1-lecture. Normed space. Banach space lesson Plan

Theorem 1.2.2(Completion). Let be a normed space. Then there is a Banach space and an isometry from onto a subspace of which is dense in . The space is unique, except for isometries.


1.3. FINITE DIMENSIONAL NORMED SPACES AND SUBSPACES
Are finite dimensional normed spaces simpler than infinite dimensional ones? In what respect? These questions are rather natural. They are important since finite dimensional spaces and subspaces play a role in various considerations (for instance, in approximation theory and spectral theory). Quite a number of interesting things can be said in this connection. Hence it is worthwhile to collect some relevant facts, for their own sake and as tools for our further work. This is our program in this section and the next one.

A source for results of the desired type is the following lemma. Very roughly speaking it states that in the case of linear independence of vectors we cannot find a linear combination that involves large scalars but represents a small vector.

Lemma 1.3.1 (Linear combinations). Let be a linearly independent set of vectors in a normed space (of any dimension). Then there is a number such that for every choice of scalars we have


. (1.3.1)


Proof. We write . If , all are zero, so that (1.3.1) holds for any . Let . Then (1) is equivalent to the inequality which we obtain from (1.3.1) by dividing by and writing , that is,


(1.3.2)

Hence it suffices to prove the existence of a such that (1.3.2)holds for every -tuple of scalars with .

Suppose that this is false. Then there exists a sequence of

such that



Now we reason as follows. Since , we have . Hence for each fixed the sequence

is bounded. Consequently, by the Bolzano-Weierstrass theorem, has a convergent subsequence. Let denote the limit of that subsequence, and let denote the corresponding subsequence of . By the same argument, has a subsequence for which the corresponding subsequence of scalars converges; let denote the limit. Continuing in this way, after steps we obtain a subsequence of whose terms are of the form

with scalars satisfying as . Hence, as ,

where , so that not all can be zero. Since is a linearly independent set, we thus have . On the other hand, implies , by the continuity of the norm. Since by assumption and is a subsequence of , we must have . Hence , so that by (N2) in Sec. 2.2. This contradicts , and the lemma is proved.∎

As a first application of this lemma, let us prove the basic