How to prove this is a metric?

Question

I have 2 distance functions $d(x,y)=|x^2-y^2|$ and $d(x,y)=|x^3-y^3|$ and I am trying to prove that they are metrics on $\mathbb R$, or give a counterexample that they are not metrics on $\mathbb R$.

I managed to prove the first 3 properties for the 2nd function namely:

1. $d(x,y)\ge0$

2. $d(x,y)=0$ iff $x=y$

3. $d(x,y)=d(y,x)$

and I now have shown that the 1st function is not a metric.

But for the triangle inequality property, $d(x,y)\le d(x,z) + d(z,y)$, Im stuck on it. How would go about proving it for $d(x,y)=|x^3-y^3|$? Thanks.

Answer 1

You can think about it in two ways, both following from Lopsy's hint. I'm not writing any new mathematics here, it's just a longer comment.

You can note that the function $f(x) = x^3$ is a homeomorphism of $\mathbb{R}$. Actually all you need is that it is injective, but using the fact that it is homeo one can show that the new metric leads to the same topology as the standard metric. Anyway - if one has a metric $d_Y$ on $Y$ and and injective function $g:X \to Y$ then the function $d_X:X\times X \to \mathbb{R}$ defined by $d_X(x,y) = d_Y(g(x), g(y))$ is a metric on $X$. It is an easy excercise and you don't have to worry about any special properties of the cubic function. In our case: $d_\mathbb{R}(x,y) = |x^3 - y^3|$ is taking the metric from $Y=f(\mathbb{R})$ to $X=\mathbb{R}$ with function $g=f$: $$d_\mathbb{R}(x,y) = d_{f(\mathbb{R})}(f(x), f(y)),$$ where $d_{f(\mathbb{R})}(a,b) = |a-b|$.

The second way - identical from the mathematical point of view but maybe useful for imagination - is looking at $d(x,y)$ as a metric on the graph of function $f$ defined as above. One projects the graph on the image and takes the metric from the image. It is important that the projection does not glue any points (because $f$ is injective).

Using this language - the first function $d(x,y)=|x^2-y^2|$ is not a metric because $f$ is not injective and glues $a$ and $-a$. When you thought about positive halfline then it was injective and everything was OK.

Answer 2

I'll write up an answer, so that it is not on the Unanswered Category:

A function $d: X \times X \to \mathbb R$ is said to be a metric on $X$ if:

1. (Non-negativity) $d(x,y) \ge 0$ for all $x,y \in X$
2. (Definiteness) $d(x,y) =0 \iff x=y$
3. (Symmetry) $d(x,y)=d(y,x)$ for all $x,y \in X$
4. (Triangle Inequality) $d(x,y) \le d(x,z)+d(z,y)$ for all points $x, y, z \in X$

Consider the following functions $d$ as metrics on $\mathbb R$.

• $d(x,y)=|x^2-y^2|$

It is NOT a metric. Note that "definiteness axiom" fails here: $d(1,-1)=0$ but obviously, $1 \neq -1$

• $(d(x,y)= |x^3-y^3|$

It is a metric. Note that the triangle inequality follows from that on the Euclidean Metric on $\mathbb R$.