MATH 413 FINAL EXAM
Math 413 final exam, 13 May 2008. The exam starts at 9:00 am and you have 150 minutes.
No textbooks or calculators may be used during the exam.
This exam is printed on both
sides of the paper.
Good luck!
(1)
(20 marks)
Let
X
= (0
,
1]
⊂
R
.
State whether each of the following statements
about
X
is true or false, giving a brief reason for each answer.
(a)
X
is bounded.
True. For all
x
∈
X
,

x
 ≤
1
.
(b)
X
can be written as a countable union of open sets.
False. Any union of open
sets is open, but
X
is not open.
(c)
X
is compact.
False.
X
is not closed (it does not contain the cluster point
0
),
so by the HeineBorel Theorem cannot be compact.
(d) There is a point
x
0
∈
X
at which the function
f
(
x
) = log(
x
) +
x
5

8
x
4

3
achieves its supremum on
X
(that is,
f
(
x
0
) = sup
{
f
(
x
) :
x
∈
X
}
).
True. Since
log(
x
)
→ ∞
as
x
→
0
, we have also that
f
(
x
)
→ ∞
as
x
→
0
. So there exists
n
∈
N
such that
f
(
x
)
< f
(1)
if
x <
1
n
. Therefore,
sup
x
∈
X
f
(
x
) = sup
x
∈
[
1
n
,
1]
f
(
x
)
and this is attained by
f
since
[
1
n
,
1]
is a compact set.
(2)
(20 marks)
Let
A
⊂
R
.
Recall that a function
f
:
A
→
R
is said to satisfy a
Lipschitz condition on
A
if there is some
M
∈
R
such that

f
(
x
)

f
(
y
)
 ≤
M

x

y

for all
x, y
∈
A
.
(a) Let
n
∈
N
. Show that the function
f
n
: [0
,
1]
→
R
defined by
f
n
(
x
) =
q
x
+
1
n
satisfies a Lipschitz condition on [0
,
1].
(Hint: you may wish to use the fact that for all
a, b >
0, (
√
a
+
√
b
)(
√
a

√
b
) =
a

b
.)
We need to show that there exists
M
such that

f
n
(
x
)

f
n
(
y
)
 ≤
M

x

y

1
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for all
x, y
∈
[0
,
1]
.
Using the hint, if
x
≥
y
, we have

f
n
(
x
)

f
n
(
y
)

=

q
x
+
1
n

q
y
+
1
n

=
(
x
+
1
n
)

(
y
+
1
n
)
√
x
+
1
n
+
√
y
+
1
n
. Since
x, y
≥
0
, we get

f
n
(
x
)

f
n
(
y
)
 ≤
(
x

y
)
/
(2
/
√
n
) =
M

x

y

where
M
=
√
n/
2
.
(b) Show that the sequence of functions
{
f
n
}
converges uniformly on [0
,
1] to the
function
f
(
x
) =
√
x
.
We need to show that for all
ε >
0
there exists
N
∈
N
such that if
n > N
then

f
n
(
x
)

f
(
x
)

< ε
for all
x
∈
[0
,
1]
. We have

f
n
(
x
)

f
(
x
)

=
q
x
+
1
n

√
x
=
x
+
1
n

x
√
x
+
1
n
+
√
x
≤
1
n
/
1
√
n
=
1
√
n
. So given
ε >
0
, we choose
N
with
1
/
√
N < ε
, and
then if
n > N
and
x
∈
[0
,
1]
, we have

f
n
(
x
)

f
(
x
)
 ≤
1
√
n
<
1
√
N
< ε
as required.
(c) Show that
f
(
x
) =
√
x
does
not
satisfy a Lipschitz condition on [0
,
1].
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 Spring '08
 FLANAGAN
 Physics, Power Series, Continuous function, Metric space, Riemann

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