Question

Suppose we have a normed vector space $V$ and its dual $V^*$, and suppose that $X \subseteq V^*$ has the property that for every $v \in V$, there is some $\phi \in X$ with $\Vert \phi \Vert = 1$ such that $\phi(v) = \Vert v \Vert$. Is $X$ dense in $V^*$ (in the operator norm)? Note that this is a stronger property than $\Vert v \Vert = \sup_{\phi\in X} \frac{\phi(v)}{\Vert \phi \Vert}$, since we are assuming that the supremum is realized.

I think the answer is probably "no." A nice example (passed to me originally made up by Terry Tao) showing that the second condition (the supremum over $X$ gives the norm) does not imply dense is the following: consider $l^1$ and $(l^1)^* = l^\infty$. Then the space of eventually zero sequences in $l^\infty$ is sufficient for the norm: given $f\in l^1$, let $\phi_n$ be a truncation of the sign function of $f$ to the first $n$ indices. Then $\lim_{n\to \infty} \phi_n(f) = \Vert f \Vert$. However (for $f$ with infinite support), there is no finite sequence $\phi$ with $\phi(f) = \Vert f \Vert$.

Answer 1

The answer is negative. Since the linear span of the Dirac masses is not a dense subspace of the dual of $C[0,1]$.

Answer 2

Far from a complete answer, but the answer is yes if $V=\ell^1$: For then $X$ must contain every sequence $x\in\ell^\infty$ with $|x_n|=1$ for all $n$ (consider $v\in\ell^1$ given by $v_n=\bar x_n/n^2$), and the space of linear combinations of such sequences is dense in $\ell^\infty$. To see the latter, merely note that any complex number $z$ with $|z|\le2$ can be written as $x+y$ with $|x|=|y|=1$. (Over the reals, you need to work a tiny bit harder.)