# Showing $d_1(x,y)=|x-y|$ and $d_2=\sqrt{|x-y|}$ for $x,y\in\mathbb{R}$ define equivalent metrics

This isn't homework, I am just trying to better understand how to show two functions define the same metric.

I would like to show that $d_1(x,y)=|x-y|$ and $d_2=\sqrt{|x-y|}$ for $x,y\in\mathbb{R}$ define equivalent metrics. I know that if there exist positive real numbers $a,b$ such that $ad_1(x,y)\leq d_2(x,y)\leq bd_1(x,y)$, that is sufficient to show they are equivalent, but I can't seem to find such constants.

Do constants $a$ and $b$ exist to satisfy the above inequality, or is this a case where one must look at the open balls generated by $d_1$ and $d_2$? Thanks for your help.

• No. $\:$ (i.e., such constants do not exist, and this is not a case where one must look at the open balls.) $\hspace{.21 in}$ One should show that both change-the-metric maps are Hölder. $\;\;\;\;$ – user57159 Nov 12 '15 at 21:36
• The metrics generate the same topology but they are not equivalent metrics in the sense described above. – copper.hat Nov 12 '15 at 21:38
• @copper.hat equivalent metrics means that they generate the same topology. The constants condition means they are Lipschitz equivalent which is stronger. – Henno Brandsma Nov 12 '15 at 21:39
• @HennoBrandsma: Thanks, I have seen various definitions, I edited my comment to clarify what I meant. – copper.hat Nov 12 '15 at 21:39
• With the constants, that is often called "strongly equivalent". – zhw. Nov 12 '15 at 21:40