Proving Equivalence of Metrics Using Concavity

[MathematicalPhysicist]

I need to prove the following:
let p be a metric on X, and f:[0,infinity)->[0,infinity) s.t:
1.f(0)=0.
2.f is monotonically increasing.
3. f satisfy: f((a+b)/2)>=(f(a)+f(b))/2
prove that: f(p(x,y)) is a metric on X, and that that f(p(x,y)) and p(x,y) are equivalent, i.e that there exists reals: b>a>0 s.t a<=f(p(x,y))/p(x,y)<=b.

now to prove the first two consitions for metric is quite easy and i did it, but i find it a bit difficult to prove the triangle inequality, i have a feeling that 3's sign should <=, this way we do get the triangle inequality, am i right?
and concerning equivalence of metrics, basically if it's f((a+b)/2)<=(f(a)+f(b))/2, then
f(p(x,y))/p(x,y)<=2f(p(x,y)/2)/p(x,y)<=...<=2^nf(p(x,y)/2^n)/p(x,y), so for p(x,y) we get the maximum of the ratios, so i think that basically f(p(x,y))/p(x,y)<=f(1), don't know about the left inequality.

any hints?

as always your help is appreciated.

 

[MathematicalPhysicist]

anyone?

 

[gel]

3 seems fine as it is. It is just saying that f is concave. That is

$$f(pa+(1-p)b)\ge pf(a)+(1-p)f(b)$$

for any a,b and 0 < p < 1.
Then, if x,y > 0 you can set a = 0, b = x+y and p = x/(x+y)

$$f(y)\ge \frac{y}{x+y} f(x+y)$$

similarly, exchange the roles of x and y to get f(x) >= (x/(x+y))f(x+y) and add these inequalities to get f(x)+f(y)>=f(x+y). Then you can apply this to the triangle inequality.

btw, as you needed to prove f(x+y)<=f(x) + f(y) the easiest approach is probably to draw or imagine a graph of a concave function to see what this means, then formulate a rigorous argument based on the intuition gained (which is what I did).

For the second bit, its not true. The metrics won't be equivalent in the sense you state unless f has bounded derivative, which is false for $f(x)=\sqrt{x}$.

 

[MathematicalPhysicist]

thanks gel.

 

[MathematicalPhysicist]

btw, can you prove this by using the first definition of concavity that iv'e given (i know that they are equivalent but still I would like to see also a proof with the first definition).

thanks in advance.