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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    $\begingroup$ $\LaTeX$ tip: \parallel is a relation symbol, so it includes space on both sides. You want \lVert and \rVert for left and right delimiters, so that there is space on the "outside", but not on the "inside". $\endgroup$ – Arturo Magidin Jul 4 '12 at 2:33
  • $\begingroup$ @ArturoMagidin Thank you very much sir. $\endgroup$ – srijan Jul 4 '12 at 2:35
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    $\begingroup$ Related: math.stackexchange.com/a/38638/5798 $\endgroup$ – Rudy the Reindeer Jul 4 '12 at 5:07
  • $\begingroup$ @MattN. Thank you very much. That was helpful to me. $\endgroup$ – srijan Jul 4 '12 at 5:14

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 $x \in V$.

  • $\begingroup$ Thank you very much. I understand now. $\endgroup$ – srijan Jul 4 '12 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: $d(\cdot,\cdot)\to\|\cdot\|\to d(\cdot,\cdot)$ $\endgroup$ – C-Star-W-Star Sep 29 '14 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 '15 at 0:16

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.

  • $\begingroup$ It was nice explanation to me. I had forgot to thank you. :) $\endgroup$ – srijan May 25 '13 at 9:21

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

  • $\begingroup$ Thank you sir. I am fully satisfied with two answers given by you and Henry.:) $\endgroup$ – srijan Jul 4 '12 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 '12 at 6:27
  • $\begingroup$ @ThomasE. Sorry, you are right, I muddled up the words. Thanks. $\endgroup$ – Ragib Zaman Jul 4 '12 at 8:43

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: $$\|x\|'=\|x-0\|=\|x\|$$

  • $\begingroup$ @ Thanks for the answer. $\endgroup$ – srijan Feb 15 '16 at 4:15

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 '16 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 '18 at 8:19
  • $\begingroup$ Yes, why not? :) $\endgroup$ – AIM_BLB Oct 23 '18 at 15:33

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 > 1 $ which is a contradiction.


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