Convex hull of $\{ \Vert x \Vert = 1 \}$ is closed in strictly convex space

Question

I'm trying to show that the convex hull of $\{ \Vert x \Vert = 1\}$ is closed in a strictly convex Banach-space.

I don't know how to tackle the problem. Are there any nice characterizations for a convex hull to be closed?

Thanks a lot!

Answer 1

Let us assume that the Banach space $X$ is not the trivial Banach space $\{0\}$.

Let us set $A = \{x \in X : \|x\| = 1\}$ and $B = \{x \in X : \|x\| \le 1\}$. Obviously, $A \subset B$ and $B$ is closed. Further, every $x \in B$ can be written as a convex combination of two elements from $A$:

• If $x = 0$, we have $x = \frac12 \, y + \frac12 \, (-y)$, where $y \in A$ is arbitrary (such $y$ exists, since $X \ne \{0\}$.
• If $x \ne 0$, it is a suitable convex combination of $x / \|x\|$ and $-x/\|x\|$.

Hence, $B$ is the convex hull of $A$ and it is closed.