# Normed Spaces

Definition 16.1.1 (Norm)   Let be a field with valuation and let be a vector space over . A real-valued function on is called a norm if
1. for all nonzero (positivity).
2. for all (triangle inequality).
3. for all and (homogeneity).

Note that setting for all does not define a norm unless the absolute value on is trivial, as . We assume for the rest of this section that is not trivial.

Definition 16.1.2 (Equivalent)   Two norms and on the same vector space  are equivalent if there exists positive real numbers and such that for all

and

Lemma 16.1.3   Suppose that is a field that is complete with respect to a valuation and that is a finite dimensional  vector space. Continue to assume, as mentioned above, that is complete with respect to . Then any two norms on are equivalent.

Remark 16.1.4   As we shall see soon (see Theorem 17.1.8), the lemma is usually false if we do not assume that  is complete. For example, when and is the -adic valuation, and is a number field, then there may be several extensions of to inequivalent norms on .

If two norms are equivalent then the corresponding topologies on  are equal, since very open ball for is contained in an open ball for , and conversely. (The converse is also true, since, as we will show, all norms on  are equivalent.)

Proof. Let be a basis for . Define the max norm by

It is enough to show that any norm is equivalent to . We have

where .

To finish the proof, we show that there is a such that for all ,

We will only prove this in the case when is not just merely complete with respect to but also locally compact. This will be the case of primary interest to us. For a proof in the general case, see the original article by Cassels (page 53).

By what we have already shown, the function is continuous in the -topology, so by local compactness it attains its lower bound on the unit circle . (Why is the unit circle compact? With respect to , the topology on is the same as that of a product of copies of . If the valuation is archimedean then or with the standard topology and the unit circle is compact. If the valuation is non-archimedean, then we saw (see Remark 15.1.7) that if  is locally compact, then the valuation is discrete, in which case we showed that the unit disc is compact, hence the unit circle is also compact since it is closed.) Note that by part 1 of Definition 16.1.1. Also, by definition of , for any there exists such that (just take the max coefficient in our basis). Thus we can write any as where and with . We then have

Thus for all  we have

where , which proves the theorem.

William Stein 2012-09-24