Metric function composed with concave function

[shoescreen]

Hi,

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 d, and an increasing, concave function f:\mathbb{R} \rightarrow \mathbb{R} so that f(0)=0, show that f\circ d is a metric.

Of course, only the triangle inequality is nontrivial.

 

[morphism]

Hint: First show that f(tx) >= tf(x).

 

[trickycheese1]

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?

 

[micromass]

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?

Did you first show that mathman's hint is correct??

 

[trickycheese1]

Did you first show that mathman's hint is correct??

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.

 

[micromass]

Now write

f(a)+f(b)=f\left((a+b)\frac{a}{a+b}\right)+f\left((a+b)\frac{b}{a+b}\right)

Apply the hint with t = the fractions.