18.100B/C: Fall 2008
Solutions to Homework 12
1.(a)
(Rudin page 165 problem 2) Let
f
and
g
be the functions such that
f
n
→
f
and
g
n
→
g
.
Choose
ε >
0. Then there is some
N
1
∈
N
such that
n
≥
N
1
implies

f
n
(
x
)

f
(
x
)
 ≤
ε/
2 for
all
x
and some
N
2
∈
N
such that
n
≥
N
2
implies

g
n
(
x
)

g
(
x
)
 ≤
ε/
2 for all
x
. Then with
N
= max
{
N
1
, N
2
}
we have
n
≥
N
implies

(
f
n
+
g
n
)(
x
)

(
f
+
g
)(
x
)
 ≤ 
f
n
(
x
)

f
(
x
)

+

g
n
(
x
)

g
(
x
)

≤
ε
2
+
ε
2
=
ε
for all
x
, so (
f
n
+
g
n
) converges uniformly to
f
+
g
.
If
f
n
and
g
n
are bounded for all
n
then we have by an argument similar to problem 4b from the
last problem set that there is some
M
∈
R
such that

f
(
x
)
 ≤
M
and

g
(
x
)
 ≤
M
and

f
n
(
x
)
 ≤
M
and

g
n
(
x
)
 ≤
M
for all
x
. Fix
ε >
0, then we can choose
N
∈
N
such that
n
≥
N
implies

g
n

g
 ≤
ε
/
2
M
and

f
n

f
 ≤
ε/
2
M
for all
x
. Then

f
n
g
n

fg

=

(
f
n

f
)
g
n
+
f
(
g
n

g
)

≤ 
f
n

f
 · 
g
n

+

f
 · 
g
n

g

≤
ε
2
M
+
ε
2
M
=
ε
for all
x
, so (
f
n
g
n
) converges uniformly to
fg
.
(b)
(Rudin page 165 problem 3) Let
f
n
(
x
) =
x
and
g
n
(
x
) = 1 +
1
n
for all
n
∈
N
, x
∈
R
.
These converge uniformly to
f
(
x
) =
x
(since
f
n
=
f
is actually independent of
n
) resp.
g
n
(
x
) = 1
(since
g
n
converges pointwise and does not actually depend on
x
). Also, we clearly have pointwise
convergence
f
n
·
g
n
→
x
. Now ﬁx any
ε >
0 and any
n
, and choose some
x > nε
. Then
f
n
(
x
)
g
n
(
x
) =
x
·
(1 + 1
/n
)
> f
(
x
)
g
(
x
) +
ε
, so convergence is not uniform.
2.
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 Fall '10
 Prof.KatrinWehrheim
 Calculus, Compact space, FN, Uniform convergence, Pointwise convergence, Modes of convergence

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