Another way to prove $d(x,y)$ is a metric

by DavidS   Last Updated October 16, 2018 21:20 PM

Today on an exam, I was tasked with proving that for all $x,y$ in $R_+^n$, $d(x,y)=\sqrt{\sum_{i=1}^n(\sqrt{x_i}-\sqrt{y_i})^2 }$ is a metric. It's pretty clear that $d(x,y)=0$ iff $x=y$, and also that $d(x,y)=d(y,x)$, but when it came to verifying the triangle inequality, I did not know what to do.

In the end, I resorted to the fact that if $d(x,y)$ is a metric and $f(x)$ is an injective function over the domain of $x$, then $d(f(x),f(y))$ is a metric as well. In this case, the metric I was told to prove is similar in form to the Euclidean metric, where $f(x)=\sqrt{x}$ is the injective function.

This may seem totally fine, except for the fact that we were never taught this theorem in class, so I doubt that we were expected to use it.

Can someone show me another way to prove that $d(x,y)$ is a metric other than using the aforementioned theorem? If not, a proof of the theorem I used would also be greatly appreciated.



Answers 1


You can do it as follows:\begin{align}d\bigl((x_1,\ldots,x_n),(z_1,\ldots,z_n)\bigr)&=\sqrt{\sum_{k=1}^n\left(\sqrt{x_k}-\sqrt{z_k}\right)^2}\\&=\left\|\left(\sqrt{x_1},\ldots,\sqrt{x_n}\right)-\left(\sqrt{z_1},\ldots,\sqrt{z_n}\right)\right\|_2\\&\leqslant\left\|\left(\sqrt{x_1},\ldots,\sqrt{x_n}\right)-\left(\sqrt{y_1},\ldots,\sqrt{y_n}\right)\right\|_2+\\&\phantom{=}+\left\|\left(\sqrt{y_1},\ldots,\sqrt{y_n}\right)-\left(\sqrt{z_1},\ldots,\sqrt{z_n}\right)\right\|_2\\&=\sqrt{\sum_{k=1}^n\left(\sqrt{x_k}-\sqrt{y_k}\right)^2}+\sqrt{\sum_{k=1}^n\left(\sqrt{y_k}-\sqrt{z_k}\right)^2}.\end{align}

José Carlos Santos
José Carlos Santos
October 16, 2018 21:16 PM

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