When is a subspace of $V^*$ dense, where $V$ is a Banach space?

Question

Let $V$ be a Banach space. Suppose $\alpha_i$ is a collection of bounded linear functionals $V\to \mathbb{C}$ such that:

$$ \bigcap_i \ker(\alpha_i) = 0. $$

Does this imply the $\alpha_i$ span a dense subset of the dual $V'$?

I'm motivated by the following special case: $V = C([0, 1])$, the space of all continuous $[0, 1] \to \mathbb{C}$. Now for $x\in [0, 1]$, the "evaluation function" $e_x : V \to \mathbb{C}$ simply takes $v \mapsto v(x)$. I'd like to prove that the set of evaluation functions spans a dense subset of $V'$ (the usual way is to use Riesz representation theorem to describe $V'$ as the space of complex measures on $[0, 1]$, but I'm trying not to use that). Note that in our case, we clearly do have $\bigcap_i \ker \alpha_i = 0$.

Answer 1

This is not in fact true in the case that you are interested in. For instance, consider the functional $I:C([0,1])\to\mathbb{C}$ given by $I(f)=\int f d\mu$, where $\mu$ is Lebesgue measure. It is easy to see that any finite linear combination of evaluation functionals has distance $\geq 1$ from $I$ (because you can find an element of $C([0,1])$ of norm $1$ which vanishes at the finitely many points you're evaluating at but nevertheless has integral $1-\epsilon$).

Answer 2

Not in general: Take $V = \ell^1$, and for each $i\in \mathbb{N}$, let $\alpha_i$ denote the "evaluation" map $$ \alpha_i ((x_n)) := x_i $$ Then clearly $$ \bigcap_i \ker(\alpha_i) = 0 $$ However, consider the dual space pairing $$ \ell^{\infty} \to (\ell^1)^{\ast} $$ then the $\alpha_i$ correspond to the elements $$ e_i = (0,0,0, \ldots, 0, 1,0,\ldots) \in \ell^{\infty} $$ In particular, $$ \text{span}(\alpha_i) \subset c_0 \neq \ell^{\infty} $$ However, what you say is true if $V$ is reflexive : Since $$ \bigcap_i \ker(\alpha_i) = 0 $$ it follows from reflexivity that for any $T \in V^{\ast\ast}$, $$ T(\alpha_i) = 0\quad\forall i \Rightarrow T \equiv 0 $$ It is now a consequence of Hahn-Banach that this implies $$ \overline{\text{span}(\alpha_i)} = V^{\ast} $$ (else you could construct a non-zero linear functional on $V^{\ast}$ that annihilates all the $\alpha_i$)