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Thread: Proving a function is a metric

  1. #1
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    Proving a function is a metric

    Hello.

    I have the following function:

    $\displaystyle d_1(x,y)=d(x,y)/(1+d(x,y))$

    where $\displaystyle d(x,y)$ is already a metric on some set $\displaystyle X$ and I am asked to prove that $\displaystyle d_1(x,y)$ is also a metric on $\displaystyle X$.

    It is easy enough to show that $\displaystyle d_1(x,y)$ is symmetric and positive definite, however, I am having problems with showing that $\displaystyle d_1(x,y)$ satisfies the triangle identity.

    I have tried a couple of different things (lots of algebra, transforming into a geo series, etc...) but either I keep on making mistakes or I am heading down the wrong path because I can never get it to reduce the way I want it too.

    Any hints would be very much appreciated.

    Thanks.
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  2. #2
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    Quote Originally Posted by mpetnuch View Post
    Hello.

    I have the following function:

    $\displaystyle d_1(x,y)=d(x,y)/(1+d(x,y))$
    We want to show,
    $\displaystyle d_1(x,y)\leq d_1(x,z)+d_1(z,y)$
    Iff,
    $\displaystyle \frac{d(x,y)}{1+d(x,y)} \leq \frac{d(x,z)}{1+d(x,z)}+\frac{d(y,z)}{1+d(y,z)}$
    Iff,
    $\displaystyle d(x,y)(1+d(x,z))(1+d(y,z))$$\displaystyle \leq d(x,z)(1+d(x,y))(1+d(y,z))+d(y,z)(1+d(x,y))(1+d(x, z))$
    Let $\displaystyle A=d(x,y), \ B=d(x,z) , \ C=d(y,z)$.
    Thus,
    $\displaystyle A(1+B)(1+C)\leq B(1+A)(1+C)+C(1+A)(1+B)$
    Multiply,
    $\displaystyle A+AB+AC+ABC\leq B+BC+AB+ABC+C+AC+BC+ABC$
    Thus,
    $\displaystyle A\leq (B+C) + (2BC+ABC)$
    This is true because,
    $\displaystyle A\leq B+C$ by triangle inequality.
    And you are adding $\displaystyle 2BC+ABC$ a non-negative quantity thus it does not decrease.
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  3. #3
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    Thanks! I originally tired that but for some reason I didn't think to reuse the original T.I.!
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