The function
g
(
x
) is continuous, hence by the Intermediate Value Theorem
there is a
a
∈
(0
,
1) such that
g
(
a
) = 3
/
5. By Rolle’s theorem there is an
x
∈
(
a,
2) such that
g
0
(
x
) = 0. Then
f
0
(
x
) = 1
/
5.
5
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Problem 6. 30 points.
Consider the sequence of functions
f
n
(
x
) =
x

x
n
defined on [0
,
1].
1. Show that
f
n
(
x
) converges pointwise to a function
f
(
x
) on [0
,
1], and
find a formula for
f
(
x
).
2. Does
f
n
(
x
) converge to
f
(
x
) uniformly on [0
,
1]? Justify.
3. Does
f
n
(
x
) converge to
f
(
x
) uniformly on [0
, a
] for
a <
1? Justify.
Solution
1. We have
f
(
x
) = lim
f
n
(
x
) =
x,
x <
1
0
,
x
= 1
2. The limit function
f
(
x
) is not continuous, hence the convergence is not
uniform.
3. On [0
, a
], the function
f
(
x
)

f
n
(
x
) =
x
n
increases from 0 to
a
n
. Hence
lim
sup
x
∈
[0
,
1
/
2]

f
n
(
x
)

f
(
x
)

= lim
a
n
= 0
,
so the convergence is uniform.
6
Problem 7.
30 points.
Suppose that
f
(
x
) is a continuous function on
[
a, b
], suppose that
f
(
x
)
≥
0 for all
x
∈
[
a, b
], and suppose that
Z
b
a
f
(
x
)
dx
= 0
.
Prove that
f
(
x
) = 0 for all
x
∈
[
a, b
]. (Hint: suppose that
f
(
c
)
6
= 0 for some
c
∈
[
a, b
], and draw a picture).
Proof.
Suppose that
f
(
c
)
>
0 for some
c
∈
[
a, b
].
Then for any
ε >
0
there exists a
δ >
0 such that

x

c

< δ, x
∈
[
a, b
] implies

f
(
x
)

f
(
c
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 Spring '08
 Staff
 Continuous function, Natural number, Ks, lim sn

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