Show the following is a metric

[chipotleaway]

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]

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.

 

[chipotleaway]

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]

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?

 

[chipotleaway]

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]

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

[Edit] Fixed typo missing log