Prove that the metric $d(x,y) = ||x-y||/(1+||x-y||)$ is not induced by any norm. Working in $\mathbb R^n$, prove that the metric $d(x,y) = |x-y|/(1+|x-y|)$ is not induced by any norm.
I'm confused by this question: isn't the definition of a metric induced by a norm a distance function $d(x,y) = ||x-y||$. Of course then $||x-y||/(1+||x-y||)$ isn't induce by any norm because $||x-y||/(1+||x-y||) \neq ||x-y||$. Where is the confusion?
My book literally states: "if we start with any vector space $V$ with a norm $||x||$ then $V$ becomes a metric space with the distance function $d(x,y) = ||x-y||$. This metric is said to be the metric induced by the norm."
Edit: How's this proof? We're working in $\mathbb R^n$. Suppose to the contrary such a norm does exist, so $d(x,y) = |x-y|_2 = \frac{|x-y|_1}{1+|x-y|_1}$. Let's take a look at $d(ax,ay) = |a||x-y|_2 = |a|d(x,y) = |a|\frac{|x-y|_1}{1+|x-y|_1} \neq \frac{|a||x-y|_1}{1+|a||x-y|_1}$. Hence, a contradiction.
By the way, my textbook uses both $|x|$ and $||x||$ for norm, I'm not sure why.
 A: Let $V$ be vector space. Given any norm $n$ on $V$, you can construct a metric $d_n$ on $V$, defined as
$$d_n(x,y)=n(x-y).$$
We call a metric induced by a norm, if it is of the $d_n$ for some norm $n$. The point is that not every metric on $V$ is induced by a norm.
Therefore you have to prove that there does not exists any norm $n$ on $V$, such that
$$n(x,y)=|x-y|/(1+|x-y|). $$
A: You're starting with a norm $\|\cdot\|_1$ on a vector space $V$. You want to show that there can't exist some other norm $\|\cdot\|_2$ on $V$ such that $\|x-y\|_2 = d(x,y) = \|x-y\|_1/(1+\|x+y\|_1)$.
A: Let $E$ be a real normed space, of non-zero dimension.
The distance induced by the norm is defined by : $d(x,y)=\Vert x-y\Vert$.
Obviously, if $u\in E$ is such that $\Vert u\Vert=1$, then we have for any $t>0$ : $d(tu,0_E)=t$, which shows that $d$ is not bounded from above.
On the contrary, the distance defined by $\delta(x,y)=\dfrac{\Vert x-y\Vert}{1+\Vert x-y\Vert}$ is bounded by $1$. Hence, it cannot be induced by any norm.
