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In full:

Show that if $X^*$ is separable then any bounded sequence $(x_n) \in X$ has a subsequence $x_{n_k}$ such that $\lim_{k \to \infty}f(x_{n_k})$ converges for every $f \in X^*$

What I have so far:

let $\{f_i\}$ be a countable dense subset of $X^*$, then we can use a diagonal argument to show that $\exists x_{n_k}$ such that $$\lim_{k \to \infty}f_i(x_{n_k})$$ exists for every $f_i$ in our dense set.

Now given any $f \in X^*$, pick $f_{m_j}$ such that $||f - f_{m_j}|| < 1/j$, then

$$\lim_{k \to \infty}f(x_{n_k}) = \underbrace{\lim_{k \to \infty}(f(x_{n_k}) - f_k(x_{n_k}))}_{0} + \lim_{k \to \infty}f_k(x_{n_k})$$

but I can't see if the right hand side converges.

Christian Fieldhouse
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  • Can You specify what $X$ is? I suppose it is a normed vector space over the reals or complexes. – m.s May 10 '18 at 15:09

1 Answers1

1

Let $K$ be a constant such that $||x_{n}|| \leq K$ for all $n$. Let $f \in X^{\ast}$ be arbitrary. Let us try to show that the sequence of complex numbers $(f(x_{n_k}))_{k \geq 1}$ is a Cauchy sequence. Let $\varepsilon > 0$ be given. Choose $i$ such that $||f-f_i|| < \varepsilon$. Since $(f_i(x_{n_k}))_{k \geq 1}$ is Cauchy, there is $k_0$ such that $|f_i(x_{n_k}) - f_i(x_{n_{\ell}})| < \varepsilon$ fro all $\ell, k \geq k_0$. Hence

$$ |f(x_{n_k}) - f(x_{n_{\ell}})| \leq ||f_i-f||\, ||x_{n_k}|| + \varepsilon + ||f_i-f||\, ||x_{n_\ell}|| \leq K \varepsilon + \varepsilon + K \varepsilon $$

for all $\ell, k \geq k_0$

m.s
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