Basic Analysis – Gently Done Topological Vector Spaces
Basic Analysis – Gently Done Topological Vector Spaces
Basic Analysis – Gently Done Topological Vector Spaces
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9: The Dual Space of a Normed Space 107<br />
form on X is norm continuous. Nonetheless, the weak and norm topologies on X<br />
are not (always) equal, as we shall now show.<br />
Proposition 9.16 Every weak neighbourhood of 0 in an infinite dimensional<br />
normed space contains a one dimensional subspace.<br />
Proof Let U be any weak neighbourhood of 0 in the infinite dimensional normed<br />
space X. Then there is r > 0 and elements ℓ 1 ,...,ℓ n in X ∗ , the dual of X, such<br />
that<br />
V(0,|ℓ 1 |,...,|ℓ n |;r) ⊆ U.<br />
Since X is infinite dimensional, so is X ∗ . (If X ∗ were finite dimensional, then<br />
X ∗∗ would also be finite dimensional. However, this is impossible because we<br />
have seen that X is linearly isometrically isomorphic to a subspace of X ∗∗ .) By<br />
Proposition 5.16, there is z ∈ X such that z �= 0 and ℓ i (z) = 0 for all 1 ≤ i ≤ n.<br />
Thus tz ∈ U for all t ∈ K.<br />
Theorem 9.17 The weak topology on an infinite dimensional normed space X is<br />
strictly weaker than the norm topology.<br />
Proof Every element of X ∗ is continuous when X is equipped with the norm<br />
topology. The weak topology is the weakest topology on X with this property, so<br />
it is immediately clear that the weak topology is weaker than the norm topology.<br />
Now suppose that X is infinite-dimensional. We shall exhibit a set which is<br />
open with respect to the norm topologybut not with respect to the weak topology.<br />
Consider the ‘open’ unit ball<br />
G = {x ∈ X : �x� < 1}.<br />
Then clearly G is open with respect to the norm topology on X. We claim that<br />
G is not weakly open. If, on the contrary, G were weakly open, then, since 0 ∈ G,<br />
there would be some z ∈ X with z �= 0 such that tz ∈ G for all t ∈ K, by<br />
Proposition 9.16. Clearly this is not possible for values of t with |t| ≥ 1/�z�. We<br />
conclude that G is not weakly open and the result follows.<br />
Remark 9.18 We have shown that no weak neighbourhood of 0 can be norm<br />
bounded. Since boundedness is preserved under translations, we conclude that no<br />
non-empty weakly open set is norm bounded.<br />
We now turn to a discussion of a topology on X ∗ , the dual of the normed space<br />
X. The idea is to consider X as a family of maps : X ∗ → C given by x : ℓ ↦→ ℓ(x),<br />
for x ∈ X and ℓ ∈ X ∗ —this is the map ϕ x defined earlier.