Factor space norm calculation when the subspace is finite-dimensional

Question

Let $(X,\|\;\|_X)$ be a normed vector space and let $M$ be a closed finite-dimensional subspace of $X$. I want to prove that: $$ \forall x\in X\;\;\exists\;m\in M\text{ s.t. } \|[x]_M\|_{X/M}=\|x+m\|_X $$ I'm really lost with this one since I don't get at all what it says. The only thing I managed to do is that:
$\big($If any doubt, $X/M=\{[x]_M:x\in M\}$ and $[x]_M=\{y\in X:y\sim x\}$, where $y\sim x \Leftrightarrow\;y-x\in M$$\big)$ $$ \|[x]_M\|_{X/M}=\inf\{\|y\|_X:y\sim x\}=\inf\{\|y\|_X:y-x\in M\}=\inf\{\|y\|_X:y-x=m\;\text{ for }m\in M\}=\inf\{\|y\|_X:y=x+m\;\text{ for }m\in M\}=\inf\{\|x+m\|_X:\text{ for }m\in M\} $$ and got stucked here. Any ideas, comments and suggestions would be appreciated.

Answer 1

Let $x\in X$ be arbitrary. Then $\|x-m\|\ge 0,\,\forall m\in M\Rightarrow \inf\limits_{m\in M}{\|x-m\|_X}$ exists. Let $\{m_n\}$ be a minimizing sequence, i.e $\|x-m_n\|\to \inf\limits_{m\in M}{\|x-m\|_X}$. Because the functional $J(m)=\|x-m\|$ is coercive over $M$ it follows that $\{m_n\}$ is bounded in $M$ and therefore contains a convergent subsequence $\{m_{n_k}\}$ to some element $m_0\in M$ (because $M$ is finite dimensional and closed). For this subsequence you get $$\|x-m_0\|=\lim\limits_{k\to\infty}{\|x-m_{n_k}\|}=\inf\limits_{m\in M}{\|x-m\|_X}=\|[x]\|_{X/M}$$ In the above equality, we used the fact that the norm is continuous.