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
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 Spring '09
 LARSON
 Vector Space, Empty set, Topological vector space, greatest norm

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