18.100B/C: Fall 2008
Solutions to Homework 8
1. (a)
Dividing the given equation through by

x

y

gives
0
≤

f
(
x
)

f
(
y
)


x

y

≤
C
· 
x

y

α

1
.
As
y
→
x
, the righthand term in this inequality approaches 0, so the middle term must approach
0 as well and we have
lim
y
→
x

f
(
x
)

f
(
y
)


x

y

= 0
.
Since
f
(
x
)

f
(
y
)
x

y
=
±

f
(
x
)

f
(
y
)


x

y

for every
x
6
=
y
, this implies that
f
0
(
x
) exists for each
x
and that
f
0
(
x
) = 0 for all
x
. Then by Rudin 5.11(a),
f
(
x
) must be a constant function, as desired.
(b)
No. For example, for any
α
∈
(0
,
1], consider the function
f
α
(
x
) =

x

α
. This function
satisﬁes

f
α
(
x
)

f
α
(
y
)
 ≤ 
x

y

α
for all
x,y
∈
R
but is not diﬀerentiable at 0 for any
α
≤
1.
2.(a)
By considering the closed interval [
x,y
], it follows immediately from the Mean Value
Theorem that if

f
0
(
x
)
 ≤
M
for all
x
then

f
(
x
)

f
(
y
)
 ≤
M
·
x

y

for all
x,y
(i.e., functions with
bounded derivative are Lipschitz). We will show that Lipschitz functions take Cauchy sequences to
Cauchy sequences: suppose (
x
n
)
n
∈
N
is a Cauchy sequence. Then for any
ε >
0 we can ﬁnd
N
∈
N
such that
m,n
≥
N
implies

x
m

x
n

<
ε
M
. Then it follows immediately that for all
m,n
≥
N
we
have

f
(
x
m
)

f
(
x
n
)

< M
·
ε
M
=
ε
, so (
f
(
x
n
))
n
∈
N
is Cauchy. In particular,
(
f
(
1
n
)
)
n
∈
N
is Cauchy
and so converges in
R
, as needed.
(b)
As mentioned in part (a), we have by the Mean Value Theorem that if

f
0
(
x
)
 ≤
α
for all
x
then
f
is Lipschitz with constant
α
. If
α <
1 then by problem set 6, problem 7
f
has a unique
ﬁxed point, as desired.
(c)
No. Consider for example the function
f
(
x
) =
√
x
2
+ 1. We have
f
(
x
)
> x
for all
x
∈
R
, so
f
has no ﬁxed points, but
f
0
(
x
)
∈
(

1
,
1) for all
x
∈
R
. (This function is half of a hyperbola with
asymptotes
y
=
x
and
y
=

x
.)
3.
(Rudin page 114 problem 2) Suppose