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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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$\LaTeX$ 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". –  Arturo Magidin Jul 4 '12 at 2:33
    
@ArturoMagidin Thank you very much sir. –  srijan Jul 4 '12 at 2:35
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Related: math.stackexchange.com/a/38638/5798 –  Matt N. Jul 4 '12 at 5:07
    
@MattN. Thank you very much. That was helpful to me. –  srijan Jul 4 '12 at 5:14
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3 Answers

up vote 20 down vote accepted

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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Thank you very much. I understand now. –  srijan Jul 4 '12 at 2:42
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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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It was nice explanation to me. I had forgot to thank you. :) –  srijan May 25 '13 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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Thank you sir. I am fully satisfied with two answers given by you and Henry.:) –  srijan Jul 4 '12 at 2:44
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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..? –  Thomas E. Jul 4 '12 at 6:27
    
@ThomasE. Sorry, you are right, I muddled up the words. Thanks. –  Ragib Zaman Jul 4 '12 at 8:43
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