Basic Analysis – Gently Done Topological Vector Spaces

6: Topological Vector Spaces 61
Remark 6.22 In a normed space, any neighbourhood of 0 contains a ball {x ∈ X :
�x� < r} for some r > 0. Thus property (iii) of the theorem is equivalent to the
property that for some K > 0, there is r > 0 such that
|Λ(x)| < K for x ∈ X with �x� < r.
If x �= 0, then y = rx/2�x� has norm less than r so that |Λ(y)| < K, i.e.,
|Λ(x)| < 2K 2K �x�. Setting = M and allowing for x = 0, we see that (iii) is
r r
equivalent to
|Λ(x)| ≤ M�x�, x ∈ X.
This is the property that Λ be a bounded linear functional—in which case we have
�Λ� ≤ M.
Theorem 6.23 Every finite dimensional Hausdorff topological vector space
(X,T) over K with dimension n is linearly homeomorphic to K n . In particular,
any two norms on a finite dimensional vector space are equivalent.
Proof Suppose thatdimX = n andthatx 1 ,...,x n isabasisfor X. Let e 1 ,...,e n
be the standard basis for K n , i.e., e i = (δ ij ) n j=1 ∈ Kn , and let ϕ : K n → X be the
linear map determined by the assignment ϕ(e i ) = x i , so that
ϕ((t 1 ,...,t n )) = ϕ(t 1 e 1 +···+t n e n ) = t 1 x 1 +···+t n x n .
It is clear that ϕ : K n → X is a (linear) isomorphism between K n and X. Furthermore,
ϕ is continuous, by Corollary 6.9. We shall show that the linear map
ϕ −1 : X → K n is continuous.
Let K be the surface of the unit ball in K n ,
K = {ζ ∈ K n : �ζ� = 1}.
Then K is compact in K n and so ϕ(K) is compact in X, because ϕ is continuous.
Since X is Hausdorff, ϕ(K) is closed in X. Now, 0 /∈ K so that 0 = ϕ(0) is not an
element of ϕ(K). In other words, 0 is an element of the open set X \ϕ(K). By
Proposition 6.18, there is a balanced neighbourhood W of 0 with
W ⊆ X \ϕ(K).
In particular, W ∩ ϕ(K) = ∅. Since ϕ −1 is linear, it follows that ϕ −1 (W) is a
balanced set in K n such that ϕ −1 (W)∩K = ∅. We claim that �ζ� < 1 for every
ζ ∈ ϕ −1 (W). Indeed, if ζ ∈ ϕ −1 (W) and if �ζ� ≥ 1, then �ζ� −1 ζ ∈ K and also
belongs to the balanced set ϕ −1 (W). This contradicts ϕ −1 (W)∩K = ∅ and we
conclude that every ζ ∈ ϕ −1 (W) satisfies �ζ� < 1, as claimed.