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I'm to check if:

$$p(x,y) = d(x, y) \cdot r(x,y)$$

is a metric, where $d(x, y)$, $r(x, y)$ are metrics and $x, y \in X$.

It is easy to prove for all $x, y \in X$ that:
1) $p(x,y) \ge 0$
2) $p(x,y) = 0 \iff x = y$
3) $p(x, y) = p(y, x)$

However it's not so easy to check the triangle inequality:

$$ p(x, y) \le p(x, z) + p(z, y)\;\text{ for }x, y, z \in X $$

I would appreciate any help!

hardmath
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Hendrra
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    I think maybe it is false, and you can disprove with a counterexample. Take $d$ to be the usual Euclidean metric, and $r$ the "taxi-cab" metric. Try to find three points so that the triangle inequality doesn't work for $p$. – Nick Mar 14 '17 at 19:42

1 Answers1

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let $X$ be $\mathbb{R}$ and $d(x,y) = r(x,y) = |x-y|$. Suppose that $p(x,y) \leq p(x,z) +p(z,y) \implies d(x,y)^2 \leq d(x,z)^2 + d(z,y)^2 $. Choose $x=1, y=0$ and $z=\frac{1}{2}$. So we have $1^2 \leq (\frac{1}{2})^2 + (\frac{1}{2})^2 = \frac{1}{2}$ that is a contradiction. So $p$ is not necessarily a metric.

Danuso Rocha
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