What conclusion can we get from a closed unit ball of a subspace $M$ of a norm space $X$ is closed in $X$?

Question

Let $X$ be a normed space and $M$ be a linear subspace of $X$. Let $B_M$ be a closed unit ball in $M$ such that $B_M$ is closed in $X$. Then can we get $M$ is closed in $X$? Or other conclusions we can get from it.

Thank you in advance!

Answer 1

Here's a proof using a nice norm identity: Suppose $(x_n)$ is a sequence in $M$ convergent to some $x\in X$. Without loss of generality, assume $x\neq0$, so we may assume $x_n\neq0$ for all $n$. By continuity of the norm, $\|x_n\|\to\|x\|$, so by the identity $$\left\|\frac{z}{\|z\|}-\frac{y}{\|y\|}\right\|=\frac{1}{\|z\|}\left\|(z-y)+(\|y\|-\|z\|)\frac{y}{\|y\|}\right\| \qquad (y,z\in X\setminus\{0\}),$$ it follows that $\frac{x_n}{\|x_n\|}\to\frac{x}{\|x\|}$. Since $B_M$ is closed, it follows that $\frac{x}{\|x\|}\in B_M$, whence $x\in M$ and therefore $M$ is closed.