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Metrics math problem help

  1. Jan 24, 2008 #1
    I need to prove the following:
    let p be a metric on X, and f:[0,infinity)->[0,infinity) s.t:
    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.
  2. jcsd
  3. Jan 26, 2008 #2
  4. Jan 26, 2008 #3


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    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 [itex]f(x)=\sqrt{x}[/itex].
  5. Jan 27, 2008 #4
  6. Jan 27, 2008 #5
    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.
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