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The proof for why $l^p$ is a metric space doesn't ever use any property of p being greater than 1. On top of that there is no counterexample for the metric, like there is for the norm. The metric is defined as

$d_p=\|x-y\|_p=\left(\sum_{i=1}^n|x_i-y_i|^p \right)^{1/p}$

Where $n$ is finite, (obviously this wont hold if we permit infinite sequences).

Is there a proof/counterexample that this is a metric/not?

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    $\begingroup$ You can get proper norm bars by using \| instead of ||. $\endgroup$ – joriki Sep 15 '18 at 6:25
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$d_p=\|x-y\|_p$ doesn't define a metric basically for the same reason it doesn't define a norm: Counterexamples to the subadditivity of the potential norm also afford counterexamples to the triangle inequality for the potential metric; e.g.

$$ \left(1^{\frac12}+1^{\frac12}\right)^2\gt\left(1^{\frac12}\right)^2+\left(1^{\frac12}\right)^2 $$

shows that the triangle inequality is violated for the canonical unit vectors in two dimensions.

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    $\begingroup$ +1. Just to spell it out a little more: Subadditivity of two vectors $x$ and $y$, $$\|x+y\|\le\|x\|+\|y\|,$$ is equivalent to the triangle inequality between $0$, $x$, and $-y$, $$d(x,-y)\le d(0,x)+d(0,-y).$$ $\endgroup$ – Rahul Sep 15 '18 at 7:02
  • $\begingroup$ @Rahul: Thanks, I think that's helpful. $\endgroup$ – joriki Sep 15 '18 at 7:16
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    $\begingroup$ Oops, I meant to write $d(x,0)$ as the first term on the right. Too late to edit. $\endgroup$ – Rahul Sep 15 '18 at 7:37
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    $\begingroup$ Thanks for the replies. Yes that actually makes much more sense. In case anyone was wondering it becomes a metric only when you raise to the p-th power $d_p=\| x-y\|^p$ $\endgroup$ – user593295 Sep 15 '18 at 20:12
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The norm is equivalent to its associated metric applied when one vector is the zero vector (or one point is the origin, if you prefer).

So any property the metric has, the norm has, and contrapositively, any property the (candidate) norm does not have, the (candidate) metric also does not have.

You can figure out what that means for this case.

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  • $\begingroup$ This is correct for (candidate) metrics defined in terms of (candidate) norms; but it's perhaps worth mentioning that this does not imply a one-to-one corresponence between metrics and norms: There are metrics that aren't defined in terms of a norm and don't correspond to a norm (and in fact, unlike norms, metrics aren't restricted to vector spaces); a metric on a vector space corresponds to a norm exactly if it's homogeneous and translation invariant (see e.g. Wikipedia). $\endgroup$ – joriki Sep 15 '18 at 8:30
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    $\begingroup$ @joriki : Tweaked. $\endgroup$ – The_Sympathizer Sep 15 '18 at 8:36

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