Let $(X,d_X)$ be a metric space.
Suppose that there is a one-to-one function $f:X\rightarrow V$, where $V$ is a vector space.
Is there any theorem saying that when is $f$ an isometric embedding? That is, there exists a metric $d_V$ on $V$ such that $d_X(x,y)=d_V(f(x),f(y))$.
I am thinking if I could define $d_V(u,v)\equiv d_X(f^{-1}(u),f^{-1}(v))$, then I am done. However, if $f$ is not onto, then the inverse of $f$ is not well-defined for every $v\in V$. Is there any other way I can argue the existence of such $d_V$?
Edit: I want $d_V$ on $V$ is a norm. So I should ask when there exists a norm $||\cdot||$ such that $d_X(x,y)=||f(x)-f(y)||$.
It's worse than "not onto" -- presumably you want the metric on the vector space to respect the vector space structure, i.e., you want, for $v \in V$, that $ v \mapsto d_V(v, 0)$ be a norm on the space; if the underlying field has non-two characteristic, the polarization identity then gives you an inner product, too. And at that point, you need to have things like $$ \| f(u) \| = |\alpha| \|f(v)\| $$ whenever $f(u) = \alpha f(v)$, i.e., if your mapping sends any two points of $X$ to the same line through the origin, you've got yet another constraint.
I doubt there's any simple characterization of $f$ for which this is possible.
But let's suppose that you DO have that property -- "linearity on each line through the origin of $V$" --- there's still the question of whether the metric (which is now defined on those lines) can be extended to a metric on the whole space. You have to check that the triangle inequality works, and you have to extend to the rest of the lines through the origin in a way that continues to have that property. Both of those, to me, look challenging to express in a simple way.