Math 116  Problem Set 4
(Thanks to Nick Wage for his writeup. Also, the solution to problem 4c have not been
typed up yet. A new version will be posted in the future).
1.
(a)
Let
x
0
∈
(0
,
1] and
>
0. Then take 0
< δ <
min
{
x
2
0
2
,
x
0
2
,
1

x
0
2
}
. Then if

x

x
0

< δ
we
have that
1
x

1
x
0
=
x
0

x
xx
0
≤
2(
x
0

x
)
x
2
0
<
,
so
f
(
x
) =
1
x
is continuous.
On the other hand, suppose
= 1 and
δ >
0.
We have that

f
(1
/
2)

f
(1
/
3)

= 1 =
, so if
δ >
1
/
6 we are done. Otherwise, take
x
=
δ
and
x
=
δ/
2.
Then

f
(
x
)

f
(
x
)

=
1
δ
≥
6
>
1 =
, so
δ
is not valid. Thus, no choice of
δ
works for all
x, x
∈
(0
,
1] when
= 1, and
f
(
x
) =
1
x
is not uniformly continuous on this interval.
(b)
Suppose
f
:
S
⊆
X
→
R
is a continuous function, and
S
is compact. Suppose for the
sake of contradiction that there exists some
>
0 such that for all
δ >
0 there exist
x, x
∈
S
such that

x

x

< δ
but

f
(
x
)

f
(
x
)

>
. Then, for each
n
∈
N
, pick
x
n
, y
n
∈
S
such that

x
n

y
n

<
1
2
n
but

f
(
x
n
)

f
(
y
n
)

>
. By compactness we know there exists a subsequence
of the natural numbers
a
1
< a
2
<
· · ·
such that
x
a
1
, . . .
converges to some
z
∈
S
. Further,
because the sequence
y
a
1
, . . .
is fully contained in
S
, we know there exists a subsequence of
the natural numbers
b
1
< b
2
<
· · ·
such that
y
a
b
1
, y
a
b
2
, . . .
converges to some point
w
∈
S
.
Note that
x
a
b
1
, x
a
b
2
, . . .
still converges to
z
∈
S
. For simplicity define
z
i
=
x
a
b
i
and
w
i
=
y
a
b
i
.
Then for any
>
0 we may find an
i
such that

z

w

<

z

z
i

+

z
i

w
i

+

w
i

w

<
3
/
3 =
,
so therefore
z
=
w
. Becasue
f
is continuous at
z
, we may find a
δ
such that if

ξ

z

< δ
then

f
(
ξ
)

f
(
z
)

<
/
2. However then we may find an
n
such that

z
n

z

< δ
and

w

w
n

< δ
.
1
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Then because
z
=
w
we have that

f
(
z
n
)

f
(
w
n
)
 ≤ 
f
(
z
n
)

f
(
z
)

+

f
(
w
)

f
(
w
n
)

<
2
/
2 =
,
which contradicts our choice of
z
n
and
w
n
, and thus shows
f
is uniformly continuous on
S
.
2.
(a)
Theorem 1
(The Weierstrass Approximation Theorem)
.
If
f
is a continuous real valued
function on the closed interval
[
a, b
]
, then for any
>
0
, there exists a polynomial
P
(
x
) =
∑
n
i
=0
a
i
x
i
for some
n
∈
N
and
a
i
∈
R
such that

f
(
x
)

P
(
x
)

<
for all
x
∈
[
a, b
]
.
The function is continuous on a closed and bounded interval in
R
which we know to be
compact, and thus the function is automatically uniformly continuous. Thus, we need not
require
f
to be uniformly continuous, but as a side effect of the other problem assumptions
f
must end up being uniformly continuous on the interval.
(b)
If the statement is true when
a
= 0 and
b
= 1, then a simple change of variable proves
it for all
a
≤
b
, so we may assume we are working on the interval [0
,
1].
Let
ν
and
n
be
integers such that 0
≤
ν
≤
n
. Then we define the Bernstein polynomial
b
ν,n
(
x
) =
n
ν
x
ν
(1

x
)
n

ν
.
Then for a function
f
: [0
,
1]
→
R
and positive integer
n
define
B
n,f
(
x
) :=
n
ν
=0
f
(
ν/n
)
b
ν,n
(
x
)
.
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 Fall '09
 Math, Topology, Metric space, Limit of a sequence, Cauchy sequence, Cauchy

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