If d is a norm on V, is $\frac{d(x,y)}{1+d(x,y)}$ a norm on V?

Question

Let d be a norm on a vector space V and let $\psi:V \to [0,\infty)$ be a function defined as $\psi(v)=\frac{d(v)}{1+d(v)}$. Is $\psi$ a norm on $V$?

It seems that $\psi$ does not satisfy the homogeneity property of a norm. I tried a proof by contradiction, but I think my contradiction may be incorrect. Any feedback on the correctness of the proof and alternative ways to approach the problem is appreciated.

Suppose that for all $\alpha\in\mathbb{R}$ and for all $v\in V$, $\;\psi(\alpha v)= |\alpha|\psi(v)$.
Since $\psi(\alpha v)= \frac{d(\alpha\;v)}{1+d(\alpha\;v)} = \frac{|\alpha|d(v)}{1+|\alpha|d(v)}$, we have $$\frac{|\alpha|d(v)}{1+|\alpha|d(v)}= |\alpha|\psi(v)$$$$\frac{d(v)}{1+|\alpha|d(v)}= \psi(v)$$Which is only true when $|\alpha| = 1$ (contradiction).

Answer 1

Your answer looks correct. In general, if you have something bounded it is not a norm. I'll explain what I mean:

Notice that $\frac{d}{1 + d} \le 1$. Pick any non zero $x$ and consider $\alpha x$, $\alpha \in \mathbb{R}^+.$ Assume by contradiction that it is a norm, then $$\alpha \frac{d}{1+d}(x) = \frac{d}{1+d}(\alpha x) \le 1.$$ Send $\alpha \to \infty$ to get a contradiction.

Answer 2

I agree with your answer: it is not a norm.

However, it is a distance, I believe.

Answer 3

Indeed it is not a norm, but it is worth noting that such a function does define a metric. Homogeneity of the first degree fails here, which is required for a norm, but not a metric.