A special metric in $\mathbb{R}^2$?

Question

Consider the following distance function in $\mathbb{R}^2$:

$d_L[(x_1,y_1),(x_2,y_2)]=\ln(1+|x_1-x_2|)+\ln(1+|y_1-y_2|)$.

I believe this is a metric on $\mathbb{R}^2$ since $f(x)=\ln(1+x)$ is a well-known metric-preserving function and because the sum of two metrics yields another metric. My only uncertainty is whether $|x_1-x_2|$ and $|y_1-y_2|$ are each metrics on $\mathbb{R}^2$ in their own right.

Is there an easier way to determine whether or not $d_L[(x_1,y_1),(x_2,y_2)]$ is a true metric on $\mathbb{R}^2$?

Not on the plane, for $x_1-x_2$ can be $0$ without having $(x_1,y_1)=(x_2,y_2)$. — André Nicolas, Sep 30 '15 at 17:03
I was commenting only on the part where OP wonders whether $|x_1-x_2|$ is a metric on $\mathbb{R}^2$. It isn't. — André Nicolas, Sep 30 '15 at 17:14
I can see where my argument fails. I wonder whether the distance function is nonetheless a true metric. — user275422, Sep 30 '15 at 17:16

score 0 · Answer 1 · answered Sep 30 '15 at 17:23

0

Hint: You need only prove triangle inequality. $$\ln(1+|x_1-x_2|)+\ln(1+|y_1-y_2|)=\ln((1+|x_1-x_2|)(1+|y_1-y_2|))$$ use this and $\ln x$ is an increasing function.

answered Sep 30 '15 at 17:23

R.N

4,318

Does this mean you have actually proven it to be a metric in $\mathbb{R}^2$? – user275422 Oct 03 '15 at 20:12

A special metric in $\mathbb{R}^2$?

1 Answers1