Let $(X, \lVert \cdot \rVert)$ be a separable normed space. Can we always guarantee that there is a nonempty compact set $K \subseteq B\_X$, where $B\_X$ is a closed unit ball in $X$ such that:


$$\forall \Lambda \in S\_{X^\*} \quad \sup\_{k \in K} \Lambda(k) > 0,$$


where $S\_{X^\*}$ denotes the unit sphere in the dual space $X^\*$? If not, is it possible to do so, if instead of $S\_{X^\*}$ we would take some weak\* dense subset of $S\_{X^\*}$?


In the case when $\mathrm{dim}X < \infty$ the answer is yes, since we can just take a closed unit ball for $K$, however, I'm not sure whether this result could be generalized to any separable normed space.



"Large" compact sets in separable normed space
Let $S=\{x\_n:n\in\mathbb N\}$ be a dense subset and choose
$\varepsilon\_n>0$ such that $y\_n=\varepsilon\_n x\_n\to 0$ (e.g., $\varepsilon\_n=1/n\|x\_n\|$). Then $K=\{ty\_n:n\in\mathbb N, |t|=1\}\cup\{0\}$ is compact and for every every $\Lambda\in X^\*\setminus \{0\}$ we have $\sup\{\Lambda(k):k\in K\}\ge |\Lambda(y\_n)|>0$ for some $n\in\mathbb N$ since otherwise $\Lambda$ would vanish on $S$ and hence on $X$.


