Let $K$ be a field complete wrt. an absolute value $|\cdot|$, let $L\supset K$ be a finite extension, and let $\Vert\cdot\Vert$ be a norm on the $K$-vector space $L$.

For example, if $e_i$, $i=1\dots n$, is a $K$-basis of $L$, we can choose $\Vert\sum a_ie_i\Vert=\max_i|a_i|$.

Then, at least if $K$ is locally compact, one can easily see that

$$a\mapsto \lim_{n\to\infty} \Vert a^n\Vert^{1/n}\quad(a\in L)$$ exists for every $a\in L$ and that it is an absolute value on $L$ extending the one on $K$, using the fact that an extension of $|\cdot|$ to $L$ exists and that any two norms are equivalent.

My question is: can one prove it directly, without using the existence of an extension of $|\cdot|$ to $L$, and thus actually give a construction of such an extension (as this limit)?

[You can suppose that $|\cdot |$ is non-archimedean if it simplifies things. If the terminology needs to be clarified: an absolute value on $L$ = a map $|\cdot|:L\to[0,\infty)$ s.t. $d(x,y):=|x-y|$ is a metric on $L$ and s.t. $|ab|=|a||b|$. A norm on a $L$-vector space $V$ = a map $\Vert\cdot\Vert:V\to[0,\infty)$ s.t. $d(x,y):=\Vert x-y\Vert$ is a metric on $V$ and s.t. $\Vert av\Vert=|a|\Vert v\Vert$.]

  • 1
    $\begingroup$ I have a vague recollection that the book Non-Archimedean Analysis by Bosch, Güntzer and Remmert talks a lot about these things, and in general how to turn submultiplicative norms on Banach spaces into multiplicative ones. $\endgroup$ – Torsten Schoeneberg Nov 5 '19 at 20:46
  • 1
    $\begingroup$ @TorstenSchoeneberg Thanks a lot - I looked into the book and it contains everything I wanted. $\endgroup$ – user8268 Nov 5 '19 at 23:28
  • $\begingroup$ Oh good. Do you want to add an answer here to your specific question just so it's answered, and for the common good? (I would be happy to see how it works.) $\endgroup$ – Torsten Schoeneberg Nov 6 '19 at 0:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.