h1 ≤ h, and f = g1 + h1.
Proof. Let g1 = f ∧ g so 0 ≤ g1 ≤ g. Define h1 = f − g1 ≥ 0 since
g1 ≤ f. Moreover,
g1 + h = (f + h) ∧ (g + h) ≥ f
since f + h ≥ f and g + h ≥ f. Thus, h1 = f − g1 ≤ h.
Proof of Theorem. Define for f ≥ 0,
ℓ+(f) ≡ sup(ℓ(g) | 0 ≤ g ≤ f)
Clearly (taking g = 0), ℓ+(f) ≥ 0 and
so ℓ+(f) < ∞ and
for
We claim for g, h ≥ 0,
|ℓ(g)| ≤ ℓ g ≤ ℓ f
ℓ+(f) ≤ ℓ f
ℓ+(g + h) = sup(ℓ(f) | 0 ≤ f ≤ g + h)
ℓ+(g + h) = ℓ+(g) + ℓ+(h) (2.3)
= sup(ℓ(g1 + h1) | 0 ≤ g1 ≤ g, 0 ≤ h1 ≤ h) (by the lemma)
= ℓ+(g) + ℓ+(h)
Moreover, if λ > 0 is real, ℓ+(λf) = λℓ+(f).
Now define for any f,
ℓ+(f) = ℓ+(f+) − ℓ+(f−)
Then, if
f = q+ − q−
with q+, q− ≥ 0, q+ + f− = q− + f+, so by (2.3),
In particular, since
ℓ+(f) = ℓ+(q+) − ℓ+(q−)
g + h = (g+ + h+) − (g− + h−)

4 NOTES FOR MA 110A
we find that
For λ > 0 and real,
and
ℓ+(g + h) = ℓ+(g+ + h+) − ℓ+(g− + h−)
= ℓ+(g+) + ℓ+(h+) − ℓ−(g−) − ℓ+(h−)
= ℓ+(g) + ℓ+(h)
ℓ+(λf) = ℓ+(λf+) − ℓ+(λf−) = λℓ+(f)
ℓ+(−f) = ℓ+(f−) − ℓ+(f+)
= −ℓ+(f)
Therefore, ℓ+ is a linear functional.
Clearly,
so
and ℓ = ℓ+ − ℓ−.
Note that
ℓ+(f) ≥ ℓ(f)
ℓ− ≡ ℓ+ − ℓ ≥ 0
ℓ−(f) = sup(ℓ(g − f) | 0 ≤ g ≤ f)
= sup(−ℓ(h) | 0 ≤ h ≤ f) (2.4)
by taking h = f − g.
Suppose ℓ = k+−k−. Then 0 ≤ g ≤ f implies ℓ(g) = k+(g)−k−(g) ≤
k+(g) ≤ k+(f), so taking sups, ℓ+ ≤ k+. Using (2.4), we get ℓ− ≤ k−.
Finally, given f ≥ 0 and g, h so 0 ≤ g ≤ f, 0 ≤ h ≤ f, we have
−f ≤ g − h ≤ f, so |g − h| ≤ f and so by the Banach lattice property
g − h ≤ f. Thus,
|ℓ(g) − ℓ(h)| = |ℓ(g − h)|
≤ ℓ g − h
≤ ℓ f
Taking the sup of ℓ(g) over all g and −ℓ(h) over all h, we find that
(2.2) holds.
Corollary. If ℓ ∈ C(X) ∗ , then ℓ = ℓ+ − ℓ− with ℓ± ≥ 0 and
ℓ+ + ℓ = ℓ (2.5)

NOTES FOR MA 110A 5
Proof. We only need to check (2.5). But ℓ± = ℓ±(1), so
ℓ+ + ℓ− = ℓ+(1) + ℓ−(1) ≤ ℓ
by (2.2) with f = 1. Since ℓ = ℓ+ − ℓ−,
ℓ ≤ ℓ+ + ℓ−
proving (2.5).
A signed Baire measure on X is an assignment of a real number,
µ(A), to each Baire set so that there is C < ∞ obeying if X = A1 ∪
· · · ∪ An and Ai ∩ Aj = ∅, then
n
|µ(Aj)| ≤ C (2.6)
j=1
and for any countable family of disjoint sets {Aj} n j=1 , we have µ(∪Aj) =
µ(Aj) (the sum converges absolutely by (2.6)). Given such a measure,
as usual we can form a Stieltjes-type functional ℓ on C(X) with
|ℓ(f)| ≤ f∞ so ℓ = ℓ+ − ℓ− and µ = µ+ − µ−. We set |µ| = µ+ + µ−.
Conversely, any µ+ − µ− defines a signed measure. We thus have the
Riesz–Markov theorem for signed measures which says there is a oneone
correspondence between signed measures and elements of C(X) ∗ .
3. Hahn and Jordan Decompositions
As we have seen, every signed measure µ induces an ℓ which can
be written as µ+ − µ− where µ± obeys µ+ + µ− = µ and if
µ = ν+ − ν− with ν± ≥ 0, then µ± ≤ ν±. Here we will prove two
famous theorems:
Hahn Decomposition Theorem. There exists a Baire set, A, so
that for all C,
µ+(C) = µ(A ∩ C) (3.1)
µ−(C) = −µ((X \ A) ∩ C) (3.2)
µ = µ(A) + µ(X \ A) (3.3)
Jordan Decomposition Theorem. µ+ and µ− are mutually singular.
Recall |µ| is the positive measure µ+ + µ−. We will deduce the two
theorems above from
Theorem. There exists a Baire function, g, with |g(x)| = 1 for d|µ|
a.e. x so that
f dµ = fg d|µ| (3.4)

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for all f ∈ C(X).
Proof. First, without loss of generality, we can suppose µ = 1. We
start by noting µ(f+) = µ+(f+) − µ−(f+) ≤ µ+(f+) and µ(f+) ≥
−µ−(f+) so |µ(f+)| ≤ |µ|(f+). Similarly, |µ(f−)| ≤ |µ|(f−), so
|µ(f)| ≤ |f| d|µ| (3.5)
As in von Neumann’s proof of the RN theorem, we have
|µ(f)| ≤ |f| 2 1/2 d|µ|
since d|µ| = µ = 1, so there is g ∈ L2 with
µ(f) = g(x)f(x) d|µ|(x) (3.6)
Originally, this is true for f ∈ C(X), but by use of the dominated
convergence theorem, we can define µ(f) for f any bounded Baire
function and both (3.6) and (3.5) hold.
Now let ε > 0 and let f be the characteristic function of {x | g(x) >
1 + ε} ≡ Aε. Then (3.6) says
and (3.5) says
µ(f) ≥ (1 + ε)|µ|(Aε)
|µ(f)| ≤ |µ|(Aε)
It follows that |µ|(Aε) = 0. Taking ε = 1/2 n , we see g(x) ≤ 1 for a.e.
x. Similarly, g(x) ≥ −1, so
so
|g(x)| ≤ 1 a.e. x (3.7)
From (3.6), we have for f ∈ C(X),
|µ(f)| ≤ |g| d|µ| f∞
(3.8)
1 = µ ≤ |g| d|µ| (3.9)
Since |µ(X)| = 1 and (3.7) holds, we see |g(x)| = 1 for a.e. x.
Proof of the Hahn Decomposition Theorem. Define
so
A = {x | g(x) = +1}
X \ A = {x | g(x) = −1}

For f ∈ C(X), let
ν+(f) =
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fχA d|µ| ν−(f) =
fχX\A d|µ|
so ν± ≥ 0 and ν+ − ν− = µ. Thus ν± ≥ µ±. But ν+(1) + ν−(1) =
|µ|(X) = 1 = µ+(1) + µ−(1) implies ν+ − µ+ = µ− − ν− = 0. Thus
µ+ is the measure associated to χAd|µ|, that is, µ+(B) = |µ|(B ∩ A) =
µ(B ∩ A) since µ−(B ∩ A) = |µ|((X \ A) ∩ (B ∩ A)) = 0.
Proof of the Jordan Decomposition Theorem. By the Hahn theorem,
µ+ and µ− are mutually singular.
By combining this analysis with the von Neumann proof, one gets
Theorem. Let dµ be a signed Baire measure and dν a positive Baire
measure. Then
dµ = dµac + dµsing
where
dµac = f(x) dν(x)
for f ∈ L 1 (R, dν) and dµsing is singular with respect to dν in the sense
that there is A with ν(X \ A) = 0 and µ(B) = 0 for all B ≤ A
(equivalently, |µ|(A) = 0).
Proof. We have d|µ| = h dν + d|µ|sing, so dµ = g dν + g d|µ|sing.