# Extending an absolute value via a limit of powers

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. $$\Vert av\Vert=|a|\Vert v\Vert$$.]

• 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. – Torsten Schoeneberg Nov 5 '19 at 20:46
• @TorstenSchoeneberg Thanks a lot - I looked into the book and it contains everything I wanted. – user8268 Nov 5 '19 at 23:28
• 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.) – Torsten Schoeneberg Nov 6 '19 at 0:42