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Let $X$ be a real NLS such that for every proper subspace $Y$ of $X$ , $\exists x \in X$ such that $||x||=1$ and $dist (x,Y)=1$ ; then is $X$ finite dimensional ?

  • @Jean Marie : I am not sure whether $X$ should be really finite dimensional or not . From your edit , it appears you are sure ... –  Apr 19 '16 at 06:44
  • I correct my edit. – Jean Marie Apr 19 '16 at 06:51
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    In an infinite dimensional space $X$, take $Y$ to be the kernel of a discontinuous linear functional. This is dense in $X$ and proper; so, there is no $x\in X$ with $\text{dist},(x,Y)=1$. – David Mitra Apr 19 '16 at 16:10
  • @DavidMitra : You should write that as an answer , that's exactly what I wanted –  Apr 20 '16 at 15:19

2 Answers2

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Edit: Here is a complete modification of my first answer.

I had at first leaned (without having any proof in mind) to a finite dimensional answer.

I had indicated Riesz's lemma

https://en.wikipedia.org/wiki/Riesz's_lemma

and the classical theorem (mentionned in this article): The unit ball in a Normed Linear Space is compact iff this space is finite dimensional.

In fact, the answer is that the space is infinite dimensional, for the following reason that I had already indicated in a comment:

An easy recurrence argument shows that a "tower" of subspaces

$$X_1 \subset X_2 \subset \cdots \subset X_n \subset \cdots$$

can be built with $X_1=span{x_1}=x_1\mathbb{R}$ where $x_1$ is an arbitrary ellement of $X$ and:

$$X_{n+1}:=X_n \oplus x_n\mathbb{R}$$

(where $x_n$ is the element given with norm 1 and $dist(x_n,X)=1$).

It is a direct sum, because, on the contrary, if $x_n \in X_n$, $x_n$ would be at a zero distance from $X_n$.

Thus $X$ possesses subspaces of arbitrary large dimension, therefore has infinite dimensionality.

Jean Marie
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  • I know all that , but how does it lead to answer my question ? –  Apr 19 '16 at 06:47
  • From my hypothesis , I don't think I can conclude that the closed unit ball of $X$ is compact ... –  Apr 19 '16 at 06:48
  • OK, now that it is corrected, may I say you that your "I know all that" is the first bit of information we have about your understanding of the question... – Jean Marie Apr 19 '16 at 06:55
  • I don't understand what you are saying .. ? –  Apr 19 '16 at 07:04
  • A hint: build by immediate recurrence a "tower" of subspaces $X_1 \subset X_2 \subset \cdots \subset X_n \subset \cdots$ with $dim(X_n)=n$ and $X_{n+1}=X_n + x_n\mathbb{R}$ (where $x_n$ is the element given with norm 1 and $dist(X_n,x_n)=1$). – Jean Marie Apr 19 '16 at 07:09
  • ... which proves that, in fact, such a space is infinite dimensional... And a supplementary interesting question is to prove (or disprove ?) that any infinite dimensional space has such a property. – Jean Marie Apr 19 '16 at 07:20
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Using the Axiom of Choice, one can construct in a given infinite dimensional normed linear space $X$, a discontinuous linear functional (see this, e.g.). The kernel of such a functional is a proper and dense subset of $X$ (see this). This would provide a subspace $Y$ of $X$ that is proper and such that for every $x\in X$, $\text{dist}\,(x,Y)=0$.

From this, it follows that the answer to your question is "yes".

(Incidentally, if you only consider closed subspaces, then your property characterizes reflexive spaces. This follows from a result of R. C. James: A Banach space $X$ is reflexive if and only if each $x^*\in X^*$ achieves its norm on $B(X)$.)

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