MATH 425b ASSIGNMENT 2 SOLUTIONS
SPRING 2008
Prof. Alexander
Chapter 7
(8) Since

c
n
I
(
x

x
n
)
 ≤ 
c
n

and
∑

c
n

<
∞
, the series defining
f
converges uniformly by
the Weierstrass
M
test (Theorem 7.10.) Letting
f
n
(
t
) =
∑
n
k
=1
c
n
I
(
x

x
n
), this means that
f
n
→
f
uniformly. Let
x
be a point that is not one of the
x
n
’s; then each
f
n
is continuous
at
x
, that is, lim
t
→
x
f
n
(
t
) =
f
n
(
x
). Since
f
n
(
x
)
→
f
(
x
) as
n
→ ∞
, Theorem 7.11 (applied
with
A
n
=
f
n
(
x
)) says
lim
t
→
x
f
(
t
) = lim
n
→∞
f
n
(
x
) =
f
(
x
)
,
that is,
f
is continuous at
x
.
(16) Suppose
{
f
n
}
is equicontinuous on
K
, and
f
n
→
f
pointwise.
Then
{
f
n
(
x
)
}
, being
convergent, is bounded for each
x
, so
{
f
n
}
is pointwise bounded.
Suppose (to get a contradiction) that the convergence of
f
n
to
f
is not uniform:

f
n

f

∞
6→
0. This means there exists
>
0 and a subsequence
{
f
n
k
}
with

f
n
k

f

∞
>
for
all
k
. But as a set,
{
f
n
k
}
is pointwise bounded and equicontinuous (since
{
f
n
}
is), so
{
f
n
k
}
has a uniformly convergent subsequence by Theorem 7.25, say
f
m
j
→
g
uniformly for some
g
, where
{
f
m
j
} ⊂ {
f
n
k
}
. Since
f
m
j
→
f
pointwise, we must have
g
=
f
, so

f
m
j

f

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 Fall '07
 Alexander
 Math, Convergence, Mathematical analysis, Continuous function, Uniform convergence, Pointwise convergence, Karl Weierstrass

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