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?

  • $\begingroup$ It's not (necessarily) a contradiction - you can't guarantee $d \in M$ for arbitrary $d$ but perhaps you can for $d=0$. $\endgroup$ – Ethan Bolker Aug 16 '18 at 19:27
  • $\begingroup$ @EthanBolker Which property does make that difference? $\endgroup$ – Lev Ban Aug 16 '18 at 19:32
  • $\begingroup$ 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. $\endgroup$ – Ethan Bolker Aug 16 '18 at 22:16
  • $\begingroup$ @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. math.stackexchange.com/questions/2351781/… $\endgroup$ – Lev Ban Aug 17 '18 at 1:50
  • $\begingroup$ Additionally, if you google 'Folland solution chapter 5' and look at the solution of the problem 12, you will find they use the property. $\endgroup$ – Lev Ban Aug 17 '18 at 1:51

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 $\Vert x-y\Vert >0$.

  • $\begingroup$ I was so stupid.... Thank you! $\endgroup$ – Lev Ban Aug 17 '18 at 12:43

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