# Normed vector space with a closed subspace

Suppose that $X$ is a normed vector space and that $M$ is a closed subspace of $X$ with $M\neq X$. Show that there is an $x\in X$ with $x\neq 0$ and $$\inf_{y\in M}\lVert x - y\rVert \geq \frac{1}{2}\lVert x \rVert$$

I am not exactly sure how to prove this. I believe since $M\neq X$ we can find some $z\in X\setminus M$ then if we let $\delta = \inf_{y\in M}\lVert z - y\rVert$ then we can choose some $y$ and deduce that $y\in M$.

Any suggestions is greatly appreciated.

Take some $b\in X\setminus M$. You have that $d=\inf\limits_{y\in M}{\|b-y\|}>0$. Now take $m_0\in M:\,\|b-m_0\|\leq 2d$. Then $x=\frac{b-m_0}{\|b-m_0\|}$ satisfies your condition: $$\forall m\in M:\, \|x-m\|=\|\frac{b-m_0}{\|b-m_0\|}-m\|=\|\frac{b-(m_0+m\|b-m_0\|)}{\|b-m_0\|}\|\ge \frac{d}{2d}\|x\|$$
The quotient space $X/M$ is a non-trivial Banach space with elements that are cosets of the form $x+M$. And $\|x+M\|=\inf_{m\in M}\|x+m\|$. Choose any non-zero coset $x'+M$. Then there exists $m'\in M$ such that $$\|x'+m'\| \le 2\|x'+M\|_{X/M}$$ Then $$\|(x'+m')\| \le 2\inf_{m\in M}\|x'+m'+m\| =2\inf_{m\in M}\|(x'+m')-m\|$$ Take $x=x'+m'$.