Let $X$ be a locally convex space. Following [Wilansky, "Modern Methods in Topological Vector Spaces", Problem 11-1-112] $X$ is said to have the sequential convex compactness (SCC) property if the absolutely convex hull of every null sequence is compact. Is this a meaningful property? I am asking because even the Hilbert space $X = l^2$ does not have this property: If $e_n = (0, \dots, 0, 1, 0, \dots)$ denotes the unit vector having $1$ at the $n$-th position and else $0$, then $x_n := \frac{1}{n} e_n \to 0$ in $l^2$. Set $A := \{ x_n \mid n \in \mathbb{N} \}$. Then the absolutely convex hull $\textrm{aco}(A) = \{ \sum_{k=1}^n a_k x_k \mid \sum_{k=1}^n |a_k| \leq 1 \}$ is not closed in $l^2$ since $y_n := \sum_{k=1}^n \frac{1}{2^k} x_k \in \textrm{aco}(A)$ and $y_n \to y := \sum_{k=1}^\infty \frac{1}{2^k} x_k \in l^2$ but obviously $y \not\in \textrm{aco}(A)$. (Note also that since $\textrm{aco}(A) = \textrm{aco}(A \cup \{ 0 \})$ this does not depend on how one interpretes the set of a "null sequence" being $A$ or its closure $\overline{A} = A \cup \{ 0 \}$ including the limit $0$.)
Should the SCC property be rather defined as the "closed absolutely convex hull of every null sequence is compact". If this is so, then it seems that this property is equivalent to the local completeness property (see [Perez Carreras, Bonet, "Barrelled Locally Convex Spaces", Theorem 5.1.11) which is much wider known in Functional Analysis.