- #1

chipotleaway

- 174

- 0

## Homework Statement

Given that f is continuous and strictly increasing, f:[0, ∞)->[0, ∞), f(0)=0, and d(x,y) is the standard metric on the real number line,

Is there a function f such that [itex]d'(x,y)=f(d(x,y))[/itex] is

**not**a metric on the real number line?

## The Attempt at a Solution

The standard metric firsts maps (x,y) in R

^{2}to |x-y|, which the function f then takes and maps it to a new 'distance', which preserves order since f is strictly increasing. As far as I can tell, the new metric d'(x,y) satisfies the positivity and symmetry properties of metrics as far as I can tell so I think I need to look at the triangle inequality:

Either showing d'(x,y)≤d'(x,z)+d(z,y) holds, i.e. equivalently, f(|x-y|)≤f(|x-z|)+f(|z-y|)) for all functions satisfying the conditions, or finding an example of such a function where it fails. If the function is strictly increasing - finding f where the triangle doesn't hold seems like it might be pretty tough.