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For what $p$ does $|x-y|^p$ define a metric on $\mathbb{R}$?

One can show that $d_1 = \sqrt{|x-y|}=|x-y|^\frac{1}{2}$ defineds a metric. All requirements can be easily checked. On the other hand, $d_2 = |x-y|^2$ does not define a metric. For example,

$$|1-0|^2=1\text{ , but } |1-0.5|^2+|0.5-0|^2=0.5$$

I am curious for what $p$ does $|x-y|^p$ define a metric? I suspect that if $p>1$ then it cannot be a metric since we can use the similar arguments using $0,0.5$ and $1$.

Martin Sleziak
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3x89g2
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1 Answers1

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The only thing we need to check is the triangle inequality, i.e. for what $p$, do we have \begin{align} |x-y|^p \leq |x-z|^p+|z-y|^p. \end{align} It's clear that for $p>1$ the above inequality fails because \begin{align} 2^p=|1-(-1)|^p< |1-0|^p+|0-(-1)|^p = 2. \end{align} Suppose $0<p\leq 1$, then we see the triangle iequality does hold. Let us prove it. Observe \begin{align} |x+y| \leq (|x|+|y|)= (|x|^{p/p}+|y|^{p/p}) \leq (|x|^p+|y|^p)^{1/p} \end{align} which means \begin{align} |x+y|^p \leq |x|^p+|y|^p. \end{align} Lastly, for $p<0$, we see that $1/|x-y|^{|p|}$ doesn't satisfies the condition $d(x, x) = 0$.

user2820579
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Jacky Chong
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  • How do we get $(|x|^{p/p} + |y|^{p/p})\le (|x|^p + |y|^p)^{1/p}$? – 3x89g2 Oct 16 '16 at 15:01
  • @3x89g2 The relation $(|x|^{p/p} + |y|^{p/p}) < (|x|^p + |y|^p)^{1/p}$ just states that the 1-norm is less than the p-norm, for $p\le 1$. This is true since in general, $p$-norms are decreasing with $p$, see https://math.stackexchange.com/questions/218046/ – Andreas Oct 09 '18 at 18:06