3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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: $$\forall m\in M:\, \|x-m\|=\|\frac{b-m_0}{\|b-m_0\|}-m\|=\|\frac{b-(m_0+m\|b-m_0\|)}{\|b-m_0\|}\|\ge \frac{d}{2d}\|x\|$$

$\endgroup$
0
$\begingroup$

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 $x=x'+m'$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.