# Show the following is a metric

## Homework Statement

Let (X,d) is a metric space. Show that $d_1=log(1+d)$ 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. $log(1+|x-y|) \leq log(1+|x-z|)+log(1+|y-z|)$.

It doesn't appear as though log(a+b)≤log(a)+log(b) is always true

LCKurtz
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## Homework Statement

Let (X,d) is a metric space. Show that $d_1=log(1+d)$ 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. $log(1+|x-y|) \leq log(1+|x-z|)+log(1+|y-z|)$.

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.

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Thanks, this is what I have now:

$(1+d(x,z))(1+d(z,y))=1+d(z,y)+d(x,z)+d(x,z)d(z,y)\geq 1+d(x,y)$

Then taking logs of both sides

$log((1+d(x,z))(1+d(z,y))) \geq log(1+d(x,y))$
$log(1+d(x,z))+log(1+d(z,y)) \geq log(1+d(x,y))$

LCKurtz
Science Advisor
Homework Helper
Gold Member
Thanks, this is what I have now:

$(1+d(x,z))(1+d(z,y))=1+d(z,y)+d(x,z)+d(x,z)d(z,y)\geq 1+d(x,y)$

Then taking logs of both sides

$log((1+d(x,z))(1+d(z,y))) \geq log(1+d(x,y))$
$log(1+d(x,z))+log(1+d(z,y)) \geq log(1+d(x,y))$

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?

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 $d_1=log(1+d)$ is a metric on X.

<verify first 2 properties>

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

Taking the natural logarithm of both sides gives

$log(1+d(x,z))+log(1+d(z,y)) \geq log(1+d(x,y))$
$\therefore d_1(x,z)+d_1(z,y) \geq log(x,y)$

Hence, the function $d_1$ satisfies the triangle inequality.

LCKurtz
Science Advisor
Homework Helper
Gold Member
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 $d_1=log(1+d)$ is a metric on X.

<verify first 2 properties>

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

Taking the natural logarithm of both sides gives

$log(1+d(x,z))+log(1+d(z,y)) \geq log(1+d(x,y))$
$\therefore d_1(x,z)+d_1(z,y) \geq log(x,y)$

Hence, the function $d_1$ 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$$d_1(x,z) = \log(1 + d(x,z))~...\text{string of inequalities here }...\le d_1(x,y)+d_1(y,z)$$

 Fixed typo missing log

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