Suppose there are two metric spaces $d_1$ and $d_2$ over the set $X$. For $x,y \in X$, is $d_3(x,y) =\sqrt{d_1(x,y)d_2(x,y)}$ a metric space?

I am having trouble with the triangle inequality. It is enough to show that the triangle inequality holds for $d_1(x,y)d_2(x,y)$ since the result will follow by Jensen's inequality.

Furthermore, is $(\prod_{i=1}^p d_i(x,y))^{\frac{1}{p}}$ a metric space?

  • $\begingroup$ How can two metrics be combined into a new metric space so that they equally contribute despite differences in scaling. I think this is what the geometric mean does. Would you use the p-norms? $\endgroup$ – Bobolan Jun 24 '14 at 4:44
  • $\begingroup$ The term you want is "metric," not "metric space." "Metric space" refers to the pair $(X, d)$ of a set and a metric on it. $\endgroup$ – Qiaochu Yuan Jun 24 '14 at 5:42

No. For example, let $0 < \lambda < \frac12$ and set \begin{align*} d_1(a,b) &= \lambda & d_2(a,b) &= 1-\lambda \\ d_1(b,c) &= 1-\lambda & d_2(b,c) &= \lambda \\ d_1(a,c) &= 1 & d_2(a,c) &= 1 \end{align*} (Think of $a,b,c$ as collinear points, with $b$ between $a$ and $c$.) Then the geometric mean of distances violates the triangle inequality: \begin{align*} \sqrt{d_1(a,b)d_2(a,b)} &= \sqrt{\lambda(1-\lambda)} < \tfrac12 \\ \sqrt{d_1(b,c)d_2(b,c)} &= \sqrt{\lambda(1-\lambda)} < \tfrac12 \\ \sqrt{d_1(a,c)d_2(a,c)} &= 1 \end{align*} (The inequalities are AM/GM, with strict inequality because $\lambda\ne 1-\lambda$.)

(There is a kind of reverse inequality.)


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