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I recently asked this question. Existence of x∈X such that ∥x∥=1 and ∥x+M∥=1 for a closed subspace M

And people said that when $\mathcal{X}$ is normed vector space, even if $\mathcal{M}$ is closed and $$\|x+\mathcal{M}\|=\inf_{y\in \mathcal{M}}\|x+y \|=d$$, we cannot say that there exists $y\in \mathcal{M}$ such that $\|x+y\|=d$ due to this result(Given a point $x$ and a closed subspace $Y$ of a normed space, must the distance from $x$ to $Y$ be achieved by some $y\in Y$?).

And I started to solve a problem 5.1.12 of Folland's real analysis which says

5.1.12 Let $\mathcal{X}$ be a normed vector space and $\mathcal{M}$ a proper closed subspace of $\mathcal{X}$. Then $$\|x+\mathcal{M}\|=\inf_{y\in \mathcal{M}}\|x+y \|$$ is a norm on $\mathcal{X\setminus M}$.

In order to prove it, we need to prove that $$\|x+\mathcal{M}\|=0 \iff x\in \mathcal{M}$$, which seems contradicts to the results above.

Could anyone teach me where I am thinking wrongly?

Lev Bahn
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  • It's not (necessarily) a contradiction - you can't guarantee $d \in M$ for arbitrary $d$ but perhaps you can for $d=0$. – Ethan Bolker Aug 16 '18 at 19:27
  • @EthanBolker Which property does make that difference? – Lev Bahn Aug 16 '18 at 19:32
  • I don't understand the queetion in your comment. The assertion you claim contradicts what you want to prove says that "sometimes: there is no $y$ in the set that minimizes the distance. But that "sometimes" may or may not be relevant for the particular case you want to prove. There is probably a proof for the case you care about. – Ethan Bolker Aug 16 '18 at 22:16
  • @EthanBolker I got your point. My question is then how to prove that the $$|x+\mathcal{M}|=\inf_{y\in \mathcal{M}}|x+y|$$ is a norm. I was searching the methods of the proofs people used and found that most of them are using that $|x+\mathcal{M}|=d$ then there exists $y\in \mathcal{M}$ such that $|x+y|=d$. For example, look at this. https://math.stackexchange.com/questions/2351781/norm-on-a-quotient-space – Lev Bahn Aug 17 '18 at 01:50
  • Additionally, if you google 'Folland solution chapter 5' and look at the solution of the problem 12, you will find they use the property. – Lev Bahn Aug 17 '18 at 01:51
  • I can't help you further with this. Sorry. – Ethan Bolker Aug 17 '18 at 02:09

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Suppose $x\in \mathcal{M}$ it is clear that $\Vert x+\mathcal{M} \Vert=0$. Now suppose $x\not \in \mathcal{M} $, since $\mathcal{M}$ is a closed set, and assuming we are using the normed topology, there exists an open ball centered in x, that does not intersect $\mathcal{M}$ then for every $y\in \mathcal{X}-\mathcal{B}(x)$ we have $\Vert x-y\Vert >0$.

fabian_mc
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