2
MATH 202A HOMEWORK 1 SOLUTIONS
(3)
There exists a constant
c
≥
1
such that for all
x
∈
X
,
1
c
x
1
≤
x
2
≤
c x
1
.
Proof.
Clearly (2) =
⇒
(1). We will prove that (1) =
⇒
(3). Since we are assuming
(1), we know that
∃
δ >
0 such that
∀
x
∈
X
, if
ρ
1
(0
, x
)
≤
δ
, then
ρ
2
(0
, x
)
≤
1. This
last implication converts to
x
1
≤
δ
=
⇒
x
2
≤
1
.
For any
x
∈
X
replace
x
by
δx
x
1
and apply the above implication to conclude that
δ x
2
x
1
≤
1, or
x
2
≤
1
δ
x
1
.
A similar argument shows that there exists
δ >
0 such that
δ
x
1
≤
x
2
.
Thus we can take
c
= max(
1
δ
,
1
δ
). We next prove (3) =
⇒
(2). We would like to show
that for
∀
>
0,
∃
δ >
0 such that
∀
x, y
∈
X
,
ρ
1
(
x, y
)
< δ
=
⇒
ρ
2
(
x, y
)
<
. This
implication equates to
x

y
1
≤
δ
=
⇒
x

y
2
<
. But since we are assuming
inequality in item (3), we have
x

y
2
≤
c x

y
1
≤
cδ
So we just need to take
δ
=
/c
.
Showing that for
∀
>
0,
∃
δ >
0 such that
∀
x, y
∈
X
,
ρ
2
(
x, y
)
< δ
=
⇒
ρ
1
(
x, y
)
<
is similar.
Let 1
≤
r
≤ ∞
. As in Problem 1,
(2)
x
∞
d
≤
x
r
d
Also,
x
r
r
d
=
d
j
=1

x
j

r
≤
max
1
≤
n
≤
d

x
n

r
d
j
=1
1 =
d x
r
∞
and thus
(3)
x
r
d
≤
d
1
/r
x
∞
Then (2) and (3) together show that
r
d
is uniformly equivalent to
∞
d
. Since equiva
lence of norms is transitive (in fact, it is an equivalence relation),
p
d
is equivalent to
q
d
for any 1
≤
p
=
q
≤ ∞
.
Problem 3
. By Lemma 1, it suffices to show that there does not exist a constant
c
such that for all
x
∈
X
,
(4)
x
1
≤
c x
2