Basic Analysis – Gently Done Topological Vector Spaces

9: The Dual Space of a Normed Space 105
Remark 9.11 Exactly as above, we see that the map z ↦→ ψ z is a linear isometric
mapping of ℓ 1 into the dual of ℓ ∞ . Furthermore, if λ is any element of the dual
of ℓ ∞ , then, in particular, it defines a bounded linear functional on c 0 . Thus,
the restriction of λ to c 0 is of the form ψ z for some z ∈ ℓ 1 . It does not follow,
however, that λ has this form on the whole of ℓ ∞ . Indeed, c 0 is a closed linear
subspace of ℓ ∞ , and, for example, the element y = (y n ), where y n = 1 for all
n ∈ N, is an element of ℓ ∞ which is not an element of c 0 . Then we know, from
the Hahn-Banach theorem, that there is a bounded linear functional λ, say, such
that λ(x) = 0 for all x ∈ c 0 and such that λ(y) = 1. Thus, λ is an element of the
dual of ℓ ∞ which is clearly not determined by an element of ℓ 1 according to the
above correspondence.
Theorem 9.12 Suppose that X is a Banach space and that X ∗ is separable.
Then X is separable.
Proof Let {λ n : n = 1,2,...} be a countable dense subset of X ∗ . For each n ∈ N,
let x n ∈ X be such that �x n � = 1 and |λ n (x n )| ≥ 1
2 �λ n
�. Let S be the set of finite
linear combinations of the x n ’s with rational (real or complex, as appropriate)
coefficients. Then S is countable. We claim that S is dense in X. To see this,
suppose the contrary, that is, suppose that S is a proper closed linear subspace
of X. (Clearly S is a closed subspace.) By Corollary 7.30, there exists a non-zero
bounded linear functional Λ ∈ X∗ such that Λ vanishes on S. Since Λ ∈ X∗ and
{λn : n ∈ N} is dense in X∗ , there is some subsequence (λnk ) such that λ → Λ nk
as k → ∞, that is,
as n → ∞. However,
�Λ−λ � → 0 nk
�Λ−λ nk� ≥ |(Λ−λ nk )(xnk )|, since �x � = 1, nk
= |λnk (x )|, since Λ vanishes on S,
nk
≥ 1
2 �λ nk �
and so it follows that �λ nk � → 0, as k → ∞. But λ nk → Λ implies that �λ nk � →
�Λ� and therefore �Λ� = 0. This forces Λ = 0, which is a contradiction. We
conclude that S is dense in X and that, consequently, X is separable.