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Show the following is a metric

  1. Mar 22, 2014 #1
    1. The problem statement, all variables and given/known data
    Let (X,d) is a metric space. Show that [itex]d_1=log(1+d)[/itex] is a metric space.


    3. The attempt at a solution
    (it's not stated what d is so I'm assumed d=|x-y|)
    I've checked positivity and symmetry but am having trouble with showing the triangle inequality holds. i.e. [itex]log(1+|x-y|) \leq log(1+|x-z|)+log(1+|y-z|)[/itex].

    It doesn't appear as though log(a+b)≤log(a)+log(b) is always true
     
  2. jcsd
  3. Mar 22, 2014 #2

    LCKurtz

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    You can't assume ##d(x,y)= |x-y|## or even that ##x## and ##y## are real numbers. But that doesn't really matter because ##d## satisfies the triangle inequality just like absolute values would. So if your presumed triangle inequality is written correctly it would be to show$$
    d_1(x,y) \le d_1(x,z) + d_1(z,y)$$which means$$
    log(1+d(x,y))\le log(1+d(x,z)) + log(1+d(z,y))$$Hint: Try exponentiating both sides.
     
  4. Mar 23, 2014 #3
    Thanks, this is what I have now:

    [itex](1+d(x,z))(1+d(z,y))=1+d(z,y)+d(x,z)+d(x,z)d(z,y)\geq 1+d(x,y)[/itex]

    Then taking logs of both sides

    [itex]log((1+d(x,z))(1+d(z,y))) \geq log(1+d(x,y))[/itex]
    [itex]log(1+d(x,z))+log(1+d(z,y)) \geq log(1+d(x,y))[/itex]
     
  5. Mar 23, 2014 #4

    LCKurtz

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    You have it pretty much figured out. Now how would you write up your final argument that, given that ##d(x,y)## is a metric, that implies ##d_1(x,y)## is? You want your argument in nice logical order, starting with what you are given and ending with what you wanted to prove. Do you see how to do that?
     
  6. Mar 23, 2014 #5
    This is what I'm thinking:

    Let x,y,z be points in X. Given a metric d on X, we're to show the function [itex]d_1=log(1+d)[/itex] is a metric on X.

    <verify first 2 properties>

    Consider [itex](1+d(x,z))(1+d(z,y))[/itex]. We have
    <do working to show>
    [itex](1+d(x,z))(1+d(z,y)) \geq 1+d(x,y)[/itex]

    Taking the natural logarithm of both sides gives

    [itex]log(1+d(x,z))+log(1+d(z,y)) \geq log(1+d(x,y))[/itex]
    [itex] \therefore d_1(x,z)+d_1(z,y) \geq log(x,y)[/itex]

    Hence, the function [itex]d_1[/itex] satisfies the triangle inequality.
     
  7. Mar 23, 2014 #6

    LCKurtz

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    But that last inequality isn't what you are trying to show. You are trying to show$$
    d_1(x,z)\le d_1(x,y)+ d_1(y,z)$$So your final writeup should begin$$
    d_1(x,z) = \log(1 + d(x,z))~...\text{string of inequalities here }...\le d_1(x,y)+d_1(y,z)$$

    [Edit] Fixed typo missing log
     
    Last edited: Mar 23, 2014
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