Show the following is a metric

  • #1
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Homework Statement


Let (X,d) is a metric space. Show that [itex]d_1=log(1+d)[/itex] is a metric space.


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
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


Let (X,d) is a metric space. Show that [itex]d_1=log(1+d)[/itex] is a metric space.


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

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.
 
  • #3
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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]
 
  • #4
LCKurtz
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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]

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?
 
  • #5
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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.
 
  • #6
LCKurtz
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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.

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:

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