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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

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6 Answers 6

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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 $x \in V$.

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    $\begingroup$ Thank you very much. I understand now. $\endgroup$
    – Srijan
    Jul 4, 2012 at 2:42
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    $\begingroup$ That is not enough as one needs to check that the so defined norm really induces the original metric and not another one: $d(\cdot,\cdot)\to\|\cdot\|\to d(\cdot,\cdot)$ $\endgroup$ Sep 29, 2014 at 14:36
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    $\begingroup$ This is an old post and it doesn't really matter much, but I have to be pedantic. So you say that if a metric comes from a norm on $\Bbb{R}$, then $d(ax,ay)=|a|d(x,y)$ but what if you just consider $\Bbb{R}$ as a vector space over $\{0,1\}$. Then the discrete metric IS homogeneous and translation invariant on this vector space. $\endgroup$
    – user223391
    Jun 16, 2015 at 0:16
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    $\begingroup$ @MichaelMunta If the metric is induced by a norm, then $d(x,y)=||x-y||$. So $d(x+z,y+z)=||(x+z)-(y+z)||=||x-y||=d(x,y)$. $\endgroup$
    – Jamāl
    Feb 27, 2020 at 15:30
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    $\begingroup$ You should clarify that your field $F$ is normed, otherwise the notation $|a|$ is meaningless. $\endgroup$ Jun 5, 2020 at 23:24
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Here is another interesting example: Let $|x-y|$ denote the usual Euclidean distance between two real numbers $x$ and $y$. Let $d(x,y)=\min\{|x-y|,1\}$, the standard derived bounded metric. Now suppose we look at $\Bbb{R}$ as a vector space over itself and ask whether $d$ comes from any norm on $\Bbb{R}$. Then if there is such a norm say $||.||$, we must have the homogeneity condition: for any $\alpha \in \Bbb{R}$ and any $v \in \Bbb{R}$,

$$||\alpha v || = |\alpha| ||v||.$$

But now we have a problem: The metric $d$ is obviously bounded by $1$, but we can take $\alpha$ arbitrarily large so that $||.||$ is unbounded. It follows that $d$ does not come from any norm.

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  • $\begingroup$ It was nice explanation to me. I had forgot to thank you. :) $\endgroup$
    – Srijan
    May 25, 2013 at 9:21
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As Henry states above, metrics induced by a norm must be homogeneous. You can see that they must also be translation invariant: $d(x+a,y+a)= d(x,y).$ So any metric not satisfying either of those can not come from a norm.

On the other hand, it turns out that these two conditions on the metric are sufficient to define a norm that induces that metric: $d(x,0)=\| x \|.$

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    $\begingroup$ Thank you sir. I am fully satisfied with two answers given by you and Henry.:) $\endgroup$
    – Srijan
    Jul 4, 2012 at 2:44
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    $\begingroup$ Did you mean instead that any metric not satisfying either of those can not come from a norm? And not the other way around? And also, these two conditions on the metric are sufficient to define a norm..? $\endgroup$
    – T. Eskin
    Jul 4, 2012 at 6:27
  • $\begingroup$ @ThomasE. Sorry, you are right, I muddled up the words. Thanks. $\endgroup$ Jul 4, 2012 at 8:43
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Every homogeneous metric induces a norm via: $$\|x\|:=d(x,0)$$ and every norm induces a homogeneous and translation-invariant metric: $$d(x,y):=\|x-y\|$$

The clue herein lies in wether the induced norm really represents the metric as: $$d(\cdot,\cdot)\to\|\cdot\|\to d(\cdot,\cdot)$$ which is the case only for the translation-invariant metrics: $$d'(x,y)=d(x-y,0)=d(x,y)$$ whereas the induced metric always represents the norm as: $$\|\cdot\|\to d(\cdot,\cdot)\to\|\cdot\|$$ as a simple check shows: $$\|x\|'=\|x-0\|=\|x\|$$

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  • $\begingroup$ @ Thanks for the answer. $\endgroup$
    – Srijan
    Feb 15, 2016 at 4:15
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Personal Note:

Good question, I remember wondering the same thing myself back when I new to real analysis. Here's a counter example:


Counter Example

\begin{equation} d(x,y):=I_{\{(x,x)\}}(x,y):=\begin{cases} 1 &:\, \,x=y\\ 0 &:\, \,x\neq y. \end{cases} \end{equation} This is indeed a metric (check as an exercise).

Where $I_A$ is the indicator function of the set $A$; where here the set $(x,x)\in V\times V$.

If $r\in \mathbb{R}-\{0\}$, then $d(rx,ry)\in \{0,1\}$ hence $rd(x,y)\in \{0,r\}$ which is not in the range of $d:V\times V \rightarrow \mathbb{R}$.

So, the metric $d$ fails the property that any metric induced by a norm must have, ie: \begin{equation} \mbox{it fails to have the property that: } \|rx-ry\| =r\|x-y\|. \end{equation}


Interpretation & Some Intuition:

The problem is that the topology is too fine, so all this topology can do is distinguish between things being the same or different.
As opposed to a norm topology which distinguised between object being, or not being on the same line (for some appropriate notion of line) (as well as them being different).

Hope this helps :)

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    $\begingroup$ Thank you for the explanation. :) $\endgroup$
    – Srijan
    Feb 15, 2016 at 4:15
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    $\begingroup$ Hi, is it correct to consider the metric d(x,y)=|x-y|^2018 that fails to h;ave the above property? $\endgroup$
    – Jim Art
    Oct 23, 2018 at 8:19
  • $\begingroup$ Yes, why not? :) $\endgroup$
    – ABIM
    Oct 23, 2018 at 15:33
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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 $(x_0, y_0) \in X\times X$ such that $d(x_0, y_0)\neq0$ , 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.

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