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.

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}

- Serverfault Help
- Superuser Help
- Ubuntu Help
- Webapps Help
- Webmasters Help
- Programmers Help
- Dba Help
- Drupal Help
- Wordpress Help
- Magento Help
- Joomla Help
- Android Help
- Apple Help
- Game Help
- Gaming Help
- Blender Help
- Ux Help
- Cooking Help
- Photo Help
- Stats Help
- Math Help
- Diy Help
- Gis Help
- Tex Help
- Meta Help
- Electronics Help
- Stackoverflow Help
- Bitcoin Help
- Ethereum Help