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

**Answer**

Let $V$ be a vector space over the field $\mathbb{F}$. A norm

$$\| \cdot \|: V \longrightarrow \mathbb{F}$$

on $V$ satisfies the homogeneity condition

$$\|ax\| = |a| \cdot \|x\|$$

for all $a \in \mathbb{F}$ and $x \in V$. So the metric

$$d: V \times V \longrightarrow \mathbb{F},$$

$$d(x,y) = \|x – y\|$$

defined by the norm is such that

$$d(ax,ay) = \|ax – ay\| = |a| \cdot \|x – y\| = |a| d(x,y)$$

for all $a \in \mathbb{F}$ and $x,y \in V$. This property is not satisfied by general metrics. For example, let $\delta$ be the discrete metric

$$\delta(x,y) = \begin{cases} 1, & x \neq y, \\ 0, & x = y. \end{cases}$$

Then $\delta$ clearly does not satisfy the homogeneity property of the a metric induced by a norm.

To answer your edit, call a metric

$$d: V \times V \longrightarrow \mathbb{F}$$

**homogeneous** if

$$d(ax, ay) = |a| d(x,y)$$

for all $a \in \mathbb{F}$ and $x,y \in V$, and **translation invariant** if

$$d(x + z, y + z) = d(x,y)$$

for all $x, y, z \in V$. Then a homogeneous, translation invariant metric $d$ can be used to define a norm $\| \cdot \|$ by

$$\|x\| = d(x,0)$$

for all $x \in V$.

**Attribution***Source : Link , Question Author : Srijan , Answer Author : Henry T. Horton*