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according to
C3
. In other words,
g
is bounded above. Therefore, we can apply
Lemma
4.2
and obtain
n
n
lim
n
ₒ∞
lim
j
ₒ∞
f
(
i
,
j
)
=
lim
j
ₒ∞
lim
n
ₒ∞
f
(
i
,
j
)
.
(4.3)
i
=
0
i
=
0
For any
j
∈ N
, the sequence
{
g
(
n
,
j
)
}
n
≥
0
is nondecreasing and bounded, so its limit
i
=
0
f
(
i
,
j
) exists. That is,
∞
n
lim
n
ₒ∞
f
(
i
,
j
)
=
f
(
i
,
j
)
.
(4.4)
i
=
0
i
=
0
, the limit lim
j
ₒ∞
i
=
0
f
(
i
,
j
)
In view of
C2
, we have that for any given
n
∈ N
exists and
n
n
lim
j
ₒ∞
f
(
i
,
j
)
=
lim
j
ₒ∞
f
(
i
,
j
)
.
(4.5)
i
=
0
i
=
0
By
C3
, the sequence
{
S
n
}
n
≥
0
is bounded. Since it is also nondecreasing, it converges
∞
to
lim
j
ₒ∞
f
(
i
,
j
). That is,
i
=
0
n
∞
lim
n
ₒ∞
lim
j
ₒ∞
f
(
i
,
j
)
=
lim
j
ₒ∞
f
(
i
,
j
)
.
(4.6)
i
=
0
i
=
0
Hence the left-hand side of the desired equality exists. By combining (
4.3
)-(
4.6
),
∞
∞
we obtain the result,
lim
j
ₒ∞
f
(
i
,
j
)
=
lim
j
ₒ∞
f
(
i
,
j
).
i
=
0
i
=
0
Proposition 4.3 (Bounded Continuity—General Function)
Let f
:
N × N ₒ R
be a function that satisfies the following conditions
C1.
j
1
≤
j
2
implies
|
f
(
i
,
j
1
)
|≤|
f
(
i
,
j
2
)
|
for all i
,
j
1
,
j
2
∈ N
;
C2.
for any i
∈ N
, the limit
lim
j
ₒ∞
|
f
(
i
,
j
)
|
exists;
C3.
the partial sums S
n
=
i
=
0
lim
j
ₒ∞
|
f
(
i
,
j
)
|
are bounded, i.e. there exists
some c
∈ R
≥
0
such that S
n
≤
c, for all n
≥
0
;
C4.
for all i
,
j
1
,
j
2
∈ N
j
2
and f
(
i
,
j
1
)
>
0
, then f
(
i
,
j
2
)
>
0
.
Then the following equality holds:
,ifj
1
≤
∞
∞
lim
j
ₒ∞
f
(
i
,
j
)
=
lim
j
ₒ∞
f
(
i
,
j
)
.
i
=
0
i
=
0
Proof
By Proposition
4.2
and conditions
C1
,
C2
and
C3
, we infer that
∞
∞
lim
j
ₒ∞
i
=
0
|
f
(
i
,
j
)
|=
j
ₒ∞
|
lim
f
(
i
,
j
)
|
.
(4.7)
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