If $(X, \|\cdot\|)$ is a normed vector space, then $(X\setminus\{0\}, d)$ is a metric space for $d(x, y) = \frac{\|x-y\|}{\|x\|+\|y\|}$

Question

Consider the following statement: If $(X, \|\cdot\|)$ is a normed vector space, then $(X\setminus\{0\}, d)$ is a metric space for $$d(x, y) = \frac{\|x-y\|}{\|x\|+\|y\|}$$ I'm trying to figure out whether this statement is true (and if it is, whether it's only true in a special setting, e.g. uniformly convex Banach spaces). The context is that I'm trying to define a "similarity metric" between two vectors in a general normed vector space with the invariance $d(x, y) = d(\alpha x, \alpha y)$ for $\alpha\neq 0$, among some other properties. It seems like it could maybe be a textbook problem, but I'm not quite sure if I've seen it somewhere or not. It's pretty clear that $d(x, y) = 0$ implies $x = y$ and $d(x, y) = d(y, x)$, but the triangle inequality is a bit sticky. We can rearrange $d(x, y)\leq d(x, z)+d(z, y)$ as $$\|x-y\|\leq \left(\frac{\|x-z\|}{\|x\|+\|z\|}+\frac{\|y-z\|}{\|y\|+\|z\|}\right)(\|x\|+\|y\|)$$ which feels stronger than the standard triangle inequality on $(X, \|\cdot\|)$. Does anybody have some insights on this? Am I missing a standard trick here? Thanks!

Answer 1

Consider the special case of $X=\mathbb R$. Then, $d(x,y)=\frac{|x-y|}{|x|+|y|}$ is not continuous at $(x,y)=(0,0)$. Along $x=y,\ (x,y)\to (0,0)\Rightarrow d(x,y)\to 0$ whereas along $x=0$ we have $d(x,y)\to 1.$ So even if we define $d(0,0)=0$ the triangle inequality can not be satisfied.