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$\Bbb{R^\omega}=\{(x_n)_{n\in \mathbb{N}}: x_n \in \Bbb{R}\}$

$d(x, y) =\sum_{j\in\mathbb{N}}{(a_j)} \frac{|x_j -y_j|}{1+|x_j -y_j|}$

Then $(\Bbb{R^\omega}, d) $ is a metric space.

I know in a normed space any ball is convex. And it is easy to prove.

The space $(\Bbb{R^\omega}, d) $ is not a normed space, I mean no norm on $\Bbb{R^\omega} $ can induce the metric $d$.

So, I guess in that space, It may be possible to find an open ball which is not convex.

My question :1) Can I pick any open ball to test whether it is convex or not?

  1. If no, then is there any linear metric space in which every open ball is not convex?

  2. Can we get an example of a linear metric space (not a normed space) in which every open ball is convex?

For the last question can I take $(X, \|•\|)$ be any normed space and then define a metric $d(x, y) =\sqrt\|x-y\|$. I think it works. Isn't it?

Here $d$ is not scaling equivalent, hence not induced by any norm.

$B_{d}(x_0, r) =\{x\in X : \|x-x_0\|<r^2 \}=B_{\|•\|} (x_0, r^2) $

Hence, every open ball is convex.

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    $\begingroup$ I think you should clarify what you mean by "linear metric space". I think you mean a linear space equipped with a metric that makes the linear space operations continuous. $\endgroup$
    – Rob Arthan
    Commented Dec 15, 2021 at 23:21
  • $\begingroup$ $(X, d) $ be a metric space where the underlying set is a linear space. $\endgroup$ Commented Dec 16, 2021 at 13:15
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    $\begingroup$ What you call a "linear space", people usually call a "vector space." $\endgroup$ Commented Dec 17, 2021 at 5:36
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    $\begingroup$ @Moishe Kohan, I think I am not the first one who use the term "linear space " instead of "Vector space" See this Wikipedia article en.m.wikipedia.org/wiki/Vector_space $\endgroup$ Commented Dec 17, 2021 at 6:13

3 Answers 3

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Yes, a standard example is $L^p([0,1])$ for $0 < p < 1$. It is a metric space for $$d(f,g):= \int_0^1|f(x)-g(x)|^pdx.$$

One can show that if $C$ is a convex open set in this space, then $C= L^p([0,1])$. In particular, not a single open ball is convex.

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    $\begingroup$ @SouravGhosh Would be a nice Christmas gift to actually check it. I wish you and your vector spaces a holly jolly Christmas :D $\endgroup$ Commented Dec 25, 2021 at 10:54
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The example of quantumspace is of course the classical/elegant one. Let me give another one (just for fun). We can pick $\mathbb{R}^2$ and put the "Paris"-metric on it (that is how our topology prof called it 10 years ago). Everytime you want to go anywhere, you have to go to "Paris" first and then you can continue your travel. For us Paris will be the origin. Of course it will depend from where you go to "Paris", meaning on some half-ray we put a different metric.

Let $R=\{ (x,0) \in \mathbb{R}^2 \ : \ x\geq 0 \}$, then we define the following metric $$ d(x,y) := \begin{cases} \vert x \vert + \vert y \vert,& x,y\in \mathbb{R}^2 \setminus R, \\ 2\vert x \vert + 2 \vert y \vert.& x,y \in R, \\ 2\vert x \vert + \vert y \vert,& x\in R, y \in \mathbb{R}^2\setminus R, \\ \vert x \vert + 2 \vert y\vert,& x\in \mathbb{R}^2 \setminus R, y\in R. \end{cases} $$ I guess that all balls in this metric are non-convex, if not, just add one more half-ray with a different metric.

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    $\begingroup$ Everyone who ever boarded a TGV will understand why this is called the Paris metric (or even more fittingly SNCF metric) ... $\endgroup$
    – MaoWao
    Commented Dec 16, 2021 at 9:30
  • $\begingroup$ @MaoWao As always, you are spot on :D I did not dare to comment on it myself $\endgroup$ Commented Dec 16, 2021 at 18:08
  • $\begingroup$ It's not a TVS though, is it? $\endgroup$
    – tomasz
    Commented Dec 17, 2021 at 5:38
  • $\begingroup$ @tomasz I will check this evening if I have time. Addition is continuous by the triangle inequality, will think about multiplication. $\endgroup$ Commented Dec 17, 2021 at 16:55
  • $\begingroup$ Well, the metric is definitely not invariant (which should be true for a linear metric space). Every point other than the origin is isolated, but the origin is not, so addition is not continuous, or am I missing something? $\endgroup$
    – tomasz
    Commented Dec 17, 2021 at 19:52
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A simpler example of an “$L_p$” space: $\mathbb R^2$ with the $l_p$ quasi-norm, where $p=\tfrac12$. The quasi-norm is $\|(x,y)\|_p = \sqrt{|x|}+ \sqrt{|y|}$.

I like this example because you can easily draw pictures of the open balls —- they all look like diamonds with concave sides. See here.

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  • $\begingroup$ That's a really nice example. Very neat! $\endgroup$ Commented Dec 17, 2021 at 0:56
  • $\begingroup$ Imho open balls in this metric with radius $r$ are just standard open balls with radius $r^2$, and hence convex. Do I miss something? $\endgroup$
    – daw
    Commented Dec 17, 2021 at 13:50
  • $\begingroup$ Maybe you meant $d(p,q) = \sqrt{|p_1-q_1|} + \sqrt{|p_2-q_2|} $? $\endgroup$
    – daw
    Commented Dec 17, 2021 at 13:52
  • $\begingroup$ @daw Indeed, that is what he probably meant (and I misread) $\endgroup$ Commented Dec 17, 2021 at 16:57
  • $\begingroup$ Yes, the comments are correct. I was thinking of the $\ell_p$ quasi-norm with $p=\tfrac12$, but I wrote down its definition wrong. I’ll fix the answer. $\endgroup$
    – bubba
    Commented Dec 18, 2021 at 1:02

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