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Jul 19, 2023 at 20:44 history edited Ege Erdil CC BY-SA 4.0
correcting a typo
Mar 31, 2022 at 13:16 answer added Denis Nardin timeline score: 0
Feb 16, 2022 at 18:54 answer added KConrad timeline score: 4
Aug 27, 2021 at 11:50 vote accept Ege Erdil
Aug 27, 2021 at 8:10 comment added Uri Bader @LSpice I blame my autocorrect for it. It didn't even occur to me that "accurse" is a meaningful word.
Aug 26, 2021 at 18:39 comment added LSpice Although it was clearly a typo, I like @UriBader's inadvertent suggestion to describe an unpleasant realisation as "It now accurse to me".
Aug 26, 2021 at 17:19 answer added Uri Bader timeline score: 14
Aug 26, 2021 at 7:55 comment added Uri Bader It now accurse to me that maybe I can give an existence argument using an averaging over the kernel of the norm map with respect to its Haar measure... I will give it a try later on.
Aug 26, 2021 at 7:39 comment added Uri Bader I posted an elaborated answer giving a uniform proof of the fact that $N^{1/n}$ is indeed a norm. In fact it is a unique absolute value extension of $|\cdot|$. The uniqueness part is easy and standard: if there exists an extension it must coincide with $N^{1/n}$. Thus, if you believe in existence for moral reasons or, better, have a non-constructive proof then the fact that $N^{1/n}$ is a norm follows. For me, this is the motivation to consider this specific function. Does this answer your question?
Aug 24, 2021 at 20:58 answer added Uri Bader timeline score: 7
Aug 24, 2021 at 12:19 history edited Ege Erdil CC BY-SA 4.0
added 17 characters in body
Aug 24, 2021 at 11:51 review First posts
Aug 24, 2021 at 12:43
Aug 24, 2021 at 11:46 history asked Ege Erdil CC BY-SA 4.0