Basic Analysis – Gently Done Topological Vector Spaces

86 Basic Analysis
5. The Banach space ℓ 2 is the linear space of complex sequences, x = (x n ),
satisfying
In fact, ℓ 2 has the inner product
∞�
�x�2 = (
n=1
〈x,y〉 =
and so is a (complex) Hilbert space.
|x n | 2 ) 1/2 < ∞.
∞�
n=1
x n y n
We have seen, in Theorem 6.23, that there is only one Hausdorff vector space
topology on a finite dimensional topological vector space and so, in particular,
all norms on such a space are equivalent. We can prove this last part directly as
follows.
Theorem 8.3 Any two norms on a finite dimensional vector space over K are
equivalent.
Proof Suppose that X is a finite dimensional normed vector space over K with
basis e 1 ,... ,e n . Define a map T : K n → R by
T(t 1 ,...,t n ) = �t 1 e 1 +···+t n e n �.
The inequality
�
��t 1 e 1 +···+t n e n �−�s 1 e 1 +···+s n e n � � � ≤ �(t 1 −s 1 )e 1 +···+(t n −s n )e n �
shows that T is continuous on K n . Now, T(t 1 ,... ,t n ) = 0 only if every t i = 0.
In particular, T does not vanish on the unit sphere, {z : �z� = 1}, in K n . By
compactness, T attains its bounds on the unit sphere and is therefore strictly
positive on this sphere. Hence there is m > 0 and M > 0 such that
m
� �n
k=1 |t k |2 ≤ �t 1 e 1 +···+t n e n � ≤ M
� �n
k=1 |t k |2 .
We have shown that the norm �·� on X is equivalent to the usual Euclidean norm
on X determined by any particular basis. Consequently, any finite dimensional
linear space can be given a norm, and, moreover, all norms on a finite dimensional
linear space X are equivalent: for any pair of norms � · � 1 and � · � 2 there are
positive constants µ,µ ′ such that
for every x ∈ X.
µ�x� 1 ≤ �x� 2 ≤ µ ′ �x� 1
Department of Mathematics King’s College, London