Does $d(x,y) = \log(1 + | x - y |)$ define a metric on $\mathbb{R}$? Help would be much appreciated here! I got as far as evaluating the following:


*

*$d(x,y) \geq 0$. Argument is that $(1 + |x - y|)$ doesn't produce negative values, since the $|x-y| \geq 0$ itself and, at most, it would be added to $1$. So this condition is satisfied!

*$d(x,y) = d(x,y)$. Since we have that $\log(1 + |x - y|) = \log(1 + |y - x|)$ by commutative of real numbers.
Now, here is where I am stuck. How can I demonstrate this algebraically?


*

*$d(x,y) = 0$ implies that $x = y$.


What I could do is $\log(1+|x - y|) = 0$. It seems that the only way this log can be zero, is if $1 + |x - y| = 1$. Therefore, it would $|x - y| = 0$ would rise naturally. Does it make sense?


*

*$d(x, y) \leq d(x,z) + d(z,y)$. What is left is add a zero to the inside of the modulo function and then try and break it into the triangle inequality. That is not working so well thus far, but here it goes: 
$\log(1 + |x - y|) = \log(1 + |x - z + z - y|)$ but then it turns out that:
$\log(1 + |x - z + z - y|) \leq \log(1 + |x - z| + |z - y|)$
 A: Yes, your proof that $d(x,y)=0$ implies $x=y$ makes sense. All you need is the fact that the $\log(s)$ function on $s > 0$ has only one zero, and that zero is $s=1$.  Therefore, if $\log(s) = 0$, then $s$ can only be equal to $1$, hence if $1 = s = 1 + |x-y|$, then $|x-y|=0$, which implies $x=0$.
As for the triangle inequality, ask, for a general function $f(s), s > 0$, how the condition
$$
f(a+b) \leq f(a) + f(b)
$$
relates to the concavity of that function.  Examine a few examples: e.g., the convex function $f(s) = s^2$, the concave function $f(s)=s^{1/3}$, the linear function $f(s) = 7s$.
And note that, for nonnegative $a, b$,
$$
\log(1 + a + b) \leq \log(1 + a + 1 + b).
$$
A: The $d\left(x,y\right)=0$ proof is correct. For the triangle inequality, I would start out with the right hand side. Notice that
\begin{align*}
\log\left(1+\left|x-z\right|\right)+\log\left(1+\left|z-y\right|\right)&=\log\left(\left(1+\left|x-z\right|\right)\left(1+\left|z-y\right|\right)\right)\\
&=\log\left(1+\left|x-z\right|+\left|z-y\right|+\left|x-z\right|\left|z-y\right|\right).
\end{align*}
Recall that for all $a,b>0$, $\log\left(a\right)\ge\log\left(b\right)\iff a\ge b$. By the triangle inequality, $\left|x-y\right|\le\left|x-z\right|+\left|z-y\right|$, and so
\begin{align*}
1+\left|x-y\right|&\le1+\left|x-z\right|+\left|z-y\right|\\
&\le1+\left|x-z\right|+\left|z-y\right|+\left|x-z\right|\left|z-y\right|.
\end{align*}
Thus,
\begin{align*}
d\left(x,y\right)=\log\left(1+\left|x-y\right|\right)&\le\log\left(1+\left|x-z\right|+\left|z-y\right|+\left|x-z\right|\left|z-y\right|\right)\\
&=\log\left(1+\left|x-z\right|\right)+\log\left(1+\left|z-y\right|\right)\\
&=d\left(x,z\right)+d\left(z,y\right).
\end{align*}
A: You have the right idea about how to show that $d(x,y)=0$ implies $x=y$.
Hints for the triangle inequality:


*

*\begin{aligned}d(x,z)+d(z,y)&=\log(1+|x-z|)+\log(1+|z-y|) \\&= \log(1+|x-z|+|z-y|+|x-z||z-y|)
\end{aligned}

*The function $d': \mathbb R \times \mathbb R \to \mathbb R$ given by $d'(x,y)=|x-y|$ is a metric on $\mathbb R$.

*$\log(x)$ is nondecreasing on $[1, \infty)$.
A: It is obvious that $1+a+b \leq (1+a)(1+b)$ for $a, b \geq 0$. Hence $\log(1+a+b) \leq \log(1+a)+\log(1+b)$. This and the fact that $\log$ is increasing show that for any metric $d$, $d'(x,y)=\log(1+d(x,y))$ is also a metric. 
