Convexity of unitary ball on metric vector space

Question

Let $V$ be a $F$-vector space, where $F = \mathbb{R}$ or $\mathbb{C}$, and $d$ a metric on $V$. Let $x_0\in X$, $r > 0$ and let us denote $$B_r^d [x_0] =\{x\in V : d(x,x_0)\leq r\}.$$

Definition: We say a set $A\subset V$ is convex iff for every $x,\, y\in A$ and for every $t\in[0,1]$, we have $tx+(1-t)y\in A$.

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My question is the following:

Is $B_1[0]$ convex?

I could answer it affirmatively the following particular cases:

• $d$ is induced by a norm $\lVert\cdot\rVert$.

By the triangle inequality, if $x,\, y \in B_1[0]$ and $t\in [0,1]$, we have $$d(tx,(1-t)y) = \lVert tx-(1-t)y\rVert \leq t\lVert x\rVert + (1-t)\lVert y\rVert \leq t+(1-t) = 1.$$

• $d(x,y) = 0$, if $x = y$, and $d(x,y) = r$, if $x\neq y$, for some $r > 0$.

Note that if $r \leq 1$ then $B_1 [0] = V$ and if $r > 1$, then $B_1[0] = \{0\}.$

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I have the feeling that this result is false in general, but I couldn't find any counterexample.

Answer 1

For example, on $\mathbb R^2$ consider the metric $$d(x, y) = \sqrt{(x_1 - y_1)^2 + ((x_2 + x_1^2) - (y_2 + y_1^2))^2}$$ i.e. $d(x,y) = d(F(x),F(y))$ where $F([x_1,x_2]) = [x_1,x_2 + x_1^2]$. Note that $F$ is a homeomorphism from $\mathbb R^2$ onto itself, so this is a metric. But the unit ball is not convex. For example, it contains $[0.5, -1.1]$ and $[-0.5, -1.1]$, but not $[0,-1.1]$.

Answer 2

Consider $\mathbb{R}^n$ with the metric:
$$d(\mathbf{x},\mathbf{y})=\Vert \mathbf{x}-\mathbf{y}\Vert \textrm{ if } \exists \lambda \in F \big(\mathbf{x}=\lambda \mathbf{y}\big) \textrm{ and } \Vert \mathbf{x}\Vert + \Vert \mathbf{y}\Vert \textrm{ otherwise.}$$

(I was taught that this metric was called the "streetcar metric" - the idea is to model travel within a town with radial streetcar routes coming out from the center. If you are traveling to a point on the same route (line through the origin) that you are already on, then you just hop on a streetcar and travel in a straight line. But to get to any other location, you must first travel to the center, and then board a different streetcar. However, Google tells me that this nomenclature is not widely used, so I don't know what the standard name for this metric is.)

Then if $r>\Vert\mathbf{x}\Vert>0$, the closed ball $B_{r}^{d}(x)$ looks like a ball about the origin with radius $r-\Vert\mathbf{x}\Vert$, together with a line segment connecting this ball to the point $\mathbf{x}+r\frac{\mathbf{x}}{\Vert\mathbf{x}\Vert}$. It should be obvious that this ball is not convex with respect to the vector space structure.