We know that the $l^p$ "norm" for $0<p<1$ is not a true norm. But what about the metric?

Question

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?

Answer 1

$d_p=\|x-y\|_p$ 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.

Answer 2

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.