Large ideally convex sets

Question

Let $E$ be a Banach space. A set $C \subseteq E$ is called ideally convex if for every bounded sequence $(x_n)$ in $C$ and for every sequence $(\lambda_n)$ in $[0,1]$ that sums up to $1$ the vector $\sum_n \lambda_n x_n$ is also in $C$.

(So a bounded set is ideally convex if and only if it is $\sigma$-convex in this sense. But in the following question, $C$ is automatically unbounded.)

Question. Let $C \subseteq E$ be ideally convex and dense in $E$. Does it follows that $C = E$?

Remark. Every open convex set is ideally convex (by the Hahn-Banach separation theorem). In this special case it is not particularly difficult to prove that the question has a positive answer: an open, convex, and dense set $C \subseteq E$ is automatically equal to $E$.

Answer 1

Just take your favorite decreasing sequence $\lambda_k$ of positive numbers with sum $1$ and inductively construct the vectors $x_k\in C$ such that $\left \|\sum_{k=1}^n\lambda_k x_k-x\right\|\le\lambda_{n+1}$ (then automatically $\|x_n\|\le \frac{\lambda_{n}+\lambda_{n+1}}{\lambda_n}\le 2$ for $n\ge 2$) to get a series converging to $x$. I use that the density of $C$ implies the density of $\lambda C$ for any $\lambda>0$, of course.