MAT 337 Problem Set 5 Sample Solutions
Ian Zwiers  April 14, 2009
This document is for demonstration purposes; it is not a marking key.
7.1(A)
Show that
k
(
x,y,z
)
k
=

x

+ 2
p
y
2
+
z
2
is a norm on
R
3
. Sketch the unit ball.
positivite deﬁnite
k
(
x,y,z
)
k
= 0
⇔ 
x

= 0 and 2
p
y
2
+
z
2
= 0
⇔ 
x

= 0
, y
2
= 0 and
z
2
= 0
.
That is,
k
(
x,y,z
)
k
= 0
⇔
x
=
y
=
z
= 0
⇔
(
x,y,z
) = 0.
homogeneous
Let
α
∈
R
. Then,
k
α
(
x,y,z
)
k
=

αx

+
q
(
αy
)
2
+ (
αz
)
2
=

α

±

x

+
p
y
2
+
z
2
²
=

α
k
(
x,y,z
)
k
.
triangle inequality
Note that,
(
y
+
b
)
2
+ (
z
+
c
)
2
=
(
y
2
+ 2
y b
+
b
2
)
+
(
z
2
+ 2
z c
+
c
2
)
=
(
y
2
+
z
2
)
+ 2[
y
z
]
·
[
b
c
] +
(
b
2
+
c
2
)
≤
(
y
2
+
z
2
)
+ 2
p
y
2
+
z
2
p
b
2
+
c
2
+
(
b
2
+
c
2
)
=
±
p
y
2
+
z
2
+
p
b
2
+
c
2
²
2
,
where the inequality is due to the CauchySchwarz inequality, popularly known as the
regular
R
2
dot product formula. Then, it is straightforward,
k
(
x,y,z
) + (
a,b,c
)
k
=

x
+
a

+ 2
p
(
y
+
b
)
2
+ (
z
+
c
)
2
≤
(

x

+

a

) +
±
2
p
y
2
+
z
2
+ 2
p
b
2
+
c
2
²
=
k
(
x,y,z
)
k
+
k
(
a,b,c
)
k
.
The unit ball is two open cones, each with height one and with base radius
1
2
, where the base
of one cone has been glued to the base of the other.
7.1(C)
For
f
∈ C
1
[
a,b
]
, deﬁne
ρ
(
f
) =
k
f
0
k
∞
. Show that
ρ
is nonnegative, homogeneous, and
satisﬁes the triangle inequality. Why is it not a norm?
nonnegative
ρ
is nonnegative by deﬁnition of
k·k
∞
. For
f
∈ C
2
[
a,b
], derivative
f
0
is continuous
on a compact set, and must achieve its extrema. This proves
ρ
(
f
)
<
+
∞
.
1