If A Normed Linear Space X Is Separable Then X Is Separable

If a Normed Linear Space X* is Separable then X is Separable

Theorem 1: Let $X$ be a normed linear space. If the topological dual $X^*$ is separable then $X$ is separable.
  • Proof: Let $X^*$ be separable. Consider the unit sphere $S_*$ in $X^*$:
(1)
\begin{align} \quad S_* = \{ \varphi \in X^* : \| \varphi \| = 1 \} \end{align}
  • Then $S_*$ is separable so let $\{ \varphi_n : n \in \mathbb{N} \}$ be a countable and dense subset of $S_*$. For each $n \in \mathbb{N}$ choose $x_n \in \mathbb{N}$ with $\| x_n \| = 1$ and such that:
(2)
\begin{align} \quad | \varphi_n(x_n) | > \frac{1}{2} \end{align}
  • Let $D$ be the span of $x_1, x_2, ...$:
(3)
\begin{align} \quad D = \overline{\mathrm{span} (x_1, x_2, ...)} \end{align}
  • Consider the following set:
(4)
\begin{align} \quad \bigcup_{n=1}^{\infty} \left \{ \sum_{j=1}^{n} (a_j + ib_j)x_j : a_j, b_j \in \mathbb{Q} \right \} \end{align}
  • Then this set is a countable and dense subset of $D$ and so $D$ is separable. We aim to show that $D = X$.
  • Let $\varphi \in S_*$. Since $\{ \varphi_n : n \in \mathbb{N} \}$ is dense there exists an $n \in \mathbb{N}$ such that:
(5)
\begin{align} \quad \| \varphi - \varphi_n \| < \frac{1}{2} \end{align}
  • Therefore we have that:
(6)
\begin{align} \quad | \varphi (x_n) | \geq | \varphi_n(x_n) | - | \varphi(x_n) - \varphi_n(x_n)|> \frac{1}{2} - \| \varphi - \varphi_n \| \| x_n \| = \frac{1}{2} - \| \varphi - \varphi_n \| > 0 \end{align}
  • So $D = X$. Hence $X$ is also separable. $\blacksquare$
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