# Existence of $x\in \mathcal{X}$ such that $\|x\|=1$ and $\|x+\mathcal{M}\|=1$ for a closed subspace $\mathcal{M}$

I was proving a theorem stated below.

Theorem.

Suppose that $(\mathcal{X},\|\cdot\|)$ is a normed vector space and $\mathcal{M}\leq \mathcal{X}$ is a closed proper subspace. Then for any $\epsilon>0$ there exists $x\in \mathcal{X}$ such that $\|x+\mathcal{M}\|\geq 1-\epsilon$ where $$\|x+\mathcal{M}\|=\inf_{y\in \mathcal{M}}\|x+y\|.$$

After proving, I found that, actually, we can find $x\in\mathcal{X}$ such that $\|x+\mathcal{M}\|=1$. But if it is true, then there is no need to consider $\epsilon$ in the theorem. Thus, I am not confident about my proof.

Here is my proof!

Let $z\in \mathcal{X}\setminus \mathcal{M}$ be given. Then $$d=\|z+\mathcal{M}\|=\inf_{y\in \mathcal{M}}\|z+y\|>0$$

since $\mathcal{M}$ is closed. (If $\|z+\mathcal{M}\|=0$, there exists $\{y_n\}\subset \mathcal{M}$ such that $\|y_n+z\|\rightarrow 0$ and there exists a subsequence $\{y_{n_k}\}$ such that $y_{n_k}\rightarrow -z$ which means $z\in \mathcal{M}$ since $\mathcal{M}$ is closed.)

Then, there exists $y\in \mathcal{M}$ such that $\|z+y\|=d$.

Now let $x=\frac{z+y}{\|z+y\|}$ then $\|x\|=1.$ And note that

$$\|x+\mathcal{M}\|=\frac{\inf_{a\in \mathcal{M}}\|z+y+a\|}{\|z+y\|}=1$$

since $$\|z+y\|=\inf_{b\in \mathcal{M}}\|x+b\|$$.

Thus, our $x$ has been discovered.

*Is there any defect in my proof? Any comment would be helpful for me :) *

• How do you know there exists such a $y \in \mathcal{M}$? – gd1035 Aug 14 '18 at 23:19
• The existence of a minimising $y$ is not guaranteed in a normed (or even Banach space). For example, see the answer user357515's answer. – copper.hat Aug 15 '18 at 0:53
• @copper.hat I agree.. I mistakenly though that it is Banach space... But... why it is not true even in Banach space? I cannot see that... – Lev Ban Aug 15 '18 at 1:58
• See a fairly standard counterexample in my answer mentioned below – copper.hat Aug 15 '18 at 2:02
• The space $C[0,1]$ usually means the continuous functions with the $\sup$ norm. It illustrates a closed subspace (hence convex) that has no nearest point. – copper.hat Aug 15 '18 at 2:06

Then, there exists $y∈M$ such that $∥z+y∥=d$.
This sentence is essentially asserting the truth of the statement that was to be proved. In general, no such $y$ exists. A counterexample is given in Given a point $x$ and a closed subspace $Y$ of a normed space, must the distance from $x$ to $Y$ be achieved by some $y\in Y$?