Helms L.L. Potential Theory
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248 6Energy
G
Ω
μ(x) >f(x). Then there is a neighborhood U of x such that G
Ω
μ>f
on U ∩ Γ . Letting ν = μ|
U
and 0 <≤ 1,μ− ν is a measure with support
in Γ and Φ
f
(μ) ≤ Φ
f
(μ − ν). Since
Φ
f
(μ)=
G
Ω
μdμ− 2
fdμ
≤ Φ
f
(μ − ν)
=
G
Ω
(μ − ν) d(μ − ν) − 2
fd(μ − ν)
=
G
Ω
μdμ− 2
G
Ω
μdν +
2
G
Ω
ν − 2
fdμ+2
fdν,
(f − G
Ω
μ) dν +
2
G
Ω
νdν ≥ 0.
Since G
Ω
ν ≤ G
Ω
μ and μ ∈E
+
,
G
Ω
νdν <+∞ by Theorem 6.2.4. Letting
→ 0 in the above equation,
(f − G
Ω
μ) dν ≥ 0. Since f − G
Ω
μ<0on
the support of ν, ν(U)=μ(U) = 0; that is, the support of μ is contained in
{x : G
Ω
μ(x) ≤ f }. Lastly, consider the assertion that G
Ω
μ ≥ f on Γ except
possibly for a polar set. For each n ≥ 1, let μ
Γ
n
be the capacitary distribution
corresponding to the compact set Γ
n
= Γ ∩{x : f(x) ≥ G
Ω
μ(x)+
1
n
}.If
t>0,μ+ tμ
Γ
n
is a measure with support in Γ and Φ
f
(μ) ≤ Φ
f
(μ + tμ
Γ
n
).
As above,
(G
Ω
μ − f ) dμ
Γ
n
+
t
2
G
Ω
μ
Γ
n
dμ
Γ
n
≥ 0.
Letting t → 0,
0 ≤
(G
Ω
μ − f) dμ
Γ
n
≤−
1
n
μ
Γ
n
(Γ
n
)
from which it follows that C(Γ
n
)=μ
Γ
n
(Γ
n
)=0.Thus,Γ
n
is polar for every
n ≥ 1andG
Ω
μ ≥ f on Γ except possibly for a polar set.
Putting f = 1 in the definition of Φ
f
, Gauss’s integral
Φ
1
(μ)=
(G
Ω
μ − 2) dμ
is obtained. It will be shown that the measure μ which minimizes Φ
1
(μ)is
just the capacitary distribution μ
Γ
corresponding to the compact set Γ .
Theorem 6.2.10 (Energy Principle) If μ and ν are Borel measures on
the Greenian set Ω,then0 ≤ [μ, ν]
e
≤|μ|
e
|ν|
e
.
Proof: By (iii) of Theorem 6.2.4, it suffices to prove the inequality when μ
and ν have supports in a compact set Γ ⊂ Ω.Letf be a bounded, nonneg-
ative continuous function on Ω such that f ≤ G
Ω
μ and let μ
∈E
+
be the
measure with support in Γ which minimizes Φ
f
.Sinceμ
∈E
+
and polar sets
have μ
-measure zero by Theorem 6.2.3, G
Ω
μ
= fa.e.(μ
) on the support of
μ
. It follows that
6.3 Mutual Energy 249
fdμ
=
G
Ω
μ
dμ
and
Φ
f
(μ
)=
G
Ω
μ
dμ
− 2
fdμ
= −
G
Ω
μ
dμ
= −|μ
|
2
e
.
For any α>0,Φ
f
(αν) ≥ Φ
f
(μ
)=−|μ
|
2
e
;thatis,
α
2
|ν|
2
e
− 2α
fdν+ |μ
|
2
e
≥ 0
for every α>0. Since the inequality is trivially true if α ≤ 0, it holds for
all α and
fdν ≤|ν|
e
|μ
|
e
.SinceG
Ω
μ
≤ f ≤ G
Ω
μ on the support of
μ
and G
Ω
μ
is harmonic off the support of μ
, G
Ω
μ
is finite-valued and
G
Ω
μ
≤ G
Ω
μ on Ω by the Maria-Frostman domination principle, Theo-
rem 4.4.8. By (ii) of Theorem 6.2.4, |μ
|
e
≤|μ|
e
, and therefore
fdν ≤
|ν|
e
|μ|
e
. Taking the supremum over all bounded, nonnegative continuous
f ≤ G
Ω
μ,
G
Ω
μdν ≤|μ|
e
|ν|
e
.
It is now possible to give an example of a measure μ ∈E
+
for which G
Ω
μ
cantakeonthevalue+∞.
Example 6.2.11 Let Z be a nonempty compact polar subset of the Greenian
set Ω.ThenZ has capacity zero. For each j ≥ 1, let U
j
be an open set
containing Z with compact closure U
−
j
⊂ Ω such that C(U
−
j
) < 2
−2j
,and
let μ
j
be the capacitary distribution of U
−
j
.SinceG
Ω
μ
j
≤ 1,μ
j
∈E
+
.
Moreover, G
Ω
μ
j
=1onU
j
⊃ Z. Letting μ =
∞
j=1
μ
j
,μ is a measure for
which G
Ω
μ =+∞ on Z.Since
G
Ω
μdμ =
j,k
G
Ω
μ
j
dμ
k
≤
j,k
|μ
j
|
e
|μ
k
|
e
≤
j,k
2
−2j
2
−2k
=
j
2
−2j
2
< 1,
μ ∈E
+
.
6.3 Mutual Energy
If μ and ν are two Borel measures on the Greenian set Ω, the Green potential
G
Ω
(μ−ν) is certainly not defined if both μ and ν are infinite on the same set.
To get around this, μ − ν will be regarded as a formal difference. Two such
differences μ − ν and μ
− ν
will be regarded as the same if μ + ν
= μ
+ ν
250 6Energy
and written μ−ν ∼ μ
−ν
. Strictly speaking, μ−ν should be regarded as the
ordered pair (μ, ν) in the product space E
+
×E
+
in which two pairs (μ, ν)and
(μ
,ν
) are regarded as the same if μ + ν
= μ
+ ν, written (μ, ν) ∼ (μ
,ν
).
This relation is reflexive and transitive but not symmetric. In showing that
(μ, ν) ∼ (μ
,ν
)and(μ
,ν
) ∼ (μ
,ν
) ⇒ (μ, ν) ∼ (μ
,ν
),
the equation μ+ μ
+ ν
+ ν
= μ
+ μ
+ ν
+ ν is obtained; since all measures
are assumed to be finite on compact subsets of Ω, μ+ν
= μ
+ν on compact
subsets of Ω and therefore on the Borel subsets of Ω by the assumption that
all measures are regular Borel measures.
Definition 6.3.1 Let E = E
+
×E
+
= {(μ, ν); μ, ν ∈E
+
}.Ifλ
i
=(μ
i
,ν
i
)
with μ
i
,ν
i
∈E
+
,i=1, 2, the mutual energy [λ
1
,λ
2
] is defined by
[λ
1
,λ
2
]
e
=[μ
1
,μ
2
]
e
− [μ
1
,ν
2
]
e
− [μ
2
,ν
1
]
e
+[ν
1
,ν
2
]
e
;
the energy |λ|
2
e
of λ ∈Eis defined by |λ|
2
e
=[λ, λ]
e
.
The latter part of this definition assumes that the energy of an element λ ∈E
is nonnegative. This follows from the energy principle, Theorem 6.2.10, for if
λ =(λ
1
,λ
2
), then
[λ, λ]
e
= |λ
1
|
2
e
− 2[λ
1
,λ
2
]
e
+ |λ
2
|
2
e
≥|λ
1
|
2
e
− 2|λ
1
|
e
|λ
2
|
e
+ |λ
2
|
2
e
=(|λ
1
|
e
−|λ
2
|
e
)
2
≥ 0.
It can be seen also that the definition of the mutual energy [λ
1
,λ
2
]isin-
dependent of the ordered pairs used to represent λ
1
and λ
2
as follows. If
(μ
i
,ν
i
) ∼ (μ
i
,ν
i
),i=1, 2, then
μ
i
+ ν
i
= μ
i
+ ν
i
,i=1, 2.
Using these equations, the above expression defining [λ
1
,λ
2
] can be converted
into the expression that would be obtained if the (μ
i
,ν
i
),i=1, 2, were used
instead. Addition and multiplication by real numbers is defined on E by
(μ, ν)+(μ
,ν
)=(μ + μ
,ν+ ν
)
c(μ, ν)=
(cμ, cν)ifc ≥ 0
(−cν, −cμ)ifc<0.
The zero element of E is given by 0 = (0, 0). It is easily checked that E is a
vector space over the reals. It is also easily seen that [λ
1
,λ
2
]
e
=[λ
2
,λ
1
]
e
for
all λ
1
,λ
2
∈E.
6.3 Mutual Energy 251
Theorem 6.3.2 If μ, ν ∈E,then|[μ, ν]
e
|≤|μ|
e
|ν|
e
and |μ+ν|
e
≤|μ|
e
+|ν|
e
.
Proof: Since the energy is nonnegative, [μ − αν, μ − αν]
e
≥ 0 for all α ∈ R;
that is,
|μ|
2
e
− 2α[μ, ν]
e
+ α
2
|ν|
2
e
≥ 0.
Therefore,
(−2[μ, ν]
e
)
2
≤ 4|μ|
2
e
|ν|
2
e
and so |[μ, ν]
e
|≤|μ|
e
|ν|
e
. By the preceding step,
|μ + ν|
2
e
= |μ|
2
e
+2[μ, ν]
e
+ |ν|
2
e
≤|μ|
2
e
+2|μ|
e
|ν|
e
+ |ν|
2
e
=(|μ|
e
+ |ν|
e
)
2
.
Theorem 6.3.3 The energy of μ ∈Eis zero if and only if μ ∼ 0.
Proof: According to the preceding theorem if |μ|
e
=0,then[μ, ν]
e
=0for
all ν ∈E.Ifμ =(μ
1
,μ
2
)withμ
1
,μ
2
∈E
+
,then[μ
1
,ν]
e
=[μ
2
,ν]
e
for all
ν ∈E. According to Theorem 3.5.7, for each pair x, δ such that B
−
x,δ
⊂ Ω
there is a signed measure τ
x,δ
of finite energy such that the class of functions
{G
Ω
τ
x,δ
} is a total subset of K
+
(Ω). Since
G
Ω
τ
x,δ
dμ
1
=
G
Ω
μ
1
dτ
x,δ
=[μ
1
,τ
x,δ
]
e
=[μ
2
,τ
x,δ
]
e
=
G
Ω
τ
x,δ
dμ
2
and the set {G
Ω
τ
x,δ
} is total in K
+
(Ω),μ
1
= μ
2
by Theorem 3.5.4 and
μ =(μ
1
,μ
1
) ∼ 0. Conversely, if μ ∼ 0, then the energy is zero.
From the Hilbert space point of view, E is a pre-Hilbert space in that it
is a vector space over the reals with [μ, ν]
e
serving as an inner product and
|μ|
e
as a norm but lacks the completeness property as a metric space. The
following example shows that E is not complete.
Example 6.3.4 (Cartan) Let Ω = R
3
,andforeachj ≥ 1letλ
j
= μ
j
−
ν
j
be a signed measure where μ
j
is a unit measure uniformly distributed
on a sphere with center at the origin of radius 1 − 4
−j
and ν
j
is defined
similarly on a concentric sphere of radius 1. By Gauss’s averaging principle,
Theorem 1.6.2,
[λ
j
,λ
j
]
e
=
Gμ
j
dμ
j
− 2
Gμ
j
dν
j
+
Gν
j
dν
j
=
1
4
j
− 1
.
It follows that the series
∞
j=1
|λ
j
|
e
converges. Since
p+k
j=1
λ
j
−
p
j=1
λ
j
e
=
p+k
j=p+1
λ
j
e
≤
p+k
j=p+1
|λ
j
|
e
→ 0
252 6Energy
as p →∞independently of k, the sequence of partial sums {
p
j=1
λ
j
} is a
Cauchy sequence in the energy norm. Suppose there is a λ =(λ
,λ
)inE
such that
lim
p→∞
p
j=1
λ
j
− λ
e
=0. (6.2)
By Theorem 6.3.2,
lim
p→∞
(
p
j=1
λ
j
,ν
)
e
=[λ, ν]
e
for all ν ∈E
+
;thatis,
lim
p→∞
Gνd(
p
j=1
λ
j
)=
Gνdλ
−
Gνdλ
.
In particular, this equation holds for all the potentials Gτ
x,δ
of Theorem 3.5.7
with B
−
x,δ
⊂ B = B
0,1
.IfK
+
0
denotes the set of all such potentials, then K
+
0
is a total subset of the class K
+
(B) of nonnegative continuous functions on
B having compact support in B by the same theorem. Since the Gτ
x,δ
=0
outside B
−
x,δ
,forf ∈ K
+
0
(B)
lim
p→∞
fd(
p
j=1
λ
j
) = lim
p→∞
fd(
p
j=1
μ
j
)=
fλ
−
fdλ
.
By definition of a total set, Definition 3.5.2, there is a nonnegative linear
combination f of elements of K
+
0
(B) that majorizes 1 on the closure of the
ball of radius 1 − 4
−p
with |f|≤2. Then
p
j=1
μ
j
(B) ≤
fd(
p
j=1
μ
j
)
and
+∞ = lim
p→∞
p
j=1
μ
j
(B) ≤ lim
p→∞
fd(
p
j=1
μ
j
) ≤ 2λ
(B
−
).
Thus, λ
is not a Borel measure, a contradiction. Therefore, there is no λ ∈E
satisfying Equation (6.2).
Consider a Borel measure μ and a superharmonic function u on the Gree-
nian set Ω for which G
Ω
μ ≤ ua.e.(μ) on the support of μ. Application
of the Maria-Frostman domination principle, Theorem 4.4.8, requires that
G
Ω
μ<+∞ on Ω to conclude that G
Ω
μ ≤ u on Ω. The finiteness condition
can be eliminated if μ has finite energy.
6.3 Mutual Energy 253
Theorem 6.3.5 (Cartan) If u is a superharmonic function on the Green-
ian set Ω, μ ∈E
+
,andG
Ω
μ ≤ ua.e.(μ) on the support of μ,thenG
Ω
μ ≤ u
on Ω.
Proof: The function v =min(u, G
Ω
μ)isapotentialG
Ω
ν with ν ∈E
+
since G
Ω
ν ≤ G
Ω
μ.Since(G
Ω
μ>u)=(G
Ω
μ>G
Ω
ν)andμ(G
Ω
μ>u)=0,
(G
Ω
μ − G
Ω
ν) dμ =0≤
(G
Ω
μ − G
Ω
ν) dν. It follows that
|μ − ν|
2
e
= |μ|
2
e
− 2[μ, ν]
e
+ |ν|
2
e
≤ 0.
In view of the fact that energy is nonnegative, |μ −ν|
e
=0andμ ∼ ν by the
preceding theorem. Since μ and ν are measures, this means that μ = ν;that
is, u ≥ min (u, G
Ω
μ)=G
Ω
ν = G
Ω
μ.
Given a sequence {μ
j
} in E
+
, there are three modes of convergence of the
sequence to a measure μ ∈E
+
:(i)strong convergence in the energy norm;
that is, lim
j→∞
|μ
j
− μ|
e
= 0, (ii) weak convergence in the pre-Hilbert
space E; that is, lim
j→∞
[μ
j
,ν]
e
=[μ, ν]
e
for all ν ∈E(equivalently, for all
ν ∈E
+
), and (iii) vague convergence ; that is, lim
j→∞
fdμ
j
=
fdμ
for all f ∈ C
0
0
(Ω). Strong convergence implies weak convergence since
|[μ
j
,ν]
e
− [μ, ν]
e
|≤|μ
j
− μ|
e
|ν|
e
for all ν ∈E.
Lemma 6.3.6 If {μ
j
} is a sequence in E that converges weakly to μ ∈E,
then |μ|
e
≤ lim sup
j→∞
|μ
j
|
e
.
Proof: Since
|μ|
2
e
=[μ, μ]
e
≤|[μ − μ
j
,μ]
e
+ |[μ
j
,μ]|
e
≤|[μ − μ
j
,μ]
e
|+ |μ
j
|
e
|μ|
e
,
|μ|
2
e
≤ lim sup
j→∞
|[μ
j
− μ, μ]
e
|+ |μ|
e
lim sup
j→∞
|μ
j
|
e
= |μ|
e
lim sup
j→∞
|μ
j
|
e
.
Theorem 6.3.7 Weak convergence and strong convergence are equivalent for
strong Cauchy sequences in E
+
.
Proof: As noted above, strong convergence implies weak convergence. Let
{μ
j
} beastrongCauchysequenceinE
+
. It need only be shown that weak
convergence of the sequence {μ
j
} to μ ∈E
+
implies that lim
j→∞
|μ
j
−μ|
e
=0.
Since the sequence {μ
j
− μ
k
} converges weakly to μ − μ
k
for each k ≥ 1,
|μ − μ
k
|
e
≤ lim sup
j→∞
|μ
j
− μ
k
|
e
by the preceding lemma. Since the sequence {μ
j
} is strong Cauchy, lim
k→∞
|μ − μ
k
|
e
=0.
254 6Energy
Theorem 6.3.8 If μ ∈E
+
and G
Ω
μ is the limit of a monotone sequence of
potentials {G
Ω
μ
j
},μ
j
∈E
+
,j ≥ 1, on the Greenian set Ω, then the sequence
{μ
j
} converges strongly to μ.
Proof: Suppose G
Ω
μ
j
≤ G
Ω
μ
k
.Then
|μ
k
− μ
j
|
2
e
= |μ
k
|
2
e
− 2
G
Ω
μ
k
dμ
j
+ |μ
j
|
2
e
≤|μ
k
|
2
e
− 2
G
Ω
μ
j
dμ
j
+ |μ
j
|
2
e
(6.3)
= |μ
k
|
2
e
−|μ
j
|
2
e
.
This shows that the sequence {|μ
j
|
e
} is monotone. If the sequence { G
Ω
μ
j
}
is increasing, the same is true of the sequence {|μ
j
|
e
},and
|μ
j
|
2
e
=
G
Ω
μ
j
dμ
j
≤
G
Ω
μdμ
j
=
G
Ω
μ
j
dμ ≤|μ|
2
e
,
and therefore |μ
j
|
e
≤|μ|
e
,j ≥ 1. If the sequence {G
Ω
μ
j
} is decreasing,
thesameistrueofthesequence{|μ
j
|
e
} and |μ
j
|
e
≤|μ
1
|
e
,j ≥ 1. In either
case, the sequence {|μ
j
|
e
} is a bounded Cauchy sequence. It follows from
Inequality (6.3) that the sequence {μ
j
} is a strong Cauchy sequence. By the
preceding theorem, it suffices to prove that the sequence converges weakly to
μ;thatis,
lim
j→∞
G
Ω
μ
j
dν =
G
Ω
μdν
for all ν ∈E
+
, but this follows from the Lebesgue dominated convergence
theorem.
The ordered pair λ =(μ, ν) ∈Ehas compact support if both μ and ν
have compact supports. In this case, G
Ω
μ and G
Ω
ν are both defined, but
G
Ω
(μ − ν)=G
Ω
μ − G
Ω
ν may not be defined because of the infinities of
G
Ω
μ and G
Ω
ν. The notation G
Ω
(μ−ν) will be used only when the difference
G
Ω
μ − G
Ω
ν is defined everywhere on Ω.
Theorem 6.3.9 The set of μ ∈Ehaving compact support with G
Ω
μ ∈
C
0
0
(Ω) is dense in E.
Proof: Consider any μ ∈E
+
,let{Γ
j
} be an increasing sequence of compact
subsets of Ω such that Ω = ∪Γ
j
,andletμ
j
= μ|
Γ
j
,j ≥ 1. Since each
μ
j
has compact support, G
Ω
μ
j
is a potential and {G
Ω
μ
j
} is a sequence
of potentials increasing to G
Ω
μ. By the preceding theorem, the sequence
{μ
j
} converges strongly to μ. This shows that the set E
+
0
of μ ∈E
+
having
compact support in Ω is strongly dense in E
+
. Now consider any μ ∈E
+
0
.By
taking volume averages of G
Ω
μ, a sequence {μ
j
} in E
+
0
can be constructed so
that G
Ω
μ
j
↑ G
Ω
μ with each G
Ω
μ
j
∈ C
0
(Ω). By the preceding theorem, the
sequence {μ
j
} converges strongly to μ. This shows that the set of μ ∈E
+
0
with
6.3 Mutual Energy 255
continuous G
Ω
μ is strongly dense in E
+
. Consider now an arbitrary element
of E
+
0
with continuous u = G
Ω
μ. By Corollary 2.6.31, there is an increasing
sequence {Ω
j
} of regular open sets with Ω
−
j
⊂ Ω such that Ω = ∪Ω
j
.It
can be assumed that the support of μ is contained in Ω
1
.LetH
Ω
j
u
be the
harmonic function on Ω
j
corresponding to the boundary function u|
∂Ω
j
.Now
let
u
j
=
H
Ω
j
u
on Ω
j
u on Ω ∼ Ω
j
.
By Theorem 4.6.8, u
j
↓ 0onΩ as j →∞. Letting u
j
= G
Ω
μ
j
, G
Ω
μ
j
≤
G
Ω
μ implies that μ
j
∈E
+
and, in fact, that μ
j
∈E
+
0
since u
j
= u outside
a neighborhood of the support of μ on which u is harmonic. Since u
j
=
G
Ω
μ
j
↓ 0, the sequence {μ
j
} converges strongly to the zero measure by
the preceding theorem. It follows that the sequence {μ − μ
j
} in E converges
strongly to μ.NotethatG
Ω
μ −G
Ω
μ
j
= u −u
j
=0outsideΩ
j
and therefore
G
Ω
(μ − μ
j
) ∈ C
0
0
(Ω). Therefore, any μ ∈E
+
can be approximated strongly
by a ν ∈Ehaving compact support with G
Ω
ν ∈ C
0
0
(Ω). If μ = μ
1
− μ
2
∈E
where μ
i
∈E
+
,i =1, 2, then each μ
i
can be approximated strongly by a
ν
i
∈Ehaving compact support with G
Ω
ν
i
∈ C
0
0
(Ω).
Theorem 6.3.10 Let {μ
j
} be a sequence in E
+
. Strong convergence of the
sequence implies weak convergence and norm boundedness. If the sequence
{μ
j
} is norm bounded, then weak convergence and vague convergence are
equivalent.
Proof: If the sequence {μ
j
} converges strongly to μ ∈E
+
, then norm bound-
edness follows from the inequality ||μ
j
|
e
−|μ|
e
|≤|μ
j
− μ|
e
. Suppose the
sequence {μ
j
} is norm bounded by m and converges weakly to μ ∈E
+
.Con-
sider the signed measures τ
x,δ
of Theorem 3.5.3. Since each G
Ω
τ
x,δ
is the
Green potential of a signed measure of finite energy whenever B
−
x,δ
⊂ Ω,
lim
j→∞
G
Ω
τ
x,δ
,dμ
j
=
G
Ω
τ
x,δ
dμ;
that is, lim
j→∞
fdμ
j
=
fdμ for all f in a total subset K
+
0
of K
+
(Ω).
Let Γ be any compact subset of Ω. It now will be shown that the sequence
{μ
j
(Γ )} is bounded. To do this, let f be a nonnegative linear combination of
elements of K
+
0
that majorizes 1 on Γ .Since
lim sup
j→∞
μ
j
(Γ ) ≤ lim sup
j→∞
fdμ
j
=
fdμ<+∞,
the sequence {μ
j
(Γ )} is bounded. Consider any g ∈ K
+
(Ω). If >0, choose
f ∈ K
+
0
such that |g − f| <.Sincef and g have compact supports in Ω,
there is a compact set Γ containing the supports of f and g and a constant
α such that μ
j
(Γ ) ≤ α, j ≥ 1. Since
256 6Energy
gdμ
j
−
gdμ
≤
|g −f|dμ
j
+
fdμ
j
−
fdμ
+
|f − g|dμ
≤ α +
fdμ
j
−
fdμ
+ μ(Γ )
and f ∈ K
+
0
, lim
j→∞
gdμ
j
=
gdμ; that is, the sequence {μ
j
} converges
to μ vaguely. Assume now that |μ
j
|
e
≤ m, j ≥ 1, and that the sequence
{μ
j
} converges to μ vaguely. As in the proof of Theorem 3.6.4, G
Ω
μ ≤
lim inf
j→∞
G
Ω
μ
j
. By Fatou’s Lemma, for each p ≥ 1
G
Ω
μdμ
p
≤
(lim inf
j→∞
G
Ω
μ
j
) dμ
p
≤ lim inf
j→∞
G
Ω
μ
j
dμ
p
≤ m
2
.
Therefore,
G
Ω
μdμ ≤
(lim inf
p→∞
G
Ω
μ
p
) dμ ≤ lim inf
p→∞
G
Ω
μdμ
p
≤ m
2
and μ ∈E
+
. To show that lim
j→∞
[μ
j
,ν]
e
=[μ, ν]
e
for all ν ∈E, it suffices to
prove this result for all ν in a dense subset of E. By the preceding theorem,
the set of ν ∈Ehaving compact support in Ω with Gν ∈ C
0
0
(Ω)isdenseinE.
For such ν,
lim
j→∞
[μ
j
,ν]
e
= lim
j→∞
G
Ω
νdμ
j
=
G
Ω
νdμ=[μ, ν]
e
by the vague convergence of the sequence {μ
j
} to μ.
Theorem 6.3.11 (Cartan [11]) E
+
is complete.
Proof: Let {μ
j
} be a sequence in E
+
that is strong Cauchy. For any ν in E,
the inequality
|[μ
j
,ν]
e
− [μ
k
,ν]|
e
≤|μ
j
− μ
k
|
e
|ν|
e
implies that the sequence {[μ
j
,ν]
e
} is Cauchy in R.Takingν = τ
x,δ
,where
B
−
x,δ
⊂ Ω,
L(f) = lim
j→∞
fdμ
j
exists for each f in a total subset K
+
0
of K
+
. The linear functional L(f)
can be extended to K
+
as follows. First let Γ be any compact subset of Ω
and let f ∈ K
+
0
majorize 1 on Γ . Since the sequence {
fdμ
j
} is bounded
by some m>0,μ
j
(Γ ) ≤
fdμ
j
≤ m, j ≥ 1. This shows that the sequence
{μ
j
(Γ )} is bounded for any compact subset Γ of Ω. Now consider any g ∈
K
+
having compact support Γ and a neighborhood U of Γ having compact
closure U
−
⊂ Ω.Choosem>0sothatμ
j
(U
−
) ≤ m for all j ≥ 1. For each
k ≥ 1, let g
k
∈ K
+
0
have compact support in U such that the sequence {g
k
}
converges uniformly to g on Ω.Since
6.4 Projections of Measures 257
gdμ
i
−
gdμ
j
≤
|g −g
k
|dμ
i
+
g
k
dμ
i
−
g
k
dμ
j
+
|g
k
− g|dμ
j
≤ 2m g − g
k
0
+
g
k
dμ
i
−
g
k
dμ
j
.
Given >0, choose k so that |g − g
k
| </4m.Then
gdμ
i
−
gdμ
j
≤
2
+
g
k
dμ
i
−
g
k
dμ
j
from which it follows that the sequence {
gdμ
j
} is Cauchy and
L(g) = lim
j→∞
gdμ
j
exists for all g ∈ K
+
. It follows that there is a measure μ such that
lim
j→∞
gdμ
j
=
gdμ
for all g ∈ K
+
; that is, the sequence {μ
j
} converges vaguely to μ.Sincethe
sequence is strongly bounded, the sequence converges weakly to μ by the
preceding theorem and strong convergence follows from Theorem 6.3.7.
6.4 Projections of Measures
The operation of sweeping a measure defining a potential on a Greenian set
canbeformulatedasanoperatoronthepre-HilbertspaceE by projecting a
measure μ ∈E
+
onto a convex subset of E
+
. Recall that a subset F of E is
convex if λμ +(1− λ)ν ∈Fwhenever μ, ν ∈Fand 0 <λ<1.
Lemma 6.4.1 If F is a nonempty, closed, convex subset of E
+
and μ ∈E
+
,
then there is a unique measure μ
F
∈Fsuch that |μ −μ
F
|
e
≤|μ −ν|
e
for all
ν ∈F.
Proof: Consider any two measures α, β ∈F.Thenγ =(α + β)/2 ∈Fsince
F is convex. Now
|μ − γ|
2
e
= |(μ − α)+(α − γ)|
2
e
= |(μ − α)+
1
2
(α − β)|
2
e
= |μ − α|
2
e
+
1
4
|α − β|
2
e
+[μ − α, α − β]
e
.