Why metric vector space may have few convex sets

Question

I am reading a answer by t.b. ,see here Connections between metrics, norms and scalar products (for understanding e.g. Banach and Hilbert spaces)

He(maybe she) said that "The first observation I would like to mention is that a normed vector space has an abundance of convex open sets. Indeed, the ball of radius $r$ around any point is convex. In a metric vector space, however, there is no reason that there be any convex open sets except the empty set and the space itself—this is due to the fact that the metric is not required to be homogeneous, only translation-invariant. One standard example for this is the space $L_{0}([0,1])$ of all (classes of) Lebesgue measurable functions modulo null sets, equipped with the topology of convergence in measure to be explicit, my preferred metric is $\displaystyle d(f,g) = \int \frac{|f - g|}{1 + |f-g|}$."

I can't understand the words that "In a metric vector space, however, there is no reason that there be any convex open sets except the empty set and the space itself—this is due to the fact that the metric is not required to be homogeneous, only translation-invariant."

And I don't understand his example.

Can someone show me a simple example? i.e. Construct a metric vector space, and s.t. there is no convex set in it except empty set and itself.

Thanks a lot.

Answer 1

To show that a ball $B(x,r)$ in a normed vector space $(V,\|\cdot\|)$ is convex, you will have to use at some point the fact that the norm satisfies the homogeneity property : $\|\lambda v\| = |\lambda|\|v\|$ for any $\lambda$ in the base field (equipped with the absolute value $|\cdot|$) and $v\in V$. If you now take instead of the normed vector space $(V,\|\cdot\|)$ only a metric vector space $(V,d)$ ($d$ is the metric) you can of course still define balls, but you won't be able to show that they are convex, especially because you don't have an "analogue" of the homogeneity property. (A good exercise for you would be to show that balls are convex in normed vector space and then try to show it in metric vector spaces to feel what is lacking - the homoegeity.)

As for the example, it reads clearly : you take the vector space $V = \mathbf{L}^1 ([0,1],\mu)$ (where $\mu$ is the Lebesgue measure) and you equip it with the distance (metric) $d$ as defined in your quote. Then $(V,d)$ has no other convex opens sets than $V$ and $\varnothing$.

This example is not that hard. See this for instance for an "other" example :

Existence of Non-Trivial, Convex, Open Set in $C_{\mathbb{C}}[0,1]$ Under $L^{0}$ Metric

Answer 2

But I sill have a problem. In the example mentioned above, I prove a open ball is convex in $L_{0}([0,1])$, but I don't know where the wrong is.

Let $B(f,r)$ be a open ball in $L_{0}([0,1])$, and let $g\in L_{0}([0,1])$ s.t. $d(f,g)=r$.

To show $B(f,r)$ is convex, it suffices to show for $\lambda$ defined as $tf+(1-t)g,t\in (0,1)$, it will lie in the ball $B(f,r).$

$\displaystyle d(\lambda,f)= \int \frac{|\lambda - f|}{1 + |\lambda-f|}=\int \frac{(1-t)|g - f|}{1 + (1-t)|g-f|}=\int \frac{|g - f|}{\frac{1}{1-t} + |g-f|}\lt\int \frac{|g - f|}{1 + |g-f|}=d(f,g)=r. $

So $\lambda \in B(f,r)$ , and so $B(f,r)$ is convex in $L_{0}([0,1])$.