If $0<\alpha < 1$ and $d(x,y)$ is metric, show that $((d(x,y))^{\alpha}$ is metric.

Question

If $0<\alpha < 1$ and $d(x,y)$ is metric, show that $((d(x,y))^{\alpha}$ is metric.

It remains for me to show that the triangle inequality, or $((d(x,y))^{\alpha}\leq ((d(x,z))^{\alpha}+((d(z,y))^{\alpha}$, holds.

I begin with the fact that $d(x,y)\leq d(x,z)+d(z,y)$. Then I apply the function:

$(d(x,y))^{\alpha}\leq (d(x,z)+d(z,y))^{\alpha}$

What now? I was thinking of expanding the right side but maybe it is not the best idea? I got a hint where if $a\geq 0, b\geq 0$ and $a+b=1$, then $a^{\alpha}+b^{\alpha}\geq 1$ but I don't really know how to apply this hint.

Answer 1

If you plot a graph, you can clearly see that when $a+b=1$, $a^\alpha+b^\alpha\geq 1$, because the curve $a^\alpha+b^\alpha= 1$ is "concave".

You want to prove $(d(x,y))^{\alpha}\leq (d(x,z))^{\alpha}+(d(z,y))^{\alpha}$. This can be done by noting that $(d(x,y))^{\alpha}\leq (d(x,z)+d(z,y))^{\alpha}$. Now you can let $a=\frac{d(x,z)}{d(x,z)+d(z,y)}, b=\frac{d(y,z)}{d(x,z)+d(z,y)}$, so $a+b=1$, $$ \left(\frac{d(x,z)}{d(x,z)+d(z,y)}\right)^\alpha+\left(\frac{d(y,z)}{d(x,z)+d(z,y)}\right)^\alpha=\frac{(d(x,z))^{\alpha}+(d(z,y))^{\alpha}}{(d(x,z)+d(z,y))^\alpha}\geq 1.\\ \Rightarrow (d(x,z))^{\alpha}+(d(z,y))^{\alpha}\leq (d(x,z)+d(z,y))^\alpha\leq (d(x,z)+d(z,y))^{\alpha} \leq (d(x,y))^{\alpha}, $$ as required.

Answer 2

Let $S = d(x,z) + d(z,y)$. By dividing both sides of the inequality you had by $S^\alpha$, you get $$ \frac{(d(x,y))^\alpha}{S^\alpha} \leq \left( \frac{d(x,z)}{S} + \frac{d(z,y)}{S} \right)^\alpha \leq \left( \frac{d(x,z)}{S} \right)^\alpha + \left( \frac{d(z,y)}{S} \right)^\alpha , $$ where we use the fact that $ \frac{d(x,z)}{S} + \frac{d(z,y)}{S} = 1$. Multiplying again by $S^\alpha$ gives you the result you need.

Answer 3

Basically, you have to prove that $$ x^\alpha+y^\alpha\ge (x+y)^\alpha $$ for $x,y\ge 0$. If $y=0$ this is trivial. Assume it is not and divide by $y^\alpha$. Calling $x/y=z$ you get the equivalent inequality $$ z^\alpha+1\ge (1+z)^\alpha $$ which must be true for $z\ge 0$. Consider the function $$ f(z)= z^\alpha+1-(1+z)^\alpha $$ You should prove $f(z)\ge 0$. But $f(0)=0$ and $f(\infty)=0$. You can easily check that $f$ has only one critical point on $(0,\infty)$ and it is a local maximum. Therefore, $f$ is never negative.