(members of an equicontinuous set are trivially continuous) must be continuous on [
a,b
]. Because
x
0
is in the
interior of [
a,b
], this implies that the original
f
:
R
→
R
is continuous at
x
0
.
Quickie Proof.
Let
² >
0 be given. There is
δ >
0 such that

f
n
(
x
)

f
n
(
y
)

< ²/
2 whenever

x

y

< δ
. Let
x,y
∈
R
and assume

x

y

< δ
. Then

f
(
x
)

f
(
y
)

= lim
n
→∞

f
n
(
x
)

f
n
(
y
)
 ≤
²/
2
< ².
2
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View Full DocumentNote on the quickie proof, and its variants.
A popular variant of the quickie proof was to use an
²/
3
argument, writing

f
(
x
)

f
(
y
)
 ≤ 
f
(
x
)

f
n
(
x
)

+

f
n
(
x
)

f
n
(
y
)

+

f
n
(
y
)

f
n
(
x
)

.
If one uses it, to use it correctly, given
² >
0 ﬁrst you select
δ >
0 such that if
x,y
∈
R
, then

f
n
(
x
)

f
n
(
y
)

< ²/
3
for all
n
∈
N
. Next you
ﬁx
x,y
such that

x

y

< δ
. That is,
x,y
are arbitrary with

x

y

< δ
, but will not
be moved anymore. Now you can ﬁnd
N
x
,N
y
such that
n
≥
N
x
implies

f
(
x
)

f
n
(
x
)

< ²/
3 and
n
≥
N
y
implies

f
(
y
)

f
n
(
y
)

< ²/
3. Then let
n
≥
max(
N
x
,N
y
) and conclude

f
(
x
)

f
(
y
)

< ²
, with
n
having disappeared in the
end result.
Some additional notes:
1.
I would also have accepted a much shorter proof without so much symbolism: By ArzelaAscoli, on every closed
and bounded interval the sequence
{
f
n
}
has a subsequence converging uniformly, since the limit has to be
f
, this
implies that
f
is continuous when restricted to bounded intervals, hence continuous. QED
2.
Equicontinuity on a bounded interval plus uniform boundedness at one single point implies uniform bounded
ness.
Suppose
{
f
n
}
is a sequence of real valued functions on [
a,b
],
∞
< a
≤
b <
∞
. Assume
{
f
n
}
is equicontinuous
and there is
c
∈
[
a,b
],
M
∈
R
,
M
≥
0 such that

f
n
(
c
)
 ≤
M
for all
n
. This happens, for example, if the
sequence of real numbers
{
f
n
(
c
)
}
converges. We apply equicontinuity with any
² >
0, for example, with
²
= 1.
By equicontinuity there exists
δ >
0 such that if
x,y
∈
[
a,b
] and

x

y

< δ
, then

f
n
(
x
)

f
n
(
y
)

<
1. Now every
point
x
∈
[
a,b
] is at distance at most
b

a
from
c
. So if
x
∈
[
a,b
] we can ﬁnd a sequence of at most
m
= [
b

a
]+1
points ([
b

a
] = integer part of
b

a
) leading from
x
to
c
with any two consecutive points at distance at most
1 from each other. That is, we can ﬁnd points
x
1
,...,x
k
where
k
≤
m
,
x
1
=
x
,
x
k
=
c
, and

x
i

x
i

1

<
1 for
i
= 2
,...,k
. Then, for all
n
∈
N
,

f
n
(
x
)
 ≤ 
f
n
(
x
)

f
n
(
x
2
)

+

f
n
(
x
2
)

f
n
(
x
3
)

+
···
+

f
n
(
x
k

1
)

f
n
(
c
)

+

f
n
(
c
)
 ≤
(
k

1) +
M
≤
(
b

a
) +
M
proving the sequence is uniformly bounded.
(b)
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 Spring '11
 Speinklo
 Calculus, Topology, FN, Arzela Ascoli

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