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Let be (X,S) a locally convex space, and $B \subset X$ a nonempty sequentially closed bounded and convex set such that $\hat{0} \notin clB$,(the closure of B). Define the set T:=s-clco$\{B \cup-B \}$ where s-clco denotes the sequential closure of the convex hull. This appeared in a paper that I was reading, and then the author assume that T is a balanced set. Why it is true? The correct definition of the convex hull of B is $\left\{ \displaystyle\sum_{i = 1}^n t_i b_i: \sum_{i=1}^n t_i = 1, t_i \geq 0, b_i \in B \right\}$ right?

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Assuming that the scalar field is $\mathbb R$ note first that $0=\frac {b+(-b)} 2 \in T$. Hence, for any $t \in T$ and any $c \in [0,1]$, we have $ct=ct+(1-c)0 \in T$. Also $T$ is symmetric. Hence $ct \in T$ whenever $t \in T$ and $|c| \leq 1$.

I am using the facts that the convex hull of any set is convex and the sequential closure of a convex set is also convex.

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  • $\begingroup$ In general It is not true when the field is other than $\mathbb{R}$ right? $\endgroup$ – The Student Mar 7 at 4:48
  • $\begingroup$ @TheStudent Yes, for the complex field we cannot say that $T$ is balanced. $\endgroup$ – Kavi Rama Murthy Mar 7 at 5:15

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