REAL ANALYSIS
INTERCHANGE THEOREMS
Theorem 1
Suppose
f
12
(
xy
) is continuous at (
ab
) and
∃
δ >
0 such that
f
2
(
xb
) exists for

x

a

< δ
. Then
f
21
(
ab
) exists and =
f
12
(
ab
).
Proof
Without loss of generality we may take (
ab
) = (0
,
0).
Let
ε
be given. Choose
δ
, such that 0
< δ
1
< δ
and

f
12
(
xy
)

f
12
(00)

<
ε
whenever

x

< δ
1
and

y

< δ
1
. Suppose 0
<

h

< δ
1
.
Consider
f
2
(
ho
)

f
2
(00)
h
= lim
k
→
0
Δ
hk
P
hk
Δ
hk
=
{
f
(
hk
)

f
(
h
0)
}  {
f
(
ok
)

f
(00)
}
We regard
k
as being temporarily fixed with

k

sufficiently small, and
write
F
(
h
) =
f
(
hk
)

f
(
h
0) so that
Δ
hk
hk
=
F
(
h
)

F
(0)
hk
=
F
0
(
θh
)
k
by MVT0
< θ <
1
=
f
1
(
θh, k
)

f
1
(
θh,
0)
k
=
f
12
(
θhθ
0
k
) my MVT0
< θ
0
<
1
Hence
fl
fl
fl
fl
Δ
hk
hk

f
12
(00)
fl
fl
fl
fl
< ε.
Letting
k
→
0 we have by (1)
fl
fl
fl
fl
fl
f
2
(
h
0)

f
2
(00)
h

f
12
(00)
fl
fl
fl
fl
fl
≤
ε.
Hence
f
21
(00) exists and is equal to
f
12
(00)
In the following results
R
de3notes the closed rectangle
a
≤
x
≤
b c
≤
y
≤
d
.
1
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Lemma
Let
f
(
xy
) be continuous on
R
. Then we have
φ
(
x
) =
∑
d
c
f
(
xy
)
dy
is continuous on [
ab
].
Proof
f
(
xy
) is uniformly continuous on
R
.
Hence, given
ε >
0
,
∃
δ >
0

f
(
P
)

f
(
Q
)

<
ε
d

c
whenever
P
∈
R Q
∈
R
and

PQ

< δ
.
Now if
x
1
, x
2
are each in [
ab
] and

x
1

x
2

< δ
:

φ
(
x
1
)

φ
(
x
2
)
 ≤
Z
d
c

f
(
x
1
y
)

f
(
x
2
y
)

dy < d

c
ε
d

c
=
ε
Theorem 2
Let
f
(
xy
) be continuous as a function of
y
for
c
≤
y
≤
d
relative
to this interval, for each
x
with
a
≤
x
≤
b
.
Suppose that
f
1
(
xy
) is
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 Spring '98
 pfitz

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