This preview shows pages 1–2. Sign up to view the full content.
Analysis 2
Homework 01
1. Show that the following subset
K
of
‘
2
is convex, closed and bounded, but
does
not
contain an element of
greatest
norm:
K
=
{
z
= (
z
n
)
∞
n
=1
:
∞
X
n
=1
(1 +
1
n
)
2

z
n

2
6
1
}
.
Solution
1. A ‘little’ thought suggests considering the form on
‘
2
deﬁned by
[
x

y
] =
∞
X
n
=1
(1 +
1
n
)
2
x
n
y
n
.
This form is plainly conjugated when its variables are switched; it is also
plainly conjugatelinear in the ﬁrst and linear in the second. Further, [
z

z
]
>
0 and = 0 if and only if
z
= 0: indeed,
k
z
k
2
6
[
z

z
]
6
4
k
z
k
2
.
Thus, the form [
••
] is an inner product, whose associated IP norm
 • 
(say) satisﬁes
k
z
k
6

z

6
2
k
z
k
.
(
*
)
Notice that
K
(as deﬁned in the question) is precisely the ‘closed unit ball’
for
 • 
. This implies at once that
K
is convex. The right inequality at
(*) shows that the
•
closed set is
k•k
closed. The left inequality at (*)
shows that
K
is bounded, being contained in
B
‘
2
(for
k • k
of course).
So far,
K
is convex, closed and bounded, with sup
{k
z
k
:
z
∈
K
}
6
1.
In fact, this supremum is precisely 1: by direct calculation,

n
n
+1
e
n

= 1
and
k
n
n
+1
e
n
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.
 Spring '09
 LARSON

Click to edit the document details