7.3 Lemma For all f ∈ K(G) and ǫ > 0 , there exists U ∈ ηe such that, for all x, y ∈ G,
if x ∈ U yU , then |f (x) − f (y)| < ǫ.
Proof. We consider the product topogroup G×G. Let O = {(x, y) ∈ G×G : |f (x) − f (y)| <
ǫ} . It follows from continuity of f that O is open in G × G. The subset K = {(x, x) : x ∈
G} of G×G is compact, since the mapping x 7→ (x, x) is a homeomorphism G → K . Note
that K ⊂ O . By Theorem 1.25, there exists W ∈ η(e,e) (G × G) such that W KW ⊂ O .
By the definition of product topology, there exists U ∈ ηe (G) such that U × U ⊂ W .
We show that U has the property required in the lemma. Let the points x and y of G
satisfy the condition x ∈ U yU . Then there exist u, v ∈ U such that x = uyv . We have
(x, y) = (u, e)(y, y)(v, e) and it follows, since (u, e), (v, e) ∈ U × U ⊂ W and (y, y) ∈ K ,
that (x, y) ∈ W KW ⊂ O . As a consequence, |f (x) − f (y)| < ǫ.
For a set A , we denote by |A| the cardinality of A . When A is finite, the cardinality
is simply the number of elements of A .
The following result is a version of the famous “Hall’s Marriage Theorem” from combinatorics.
7.4 Lemma Let A be a finite set, and for every a ∈ A , let Da be a subset of a set C . We
S
assume that a∈B Da ≥ |B| for every B ⊂ A . Then there exists a one–to–one mapping
φ : A → C such that φ(a) ∈ Da for every a ∈ A .
Proof. We use induction on |A|. If |A| = 1 , then the claim is trivially true. Let n be
such a natural number, that the claim has been established for sets of cardinality less than
n . We assume that |A| = n and we prove the claim in this case.
We consider two different cases. We assume that first that there exists E ⊂ A such
S
that ∅ =
6 E 6= A and a∈E Da = |E|. Since |E| < |A| = n , it follows from the inductive
hypothesis that there exists a one–to–one mapping ψ : E → C such that ψ(a) ∈ Da for
S
′
every a ∈ E . Let D = a∈E Da , and for every a ∈ A \ E , let Da = Da \ D . We
S
′ show that a∈B Da ≥ |B| for every B ⊂ A \ E . Let B ⊂ A \ E . By assumption, we
S
S
have a∈B∪E Da ≥ |B ∪ E|. Since we have |B ∪ E| = |B| + |E|, |E| = a∈E Da and
S
S
S
S
+
= , it follows from the foregoing, that
D
\
D
D
D
a
a
a
a
a∈B
a∈E
a∈E
a∈B∪E
S
S
S
S
S
S
′
a∈B Da \
a∈E Da ≥ |B|. Since
a∈B Da \
a∈E Da =
a∈B (Da \ D) =
a∈B Da ,
59
S
′ we have shown that a∈B Da ≥ |B|. Since |A \ E| < n , it follows from the foregoing
′
that we can apply the inductive hypothesis on the sets A \ E and Da , a ∈ A \ E . Hence
there exists a one–to–one mapping θ : A \ E → C such that, for every a ∈ A \ E , we have
′
θ(a) ∈ Da . Define the mapping φ : A → C by setting φ(a) = ψ(a) for every a ∈ E and
φ(a) = θ(a) for every a ∈ A \ E . Then φ is a one–to–one mapping and we have φ(a) ∈ Da
for every a ∈ A .
S
Next we assume that a∈B Da > |B| for every ∅ =
6 B 6= A . Let a0 be some element
of the set A and let c0 be some element of the non-empty set Da0 . For every a ∈ A\{a0 } ,
′
set Da = Da \ {c0 } . Then we have, for every ∅ =
6 B ⊂ A \ {a0 } , that
[
[ ′ [
Da = Da \ {c0 } ≥ Da − 1 ≥ |B| .
a∈B
a∈B
a∈B
By the inductive hypothesis, there exists a one–to–one mapping φ : A \ {a0 } → C such
′
that φ(a) ∈ Da for every a ∈ A \ {a0 } . We extend φ to A by setting φ(a0 ) = c0 ; then
the mapping φ has all the required properties.
Before we can apply the above result for the compact group G, we make the following
observation.
7.5 Lemma For every U ∈ ηe , there exists n ∈ N such that for every A ⊂ G, if |A| > n ,
then there exists a ∈ A such that A ∩ (aU ) 6= {a} .
Proof. Choose V ∈ ηe such that V −1 V ⊂ U . Since G is compact, there exists a finite set
Sn
{b1 , ..., bn} ⊂ G such that i=1 (bi V ) = G. We show that the number n has the required
Sn
property. Let A ⊂ G be such that |A| > n . Since A ⊂ i=1 (bi V ) , there exist j ≤ n and
a, ã ∈ A such that a 6= ã ja {a, ã} ⊂ bj V . Let v ∈ V and w ∈ V be such that a = bj v
and ã = bj w . Then a−1 ã = v −1 w ∈ V −1 V ⊂ U . As a consequence, ã ∈ aU . Since ã 6= a ,
we have A ∩ (aU ) 6= {a} .
7.6 Lemma For every U ∈ ηe , there exists a finite non-empty set A ⊂ G such that,
for all g, h, p, q ∈ G, there exists a one-to-one onto mapping φ : gAh → pAq such that
φ(x) ∈ U xU for every x ∈ gAh.
60
Proof. By Theorem 5.17, there exists an invariant neighborhood V of e such that V ⊂ U .
Denote by W the symmetric and invariant neighborhood V ∩ V −1 of e. Let
B = B ⊂ G : B ∩ (W b) = {b} for every b ∈ B .
The family B is non-empty, since {e} ∈ B . We note that, for every B ∈ B , every subset
of B belongs to the family B . Moreover, if B ∈ B and g, h ∈ G, then gBh ∈ B ; to see
this, note that, for every b ∈ B , we have B ∩ (W b) = {b} and hence, by invariantness of
W , we have gBh ∩ (W gbh) = {gbh} . It follows from Lemma 7.5 that there is a largest
number m in the set {|B| : B ∈ B} . Let Bm = {B ∈ B : |B| = m} .
Let H, J ∈ Bm . We show that there exists a one-to-one mapping φ : H → J such
that φ(h) ∈ U hU for every h ∈ H . For every h ∈ H , let Dh = hW ∩ J and, for every
S
I ⊂ H , let DI = i∈I Di = IW ∩ J . We show that |DI | ≥ |I| for every I ⊂ H . Let I be
a subset of H . Then I, J ∈ B and it follows, since j ∈ iW ⇐⇒ i ∈ jW ⇐⇒ j ∈ Di for
all i ∈ I ja j ∈ J that
I ∪ (J \ DI ) = I ∪ (J \ IW ) ∈ B.
From this it follows by maximality of m, that
m ≥ |I ∪ (J \ IW )| = |I| + |J \ IW | = |I| + |J \ DI | = |I| + m − |DI |
and therefore we obtain the inequality |DI | ≥ |I|. By the foregoing and Lemma 7.4, there
exists a one–to–one mapping φ : H → J such that for every h ∈ H we have φ(h) ∈ Dh
that is φ(h) ∈ hW . Since H, J ∈ Bm , the one–to–one mapping φ is onto. For every
h ∈ H we have φ(h) ∈ hW ⊂ hU ⊂ U hU .
Let now A be any member of the family Bm . Then for all x, y ∈ G we have xAy ∈
Bm , and thus it follows from the foregoing that the set A has the required property.
We are going to estimate the values of the invariant integral by averages. We shall
employ the following notation. Let A ⊂ G be a non-empty finite set . For every f ∈ K(G) ,
P
1
denote by MA (f ) the average |A|
a∈A f (a) of the values of f in the set A . Further, for
every F ⊂ K(G) , denote by MA (F ) the set {MA (f ) : f ∈ F } of complex numbers. For
every f ∈ K(G) , denote by
G fG
the subset {g fh : g, h ∈ G} of K(G) .
When D is a set of complex number, denote by diam(D) the “diameter” of D , that
is, the number sup{|d − d′ | : d, d′ ∈ D} .
61
7.7 Lemma For every U ∈ ηe , there exists a finite, non-empty set A ⊂ G such that, for
every f ∈ K(G) , we have
diam(MA (G fG )) ≤ sup{|f (x) − f (y)| : x, y ∈ G and x ∈ U yU }.
Proof. We choose for A a finite non-empty subset of G which satisfies the condition
of Lemma 7.6, and we show that then A also satisfies the condition above. Let f ∈
K(G) and g, h, p, q ∈ G. We need to establish the inequality |MA (g fh ) − MA (p fq )| ≤
sup{|f (x) − f (y)| : x, y ∈ G and x ∈ U yU } . Let φ : gAh → pAq be a one-to-one onto
mapping such that φ(gah) ∈ U gahU for every a ∈ A . Then we have
1 X
1 X
X
1
|MA (g fh ) − MA (p fq )| = f (gah) −
f (paq) = (f (gah) − f (φ(gah)))
|A|
|A|
|A|
a∈A
≤
a∈A
a∈A
1 X
|f (gah) − f (φ(gah))| ≤ sup{|f (x) − f (y)| : x, y ∈ G and x ∈ U yU } .
|A|
a∈A
We note that the following is a consequence of Lemmas 7.3 and 7.7.
7.8 Lemma For every finite F ⊂ K(G) , there exists a sequence hAn i∞
n=1 of finite nonempty subsets of G such that, for every f ∈ F , we have diam(MAn (G fG )) → 0 when
n → ∞.
7.9 Lemma Let f ∈ K(G) and let A and B be finite non-empty subsets of G. Then
diam (MA (G fG ) ∪ MB (G fG )) ≤ diam(MA (G fG )) + diam(MB (G fG )).
Proof. Let ǫ = diam(MA (G fG )) = sup{|MA (g fh ) − MA (p fq )| : g, h, p, q ∈ G} and δ =
diam(MB (G fG )) = sup{|MB (g fh ) − MB (p fq )| : g, h, p, q ∈ G} . We need to show that
|MA (g fh ) − MB (p fq )| ≤ ǫ + δ for all g, h, p, q ∈ G.
For every b ∈ B , we have |MA (fb ) − MA (g fh )| ≤ ǫ, and it follows that
!
1 X
1 X
MA (fb ) − MA (g fh ) = (MA (fb ) − MA (g fh ))
|B|
|B|
b∈B
b∈B
1 X
≤
|MA (fb ) − MA (g fh )| ≤ ǫ .
|B|
b∈B
62
A similar argument yields the inequality
!
1 X
MB (a f ) − MB (p fq ) ≤ δ.
|A|
a∈A
The desired inequality |MA (g fh ) − MB (p fq )| ≤ ǫ + δ follows now from the preceding two
inequalities, since
1 X
1 XX
1 X
MA (fb ) =
f (ab) =
MB (a f ).
|B|
|B| |A|
|A|
b∈B
b∈B a∈A
a∈A
With the help of the previous lemmas, we can determine the value of the invariant
integral for a function f .
7.10 Lemma Let f ∈ K(G) . There exists M (f ) ∈ C such that, for any sequence hBn i∞
k=1
of finite non-empty subsets of G, if diam(MBn (G fG )) → 0 , then MBn (f ) → M (f ) .
Proof. By Lemma 7.8, there exists a sequence hAn i∞
n=1 of finite non-empty subsets of G
such that diam(MAn (G fG )) → 0 . From Lemma 7.9 it follows that |MAn (f ) − MAm (f )| ≤
diam MAn (G fG ) + diam MAm (G fG ) for all n, m ∈ N . Since diam MAn (G fG ) → 0 ,
it follows from the foregoing that there exists a number M (f ) ∈ C such that MAn (f ) →
M (f ) .
If hBn i∞
n=1 is another sequence of finite non-empty subsets of G with the property
that diam MBn (G fG ) → 0 , then we have MBn (f ) → M (f ) , because it follows from
Lemma 7.9 that |MBn (f ) − MAn (f )| ≤ diam MBn (G fG ) + diam(MAn (G fG ) for every
n ∈ N.
The above result defines a mapping M : K(G) → C . Next we show that M is an
invariant integral of G.
7.11 Theorem Every compact topogroup has an invariant integral.
Proof. We define M : K(G) → C by the result of Lemma 7.10. We show that M is an
invariant integral of G.
ˆ cf, f + fˆ} . By Lemma 7.8, there exists
Let f, fˆ ∈ K(G) and let c ∈ C . Set F = {f, f,
a sequence hAn i∞ of finite non-empty subsets of G such that diam MA (G f˜G ) → 0
n=1
n
for every f˜ ∈ F . Note that we have MAn (f˜) → M (f˜) for each f˜ ∈ F .
63
To verify linearity of M , we show that M (cf ) = cM (f ) and M (f + fˆ) = M (f ) +
M (fˆ) . For every n ∈ N , we have MAn (cf ) = cMAn (f ) and MAn (f + fˆ) = MAn (f ) +
MA (fˆ) . It follows that we have
n
M (cf ) = lim MAn (cf ) = lim cMAn (f ) = c lim MAn (f ) = cM (f ) , and
M (f + fˆ) = lim MAn (f + fˆ) = lim MAn (f ) + MAn (fˆ)
ˆ .
= lim MAn (f ) + lim MAn (fˆ) = M (f ) + M (f)
Next we show that M (g fh ) = M (f ) for all g, h ∈ G. Since
G ( g fh ) G
=G fG , we have
MAn (g fh ) → M (g fh ) . Since |MAn (f ) − MAn (g fh )| ≤ diam MAn (G fG ) for every n ∈ N ,
it follows from the foregoing that M (g fh ) = M (f ) .
Clearly, we have M (1) = 1 . Using properties of averages, we see that M (f˜) ≥ 0 for
every f˜ ∈ K+ (G) . To complete the proof, we assume that f ∈ K+ (G) \ {0} , and we show
that M (f ) > 0 . Since f 6= 0 , there exists ǫ > 0 such that the set O = {x ∈ G : f (x) > ǫ}
is non-empty. By continuity of f , the set O is open. Since G is compact and the open
Sn
sets Og , g ∈ G, cover G, there exist n ∈ N and g1 , ..., gn ∈ G so that i=1 Ogi = G. For
all x ∈ G and i ≤ n , if x ∈ Ogi , then we have f (xgi−1 ) > ǫ, in other words, fg −1 (x) > ǫ;
i
˜
from this it follows that all values of the function f = f −1 + · · · + f −1 are bigger than ǫ.
g1
gn
1 ˜
f−
ǫ
As a consequence, the function
1 belongs to the collection K+ (G) . It follows that
˜ = M ( 1 f˜) ≥ M (1) = 1 . It
M ( 1ǫ f˜ − 1) ≥ 0 . Using linearity of M , we see that 1ǫ M (f)
ǫ
follows that M (f˜) ≥ ǫ. Since
M (f˜) = M (fg −1 + · · · + fgn−1 ) = M (fg −1 ) + · · · + M (fgn−1 ) = nM (f ) ,
1
it follows from the foregoing that M (f ) ≥
1
ǫ
n
> 0.
Next we prove that the invariant integral of a compact topogroup is unique. We start
with an auxiliary result.
Let f ∈ K(G) . We define a new function |f | by the formula |f | (x) = |f (x)|. Note
that it follows from continuity of the mapping z 7→ |z| that the mapping |f | is continuous
C → C . As a consequence, |f | ∈ K+ (G) .
7.12 Lemma Let L : K(G) → C be a linear functional such that L(1) = 1 and L(f ) ≥ 0
for every f ∈ K+ (G) . Then, for every f ∈ K(G) , we have |L(f )| ≤ L |f | ≤ supx∈X |f (x)|.
64
Proof. Note first that if f ∈ K(G) is real-valued, then we have L(f ) ∈ R , since we can
write f = f+ − f− , where f+ , f− ∈ K+ (G) .
Let f ∈ K(G) . We represent the complex number L(f ) in the form L(f ) = ρeiθ ,
where ρ ≥ 0 . Then |L(f )| = ρeiθ = ρ eiθ = ρ . We decompose the function e−iθ f into
its real and imaginary parts: e−iθ f = f˜ + ifˆ. We note that the real-valued functions fˆ
and f˜ belong to the family K(G) . We have
ρ = e−iθ L(f ) = L e−iθ f = L(f˜) + iL(fˆ) ,
˜ L(f)
ˆ ∈ R such that L(fˆ) = 0 and ρ = L(f˜) . For every
and it follows, since ρ, L(f),
x ∈ G, we have
f˜(x) ≤ f˜(x) + ifˆ(x) = e−iθ f (x) = |f (x)| .
By the foregoing, |f | − f˜ ∈ K+ (G) . Hence we have L(|f | − f˜) ≥ 0 , in other words,
L(f˜) ≤ L |f | . Since L(f˜) = ρ = |L(f )|, we have shown that |L(f )| ≤ L |f |. To prove
the inequality L |f | ≤ supx∈X |f (x)|, we denote by S the value of the right side of the
inequality. Then we have S − |f | ∈ K+ (G) , and thus L(S − |f |) ≥ 0 . It follows that
L |f | ≤ LS = SL1 = S .
7.13 Theorem Let M be an invariant integral of G and let L be a linear functional
K(G) → C such that we have L(1) = 1 and L(f ) ≥ 0 for every f ∈ K+ (G) .
Assume that, for every f ∈ K(G) , we have either
(i l ) L(g f ) = L(f ) for every g ∈ G
or
(i r ) L(fg ) = L(f ) for every g ∈ G.
Then L = M .
Proof. Let f ∈ K(G) and ǫ > 0 . We show that |L(f ) − M (f )| ≤ ǫ. By Lemmas 7.8 and
7.10, there exists a finite non-empty set A ⊂ G such that |MA (g fh ) − M (f )| < ǫ for all
g, h ∈ G.
Assume that f satisfies Condition (i l ). Let fˆ =
1
|A|
P
a∈A a f
− M (f ) , and note
that fˆ ∈ K(G) . For every h ∈ G, we have |fˆ(h)| = |MA (fh ) − M (f )| < ǫ. It follows
P
1
by Lemma 7.12 that |L(fˆ)| ≤ L|fˆ| ≤ ǫ. Note that L(fˆ) = |A|
a∈A L(a f ) − L(M (f )) .
Since Condition (i l ) holds, we have L(a f ) = L(f ) for each a ∈ A . It follows, since
65
L(M (f )) = M (f )L(1) = M (f ) , that L(fˆ) = L(f ) − M (f ) . By the foregoing, we have the
inequality |L(f ) − M (f )| ≤ ǫ.
Similarly, we can derive the inequality |L(f ) − M (f )| ≤ ǫ from the assumption that
Condition (i r ) holds.
It follows from the foregoing that L(f ) = M (f ) .
Now we have shown that a compact topogroup has a unique invariant integral. From
now on, we denote the invariant integral by the usual integration symbol.
7.14 Notation Let M be the invariant integral of G. For every f ∈ K(G) , we denote
R
R
the number M (f ) by G f or G f (x)dx .
Remark: Condition (i) of definition 7.2 can now be represented in the form
R
R
– G f (gxh)dx = G f (x)dx for all f ∈ K(G) and g, h ∈ G.
In case of a commutative group, the condition takes the form
R
R
– G f (g + x)dx = G f (x)dx for all f ∈ K(G) and g ∈ G.
As we have seen above, for every finite F ⊂ K(G) , there exists a sequence hAn i∞
n=1
R
of non-empty finite subsets of G such that, for every f ∈ F , we have MAn (f ) → G f .
We show now that if the neutral element of G has a countable neighborhood base, then
we can replace the finite set F above by the whole collection K(G) .
7.15 Theorem Assume that e has a countable neighborhood base in G. Then there
R
exists a sequence hAn i∞
n=1 of non-empty finite subsets of G such that MAn (f ) → G f for
every f ∈ K(G) .
Proof. Let hUn i∞
n=1 be a neighborhood base of e such that Un+1 ⊂ Un for every n ∈ N .
By Lemma 7.7 there exists, for every n ∈ N , a non-empty finite set An ⊂ G such that,
for every f ∈ K(G) , we have
diam(MAn (G fG )) ≤ sup{|f (x) − f (y)| : x, y ∈ G and x ∈ Un yUn }.
We show that sequence hAn i∞
n=1 has the required property. Let f ∈ K(G) . For every
n ∈ N , denote by ǫn the number sup{|f (x) − f (y)| : x, y ∈ G and x ∈ Un yUn } . Note
that, for every n ∈ N , we have ǫn+1 ≤ ǫn since Un+1 ⊂ Un . Since hUn i∞
n=1 is a decreasing
neighborhood base of e, it follows from Lemma 7.3 that ǫn → 0 . As a consequence, we
R
have also diam(MAn (G fG )) → 0 when n → ∞ . By Lemma 7.10, we have MAn (f ) → G f .
66
Let hUn i∞
n=1 be a decreasing neighborhood base of e. The proofs of the above theorem
and Lemma 7.7 show that a sequence hAn i∞
n=1 satisfies the condition of the theorem
provided that, for every n ∈ N and for all g, h, p, q ∈ G, there exists a one-to-one onto
mapping φ : gAn h → pAn q such that φ(x) ∈ Un xUn for every x ∈ gAn h. With the
help of this observation, we can now give some examples of invariant integrals of compact
topogroups .
7.16 Examples of invariant integrals.
1 ◦ If G is a finite topogroup, then
R
G
f = MG (f ) for every f ∈ K(G) .
2 ◦ We determine invariant integral of the circle group T. For every n ∈ N , let Un =
{t ∈ T : |t − 1| <
2π
n }
and note that the sequence hUn i∞
n=1 is a decreasing neighborhood
k
base of the neutral element 1 of T. For every n ∈ N , denote by An the set {e n 2πi :
k = 0, 1, ..., n − 1} consisting of the n th roots of unity. For every t = eθi , the translation
x 7→ tx is a rotation of T around the origin by the angle θ . It follows that for every n ∈ N
and for every t = eθi ∈ T, the set tAn divides the circle into n equal parts. If t′ An is
another such set, then its points alternate with the points of the set tAn when we move
around the circle (say, clockwise), and this gives a one-to-one mapping φ : tAn → t′ An
such that we have |φ(x) − x| <
2π
n
, that is, φ(x) ∈ xUn , for every x ∈ tAn .
It follows from the foregoing, together with commutativity of T and the remark made
R
after Theorem 7.15, that T f = limn→∞ MAn (f ) for every f ∈ K(T) .
3 ◦ Let G be a totally disconnected compact topogroup, whose neutral element e has a
countable neighborhood base. By Corollary 5.20, e has a neighborhood base hJn i∞
n=1 such
that every the set Jn is a normal open subgroup of G. For every n ∈ N , Lemma 5.21
shows that the family Jn = {gJn : g ∈ G} has only finitely many different sets. Moreover,
for all g, g ′ ∈ G, if gJn 6= g ′ Jn , then gJn ∩ g ′ Jn = ∅ . For every n ∈ N , choose a finite set
An ⊂ G such that An contains exactly one point from every set belonging to the family
Jn . Since Jn is a normal subgroup, we have {gJh : J ∈ Jn } = Jn for all g, h ∈ G. It
follows that also the set gAn h contains exactly one point from every set belonging to the
family Jn . The foregoing shows that, for all g, h, p, q ∈ G, there exists a one-to-one onto
mapping φ : gAn h → pAn q such that, for every x ∈ gAn h, the points x and φ(x) belong
67
to the same set of the family Jn . For every x ∈ gAn h, we have {x, φ(x)} ⊂ aJn for some
a ∈ G, and hence φ(x) ∈ xJn .
By the foregoing, we have
R
G
f = limn→∞ MAn (f ) for every f ∈ K(G) .
4 ◦ We use the result in 3 ◦ to determine the invariant integral of the totally disconneted
compact topogroup G = Dω = {0, 1}ω from Example 5.14 . In Example 5.14 we noted that
the (normal) open subgroups Jk = {hdn i∞
n=0 ∈ G : dn = 0 for every n ≤ k} , k ∈ N , form a
neighborhood base of the neutral element 0 in G. For every k ∈ N , denote by Ak the finite
subset {hdn i∞
n=0 ∈ G : dn = 0 for every n > k} of G and note that Ak contains exactly one
∞
point from each set of the form hbn i∞
n=0 + Jk = {hdn in=0 ∈ G : dn = bn for every n ≤ k} .
R
It follows from 3 ◦ that G f = limn→∞ MAn (f ) for every f ∈ K(G) .
To close this chapter, we consider “double integration” of a function of two variables.
7.17 Lemma Let H and J be compact topogroups and f ∈ K(H ×J) . Define a function
R
l : H → C by the formula l(h) = J f (h, y)dy . Then l ∈ K(H) .
Proof. We note first that, for every h ∈ H , the function y 7→ f (h, y) belongs the set
K(J) , since the function is a composition of the continuous functions y 7→ (h, y) and
f (h, y) 7→ f (h, y) . Hence the function l is well defined.
We show that l is continuous. Let h ∈ H and ǫ > 0 . Denote by m and n the neutral
elements of H and J , respectively. By Lemma 7.3, there exists W ∈ η(m,n) (H × J) such
that we have |f (a) − f (b)| < ǫ whenever a, b ∈ H × J and a ∈ W b . Let U ∈ ηm (H)
be such that U × {n} ⊂ W . We show that |l(h′ ) − l(h)| ≤ ǫ for every h′ ∈ U h. So
let h′ ∈ U h. For every y ∈ J , we have (h′ , y) ∈ (U × {n}) (h, y) ⊂ W (h, y) and hence
|f (h′ , y) − f (h, y)| < ǫ. It follows, by Lemma 7.12, that
Z
Z
′
(f (h , y) − f (h, y)) dy ≤
|f (h′ , y) − f (h, y)| dy ≤ ǫ .
J
Since
′
l(h ) − l(h) =
J
Z
′
f (h , y)dy −
J
Z
f (h, y)dy =
J
Z
(f (h′ , y) − f (h, y)) dy ,
J
we have |l(h′ ) − l(h)| ≤ ǫ. Since U h ∈ ηh (H) , we have shown that the function l is
continuous. Hence l ∈ K(H) .
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