Are balls around a point always symmetric about the axes?

Question

I think they are. And I have sketched the following proof so far.

Every ball around the origin is symmetric to the X-axis (generalize to all axis and around all points)

Let $\| \cdot \|$ be any norm on $\mathbb{R}^n$. Say the set $\{t \in \mathbb{R}^n : \|t\| \neq \|t_x\|\}$ is non-empty, where $t_x$ is the reflection of $t$ about the $X-$axis. Let $r$ be "a" vector in the set with the smallest norm.

Clearly, $r \neq \underline 0$, so take the line joining $r$ and $\underline 0$. The function $\|\cdot\|$ is continuous everywhere in the ball of radius $\|r\|$, and this line is inside ball by its convexity. Therefore $\|\cdot \|$ is continious on this line, and assumes $0$ at $\underline 0$ and $\|r\|$ at $r$. Therefore by $IVT$, for every large enough $m$ there is a point $r_m$ on this line such that $\|r_m\| = \|r\| - \frac{1}{m}$. By the collinearity of $r, r_m$ and $\underline 0$, we have, $$\|r-r_m\| = \|r\|-\|r_m\| = \frac{1}{m} \implies r_m \rightarrow r $$

Also, by continuity of reflection, we have $r_{m_x} \rightarrow r_x$, and by continuity of $\|\cdot\|$, we have $\|r_m\| \rightarrow \|r\|$, and $\|r_{m_x}\| \rightarrow \|r_x\|$.

Notice, by minimality of $r$, $\|r_m\| = \|r_{m_x}\|$; that is they are the same sequence, and hence must have the same limit. Hence, $\|r_x\| = \|r\|$. Contradiction! Or the set of assymetric vectors is non-empty.

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I have the following concerns.

• Does the fact really hold? If no, where did I go wrong, and what is a counterexample?
• Assuming it does, the "lines" between two points $x$ and $y$ induced by a norm and by convexity might be different. Particularly, the set $\{tx + (1-t)y: 0\leq t \leq 1\}$ is always the straight line joining $x$ and $y$. However the set $\{z : \|y - z\| + \|z - x\| = \|y - x\|\}$ needn't always be a straight line. For example, it can be "L" shaped like in the case of $\|\cdot \|_1$ norm. Does this conflict affect the proof.
• Is there a simpler way to show this?

Answer 1

Your conjecture is wrong. $\Vert(x, y)\Vert = \max(|x|, |x+y|)$ is a norm on $\Bbb R^2$, but the ball $D = \{ (x, y) \mid \Vert(x, y)\Vert \le 1 \}$ is not symmetric with respect to the $x$-axis or $y$-axis:

(Image created with Desmos.)

One problem in your argument is that the set $\{t \in \mathbb{R}^n : \|t\| \neq \|t_x\|\}$ is not closed, and therefore need not have an element with smallest norm.

Answer 2

The balls of the Manhattan metric $|x|+|y|$ (i.e. the $L^1$ norm) in $\mathbb{R}^2$ are squares parallel to the coordinate axes (edit: actually, they're diagonal, so this is a diamond), which are not symmetric with respect to most othre axes. If you rotate this norm then the balls aren't even symmetric across the coordinate axes, or apply the right skew transformation and it isn't symmetric across any line. (Of course, symmetry across axes is interpreted with respect to the usual Euclidean metric, isn't it?)