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Problem:

Suppose $d$ and $e$ are metrics on a set $X$. Let $g$ be the function $(x,y) \to min\{d(x,y),e(x,y)\}$ defined on $X × X$. Find a condition under which $g$ is a metric.

My solution: If $d(x,y) = ke(x,y)$, where $k$ is a positive constant, then $g$ will be a metric. Seems to be trivial. Is there any other possibility?

Related question here: Is the minimum of two metrics is again a metric?

gt6989b
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Vinod
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2 Answers2

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You found an (obviously very strong) sufficient condition. Can you make it weaker? Can you find a necessary and sufficient one?

HINT

To be a metric, $g$ has to map $X \times X \to [0,\infty)$ (which it obviously is) and satisfy some properties. Most of the properties, except one, will carry over directly from the definition.

On the last one, you have to think which are the least restrictive condition(s) to impose on $d$ and $e$ for $g=\min\{d,e\}$ to remain a metric still.

gt6989b
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  • My guess : e = 1/(1+d) – Vinod Jun 06 '18 at 15:15
  • @Vinod do not guess, think. This is very restrictive. Which of the properties work without additional conditions, and which don't? – gt6989b Jun 06 '18 at 15:16
  • Suppose e is a polynomial in d. – Vinod Jun 06 '18 at 15:21
  • Can we say that if $e(x,y)=1$ then $g$ is a metric? Is this condition necessary and sufficient ? – argiriskar Oct 13 '18 at 15:14
  • @argiriskar i don't think $e(x,y) \equiv 1 \ \forall x,y$ is a metric, but even if it was, clearly this could not be necessary -- OP's example certainly works for a much wider class of $e(x,y)$... – gt6989b Oct 14 '18 at 13:59
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Consider $\|\ \|_1$ on $\mathbb{R}^2$. Its unit ball is a convex hull of $$ (\pm 1,0),\ (0,\pm 1) $$ We will find two metrics $d_i$ whose infimum is $\| \ \|_1$.

Step 1 : This convex set is a union of two convex sets $A,\ B$ : $A$ is a convex hull of $$ (\pm 1,0),\ (\pm 1/2,1/2),\ (\pm 1/2,-1/2) $$

and $B$ is a convex hull of $$ (0,\pm 1) ,\ (\pm 1/2,1/2),\ (\pm 1/2,-1/2) $$

Here we have norms $d_i$ whose unit balls are $A,\ B$.

Step 2 : Then $\|\ \|_1$ is infimum of $d_i$.

HK Lee
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