Solutions for HWK # 2
Partially based on the HWKs of my former student Russell Yanofsky
September 18, 2010
Problem A
Prove
k
x
k
2
:=
q
∑
d
i
=1
x
2
i
is a norm in
R
d
.
We have to prove that
k
x
k
2
satis es the de nition of a norm.
1)
k
x
k
2
≥
0
∀
x
∈
R
d
because
x
2
i
≥
0
∀
x
i
∈
R
, and summing nonnegative reals gives
nonnegative reals.
2) If
x
=
0
then
k
x
k
2
= 0
because
q
∑
d
i
=1
0
2
= 0
.
If
k
x
k
2
= 0
then
x
=
0
. To show this, assume
∃
x
i
s.t.
x
i
6
= 0
. Then,
q
∑
d
i
=1
x
2
i
6
= 0
,
which contradicts
k
x
k
2
= 0
. Therefore, if
k
x
k
2
= 0
then
x
i
= 0
∀
x
i
, and
x
=
0
.
3)
k
α
x
k
2
=

α
k
x
k
2
because
k
α
x
k
2
=
q
∑
d
i
=1
(
αx
i
)
2
=
q
α
2
∑
d
i
=1
(
x
i
)
2
=
√
α
2
q
∑
d
i
=1
x
2
i
=

α
k
x
k
2
.
4)
k
x
+
y
k
2
≤ k
x
k
2
+
k
y
k
2
∀
x
,
y
∈
R
d
because, by the Cauchy inequality
ﬂ
ﬂ
ﬂ
ﬂ
ﬂ
d
X
i
=1
x
i
y
i
ﬂ
ﬂ
ﬂ
ﬂ
ﬂ
≤
v
u
u
t
d
X
i
=1
x
2
i
v
u
u
t
d
X
i
=1
y
2
i
,
we have

x
+
y

2
=
v
u
u
t
d
X
i
=1
(
x
i
+
y
i
)
2
=
v
u
u
t
d
X
i
=1
x
2
i
+
d
X
i
=1
y
2
i
+ 2
d
X
i
=1
x
i
y
i
≤
v
u
u
u
t
d
X
i
=1
x
2
i
+
d
X
i
=1
y
2
i
+ 2
v
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 '10
 staff
 Math, Inequalities, Yi, i=1, maxi xi

Click to edit the document details