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
=

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
,
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
=
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
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homogeneous
Let
α
∈
R
,
f
∈ C
1
[
a, b
]. Then,
ρ
(
αf
) =
(
αf
)
0
∞
=
k
α f
0
k
∞
=
sup
x
∈
[
a,b
]

α f
0
(
x
)

=

α

sup
x
∈
[
a,b
]

f
0
(
x
)

=

α
 k
f
0
k
∞
=

α

ρ
(
f
)
.
triangle inequality
Let
f, g
∈ C
1
[
a, b
]. Then,
ρ
(
f
+
g
) =
sup
x
∈
[
a,b
]
(
f
+
g
)
0
(
x
)
=
sup
x
∈
[
a,b
]

f
0
(
x
) +
g
0
(
x
)

≤
sup
x
∈
[
a,b
]

f
0
(
x
)

+

g
0
(
x
)

≤
sup
x
∈
[
a,b
]

f
0
(
x
)

+
sup
z
∈
[
a,b
]

g
0
(
z
)

=
ρ
(
f
) +
ρ
(
g
)
Note that the constant functions
y
= 1 and
y
= 2 both have
ρ
= 0. Hence,
ρ
is not a norm
because it is not positive definite.
Please Note:
This counter example is essentially the only one. If you “anchor” the functions,
you can make
ρ
a norm. For example: let
X
=
f
∈ C
2
[
a, b
] :
f
(
a
) = 0
,
then in fact
ρ
is a norm on
X
.
7.2(D)
Show that if
A
is an arbitrary subset of a normed space
V
and
U
is an open subset, then
A
+
U
=
{
a
+
u
:
a
∈
A, u
∈
U
}
is open.
Let
x
∈
A
+
U
, then
∃
a
∈
A
and
u
∈
U
such that
x
=
a
+
u
.
Since
U
is open,
∃
δ >
0 such that
B
δ
(
u
)
⊂
U
. Consider arbitrary
y
∈
B
δ
(
x
); note then that,
k
(
y

a
)

u
k
=
k
y

(
a
+
u
)
k
=
k
y

x
k
< δ,
hence
y

a
∈
B
δ
(
u
). That is,
y

a
∈
U
. Since
a
∈
A
, this proves
y
∈
A
+
U
. Since
y
was
arbitrary in
B
δ
(
x
), this proves and hence
B
δ
(
x
)
⊂
A
+
U
.
Then, since
x
was arbitrary in
A
+
U
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 Spring '09
 Trigraph, Norm, Metric space, CN, Compact space, Uniform convergence

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