I was proving a theorem stated below.


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 :) *

  • $\begingroup$ How do you know there exists such a $y \in \mathcal{M}$? $\endgroup$ – gd1035 Aug 14 '18 at 23:19
  • $\begingroup$ The existence of a minimising $y$ is not guaranteed in a normed (or even Banach space). For example, see the answer user357515's answer. $\endgroup$ – copper.hat Aug 15 '18 at 0:53
  • $\begingroup$ @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... $\endgroup$ – Lev Ban Aug 15 '18 at 1:58
  • 1
    $\begingroup$ See a fairly standard counterexample in my answer mentioned below $\endgroup$ – copper.hat Aug 15 '18 at 2:02
  • 1
    $\begingroup$ 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. $\endgroup$ – 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 $y\in Y$?


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.