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im studying for an upcoming test and doing some exercises from my book (linear operator theory in engineering and science) and came around this exercise that i can't solve:\

with $d_p$ given in exercise 1 show that $d_1$ is equivalent to $d_p$ on $R^2$ (i'll write $d_p$ so that you guys can see better). $$d_p(x,y) = \sqrt[p]{|x_1 - y_1|^p + |x_2 - y_2|^p }$$

to be honest i don't really know where to start. i've tried everything that comes to my head but the only solution that i can think of is making: $$ p = 1 $$ that is because $d_1 = |x_1 - y_1| + |x_2 - y_2|$. but i dont think that's the solution, can someone put me in the right path?. Thanks in advance

EDIT: corrected some typos due to translation errors

Eduardo Gutierrez
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  • Showing that two metrics are equivalent means that they give the same metric space. It therefore suffices to prove that open sets in the $d_1$ metric are open in the $d_p$ metric and vice versa. It also suffices that any open ball in the $d_1$ metric is a subset of an open ball in the $d_p$ metric and vice versa. Does this clarify what you have to do? – Student Nov 07 '21 at 21:53
  • Are you translating from another language? In English these concepts are usually called "the $1$-metric" and "the $p$-metric", rather than "the metric $1$". And "the same as" means something quite different from "equivalent to". – Troposphere Nov 07 '21 at 21:55
  • Yeah, i'm translating from another language, thanks for the corrections – Eduardo Gutierrez Nov 07 '21 at 22:07
  • Thanks to one of the comments i think that i found the solution to this problem. i'll check with my teacher tomorrow and if it's right i'll be answering my post. thanks ! – Eduardo Gutierrez Nov 07 '21 at 22:26

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