## Is a subspace with a certain property dense in the dual of a vector space?

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 A nice example (passed to me originally made up by Terry Tao) showing that the second condition (the supremum over $X$ gives the norm) does not imply dense is the following: consider $l^1$ and `$(l^1)^*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

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You must assume that $X$ is a subspace of $V$, else the result is trivially wrong (take $X$ to be the unit sphere of a reflexive space $V$). – Harald Hanche-Olsen May 5 2010 at 3:52
A subset $B$ of the unit sphere of the dual of a Banach space $V$ that has the property that every vector in $V$ achieves its norm at some functional in $B$ is called a boundary for $V$. Recently Hermann Pfitzner arXiv:0807.2810 solved Godefroy's boundary problem in the affirmative. The boundary problem was whether a subset of $V$ is weakly compact if it is compact in the topology of pointwise convergence on some boundary for $V$.  This was a problem of some note, with earlier related results by, including others, Grothendieck and Bourgain-Talagrand. – Bill Johnson May 6 2010 at 7:36

## 2 Answers

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

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Excellent! One of those answers which, when you see it, makes you slap your forehead and say “why didn't I think of that?” – Harald Hanche-Olsen May 5 2010 at 14:11
Thanks! This is a great simple example. – Alden Walker May 5 2010 at 17:11

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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.)

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