Show that in the space $l_1$ , $0$ is not in $conv^* \{e_i\}$ where $e_i$ 's are the standard basis vectors of $l_1$.
I was trying to solve this by contradiction but I failed..Need some help.
2026-04-13 12:02:23.1776081743
In the space $l_1$,$0$ is not in $conv^* \{e_i\}$
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I assume by $C:= conv^{\ast}(e_i)$ you mean the closed convex hull of the $\{e_i\}$. Let $\varphi \in (\ell^1)^{\ast}$ denote the linear functional $$ \varphi((x_n)) := \sum_{n=1}^{\infty} x_n $$ Then for any $z \in \text{conv}(\{e_i\})$, we have $$ \varphi(z) = 1 $$ Since $C$ is norm closed, it follows that $\varphi \equiv 1$ on $C$. In particular, $0\notin C$.