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I have two functions {d1,d2}, where d1 satisfies all metric properties, but one: triangle inequality. However, the following holds:

d1(a,b) ≤ d2(a,c) + d1(b,c)

Can I say d1 is a kind of metric? Is there a formal definition for it?

I know a super metric (or ultra metric) is an example of a function that satisfies a stronger constraint of the triangle inequality, where d(a, b) ≤ max{d(a, c),d( b, c)}. So maybe there is also a formal definition for the case involving d1.

Sergio Mergen
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  • This is not clear. Are you making any assumptions at all about $d_2$? As stated...suppose I take $d_1(x,y)=0$ if $x=y$ and $=1$ otherwise. That passes your test, no? Now take $d_2(x,y)=3$ for all $(x,y)$ Unless I am missing the point, your inequality holds but... – lulu Oct 18 '18 at 19:27
  • In your example d1 satisfies triangle inequality. Assume d1(a,b) ≤ d1(a,c) + d1(b,c) does not hold, while d1(a,b) ≤ d2(a,c) + d1(b,c) does. For simplicity, assume d2 computes the largest possible value, regardless of x and y. Not really something useful, but I am just trying to understand if this arrangement qualifies as a (constraint) metric. – Sergio Mergen Oct 19 '18 at 15:19
  • Yes, I didn't mean for $d_1$ to be a metric. Let $d_1$ be any bounded function satisfying $M≥ d_1(x,y)≥0$ and $d_1(x,y)=0\iff x=y$. Then let $d_2$ be the constant $M$. I just don't see how your inequality tells us much. – lulu Oct 19 '18 at 15:25
  • d2 (as stated in the example) is a useless function, I agree. But it is just a simplification I used to formulate the question. What I have is two distinction functions that are not are metrics in the strict sense, but can be in conjunction to create a indexing scheme that is based on the triangle inequality property. – Sergio Mergen Oct 19 '18 at 17:06

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