Let $M$ be a finite dimensional subspace of $E$. Prove that for each $x \in E\setminus M$, there exists $m_0 \in E$ such that $d(x,M) = \| x-m_0\|$.

Question

Let $M$ be a finite dimensional subspace of $E$. Prove that for each $x \in E\setminus M$, there exists $m_0 \in E$ such that $d(x,M) = \| x-m_0\|$.

I've been stuck in these exercise for some time. I've tried Riesz Lema, hoever I could not get anywhere.

Help?

Answer 1

The function $f: M \to \mathbb{R}$ given by $f(m) = \|x-m\|$ is continuous and $f(m) \ge \|m\| - \|x\|$, so if $\|m\| \ge 2 \|x\|$ then $f(m) \ge f(0) = \|x\|$. In particular, $D=\overline{B}(0,2\|x\|)$ is compact, so $f$ has a minimiser $m_0$ on $D$, that is, $f(m_0) \le f(m)$ for all $m$.