TITLE: Not every metric is induced from a norm
QUESTION [79 upvotes]: I have studied that every normed space $(V, \lVert\cdot \lVert)$ is a metric space with respect to distance function
$d(u,v) = \lVert u - v \rVert$, $u,v \in V$. 
My question is whether every metric on a linear space can be induced by norm? I know answer is  no but I need proper justification. 
Edit: Is there any method to check whether a given metric space is induced by norm ?
Thanks for help

REPLY [6 votes]: One possible way of showing that a metric does not arise from a norm is to show that it is bounded, as it then cannot be homogeneous.

Proof:
Take $d$ a bounded metric on a space $X$: i.e. $\exists D \in \mathbb{R}_+$ such that $\forall (x,y) \in X^2, d(x,y) \leq D$. Suppose now for contradiction that $d$ arises from a norm, i.e. the exists a norm $ \|\cdot\|$ such that $d(x,y) = \|x - y\|$. Recall that the distance must then must be homogeneous, for we have $d(\lambda x, \lambda y) = \|\lambda x - \lambda y\| = |\lambda| \cdot \|x - y\| = |\lambda| d(x,y)$.
Take now arbitrary  $(x_0, y_0) \in X^2$, then we must have that $d\left( \frac{D+1}{d(x_0, y_0)} \cdot x_0, \frac{D+1}{d(x_0, y_0)} \cdot y_0 \right) = \frac{D+1}{d(x_0, y_0)} \cdot d(x_0, y_0) = D + 1 > D $ which contradicts the upper bound of the metric.