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  A nice example (passed to me originally made up by Terry Tao) showing that the second condition (the supremum over $X$ gives the norm) does not imply dense is the following: consider $l^1$ and $(l^1)^* =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 $\phi(f) = \Vert f \Vert$.