Are these conditions also sufficient for a metric to be induced by a norm?

Question

Let $(X,d)$ be a metric space such that the set $X$ is also a vector space over the field $\mathbb{R}$ of real numbers or the field $\mathbb{C}$ of complex numbers. Then the following holds:

If the metirc $d$ is induced by a norm on $X$, then we must have $$d(u+x,u+y) = d(x,y) \ \mbox{ for all } \ u, x, y \in X,$$ and $$d(\alpha x, \alpha y ) = \vert \alpha \vert \cdot d(x,y) \ \mbox{ for all } \ x, y \in X \ \mbox{ and for all scalars } \ \alpha.$$

Are the above conditions sufficient also for the metric $d$ to be induced by a norm?

Answer 1

The answer here is yes. Define $\|x\| = d(0,x)$. We have:

• $\|0\| = d(0,0) = 0$
• $\|x+y\| = d(0,x+y) \leq d(0,x) + d(x,x+y) = d(0,x) + d(0,y) = \|x\| + \|y\|$
• For a scalar $\alpha$, we have $\|\alpha x\| = d(0,\alpha x) = |\alpha| \cdot d(0,x) = |\alpha| \cdot \|x\|$

Answer 2

Yes. Define $\|x\|$ to be $d(0,x)$. This is a norm (that induces $d$).