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 A nice example (passed to me originally made up by Terry Tao) showing that the second condition (the supremum over $X$ gives the norm) does not imply dense is the following: consider $l^1$ and $(l^1)^\*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



Is a subspace with a certain property dense in the dual of a vector space?
The answer is negative. Since the linear span of the Dirac masses is not a dense subspace of the dual of $C[0,1]$.


