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Metric function composed with concave function

  1. Jan 9, 2012 #1

    I have been reading about metric spaces and came across an elementary property that I am having difficulty proving. A quick search on these forums and google has also failed.

    Given a metric space with distance function [itex]d[/itex], and an increasing, concave function [itex]f:\mathbb{R} \rightarrow \mathbb{R}[/itex] so that [itex]f(0)=0[/itex], show that [itex]f\circ d[/itex] is a metric.

    Of course, only the triangle inequality is nontrivial.
  2. jcsd
  3. Jan 11, 2012 #2


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    Hint: First show that f(tx) >= tf(x).
  4. Jan 20, 2012 #3
    I have been struggling with this problem all day so I described it in a google search and found this forum.

    I have that f(d(x,y)) <= f(d(x,z)+d(z,y)) but I hit a brick wall when I try to "free" the d's out of the function, i.e. I get for example that f(d(x,z)+d(z,y)) >= (d(x,z)+d(z,y))*f(1) =d(x,z)*f(1) + d(z,y) * f(1) <=f(d(x,z)+f(d(z,y)), but that's worthless because the inequalities go back and forth.

    I also tried putting f(d(x,z)+d(z,y)) = f((a+b)(d(x,z)+d(z,y))) = f(a*(d(x,z)+d(z,y))+b*(d(x,z)+d(z,y))) >= f(a*d(x,z)+b*d(z,y)) >= a*f(d(x,z)) + b*f(d(z,y)) but that doesn't give me anything useful.

    Can anyone give another tip how I should be thinking about this problem?
  5. Jan 20, 2012 #4
    Did you first show that mathman's hint is correct??
  6. Jan 21, 2012 #5
    Yes, I put y=0 in the equation f(ax + by) >= af(x) + bf(y), and in my calculations I tried to apply the hint but it didn't get me anywhere.
  7. Jan 21, 2012 #6
    Now write


    Apply the hint with t = the fractions.
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