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Let $d,e$ be metrices on a set $X$ , then under what condition(s) the function $g:X \times X \to \mathbb R $ defined by $g(x,y):=\min \{d(x,y),e(x,y)\}$ , is a metric ?

Souvik Dey
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  • Do you recall the axioms that a metric must satisfy? – izœc Dec 04 '14 at 09:18
  • @izœc: yes .... – Souvik Dey Dec 04 '14 at 09:21
  • Ok - for example, the first is that $g(x,x) = 0$ for all $x\in X$. Do you know how you might show that [if you know that $d, e$ are metrics as well]? – izœc Dec 04 '14 at 09:23
  • @izœc: $g(x,x)$ is always zero if $d,e$ are known to be metrices , I already checked those trivial things my friend , also the symmetry .... the only thing troublesome is triangle inequality ... – Souvik Dey Dec 04 '14 at 09:27
  • Okay, well that is tricky.. in general it is not true (I think that there is a discussion of that here: https://math.stackexchange.com/questions/584155/minimum-of-two-metrics-is-metric). – izœc Dec 04 '14 at 09:42
  • However, you ask what conditions we can place that force $g$ to be a metric... it is clear that if $d = e$ we have that $g$ is a metric. I am considering whether any other weaker condition will have the same result. – izœc Dec 04 '14 at 09:43
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    Well an obvious weaker condition, which causes $g$ to be a metric is $e(x,y)\geq d(x,y)$ for all $x,y\in X$ (or $\leq$). – Dan Dec 04 '14 at 09:51
  • @izœc: $g(x,x)=\min {d(x,x),e(x,x) }=\min {0,0 }=0$ ; $g(x,y)=\min {d(x,y),e(y,x) }=\min {d(y,x),e(y,x)}=g(y,x)$ , am I wrong ??? – Souvik Dey Dec 04 '14 at 10:10
  • @SouvikDey No, what you have there is correct. – izœc Dec 04 '14 at 11:23
  • @Dan That is true of course; I was trying to think if there was something weaker in the sense of the condition not even being a global relation between the two $e$ and $d$. – izœc Dec 04 '14 at 11:25

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