7.3 Uniform convergence
115
Theorem 7.3.15
((Baby) dominated convergence thm)
.
Suppose
f
n
∈ R
[
a, b
]
for every
0
< a < b <
∞
, and suppose
f
n
unif
→
f
on every compact subset of
(0
,
∞
)
.
If
g
∈
R
[0
,
∞
)
, then

f
n
 ≤
g
=
⇒
lim
n
→∞
Z
∞
0
f
n
(
x
)
dx
=
Z
∞
0
f
(
x
)
dx.
Proof.
HW.
Theorem 7.3.16
(Stirling’s Formula)
.
lim
x
→∞
Γ(
x
+1)
(
x/e
)
x
√
2
πx
= 1
.
Proof.
HW.
Often: lim
n
→∞
n
!
(
n/e
)
n
√
2
πn
= 1, meaning that
n
!
∼
(
n/e
)
n
√
2
πn
.
7.3.6
Termbyterm differentiation
Example 7.3.11.
Recall
f
n
(
x
) =
sin
nx
√
n
. This is uniformly dominated by
1
√
n
, so converges
uniformly to
f
≡
0, but
f
0
n
(
x
)
9
f
0
(
x
)! Not even uniform convergence can save us now!
Need stronger hypothesis.
Theorem 7.3.17.
Let
f
n
∈
C
1
(
I
)
,
f
n
pw
→
f
and
f
0
n
unif
→
g
. Then
f
∈
C
1
(
I
)
and
f
0
(
x
) =
g
(
x
)
.
Proof.
Fix a point
a
∈
I
. Then FToC1 gives
f
n
(
x
)

f
n
(
a
) =
Z
x
a
f
0
n
(
t
)
dt
n
→∞
→
Z
x
a
g
(
t
)
dt.
However, we also have
f
n
(
x
)

f
n
(
a
)
→
f
(
x
)

f
(
a
), so apply FToC2 to
f
(
x
)

f
(
a
) =
Z
x
a
g
(
t
)
dt
to see that
f
∈
C
1
(
I
) with
f
0
(
x
) =
g
(
x
).
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 Fall '11
 Wong
 Calculus, Fourier Series, Compact space, FN, Uniform convergence

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