# Banach or not Banach?

In the basic theory of normed vector spaces and Banach spaces, there are a number of important theorems involving one or two such spaces, and each of them may, or may not, need to be complete (i.e., really a Banach space, and not only a normed space). Of course, at some point the statements should be engraved in memorial stone, and not require any thought, but if one hasn’t worked on the topic for a while (say, a few years), it’s easy to forget. Fortunately, it’s also easy to recover the assumptions needed by using elementary arguments to check how the theorems, if assumed to hold with all complete spaces, sometimes extend trivially to cases where one or more is not.

(1) The Hahn-Banach Theorem: here, the statement is about extending a continuous linear functional $\lambda\,:\, W\rightarrow \mathbf{C}$

(assuming everything is over the base field C for simplicity) from a subspaceW to the whole space V. Should V be complete? Well, even if we prove the theorem in that case, note that we could use it, given a non-complete space V, to extend λ to the completion Y of V, and then we can trivially restrict it again to V. So there’s no need for V to be complete! Similarly, even if W is not closed, we can extend λ by continuity to its closure, and if the theorem holds for a closed subspace, we can then extend further to V. So W need not be closed.

(2) The Banach-Steinhaus Theorem: now we have a family of continuous linear maps $T_i\,:\, V\rightarrow W$

between normed vector spaces, which are assumed to be pointwise bounded: $\sup_{i\in I}{||T_i(v)||}<+\infty$

for all v in V, and the conclusion is that the operators are uniformly bounded over the unit ball of V: $\sup_{i\in I}{||T_i||}<+\infty$

(recall the operator norm is the supremum over the unit ball). Should V, or W, be complete? Well, clearly, if the theorem holds for Banach spaces, and W is not complete, we can define a family by composition $U_i\,:\, V\rightarrow W\rightarrow Y$

where Y is the completion of W, and the pointwise bound is still trivially true for this family. The uniform conclusion for it is also the same as for the original one (for the same obvious reason that the images of the Ti and Ui are the same).
So completeness of W is clearly not necessary.

However, if we try similar arguments for V, we can say (indeed) that the Ti extend to the completion of V, but it is not obvious that the pointwise bound holds for those extensions, since they must be evaluated at more points than those of V. This can lead quickly to doubts and the conclusion that it’s certainly safer to assume V is complete; or it can lead to constructing a counter-example to confirm this impression: maybe the simplest non-complete normed vector space is the space $d=\{(x_n)\,\mid\, x_n=0\text{ for } n \text{ sufficiently large}\}$

of sequences with only finitely many non-zero coefficients, with the supremum norm of the coefficients. The coordinate mappings (which are the most obvious ones for this space!) can not serve as family of operators, since they are all of norm 1. But take $T_n(x)=nx_n,\text{ for which } ||T_n||=n\rightarrow +\infty,$

for instance. Because the coefficients of a sequence in d vanish far away, we have $T_n(x)=0\text{ for all } n \text{ sufficiently large}$

and so the pointwise bound holds, despite the operators not being uniformly bounded.

(One can also remember vaguely that the proof of the theorem uses the Baire Property somewhere, so some completeness is needed, and it must be that of V).

(3) The Open Mapping Theorem: here we have two spaces and a surjective continuous linear map $T\,:\, V\rightarrow W$

and the outcome is that T is open: it maps open subsets of V to open subsets of W. Which of the two should be Banach spaces for this to hold?

If we try to assume the result for Banach spaces and extend it, we run into problems at both ends this time: if we replace T by the map S obtained by post-composing with the embedding of W in its completion, the assumption of openness may start to fail. If we assume that W is complete and extend T by continuity to the completion Y of V, the assumptions still hold, but the conclusion we obtain is that this extension is open: it maps an open subset of Y to an open subset of W, and there is no reason that the restriction to V be open. So this leads us to suspect that both source and target spaces have to be Banach spaces for this result to have a chance to hold. This is indeed the case.

For counterexamples, one can take for instance $V=C([0,1])\text{ with } ||f||=\sup|f(t)|,\quad\quad W=C([0,1])\text{ with } ||f||_1=\int_0^1{|f(t)|dt}$

and T the identity map. Then W is not complete, and T is linear continuous and surjective but not open (that would make it an isomorphism, and then W would be complete…) This works more generally for any space V with two norms which are comparable, but only one of which is complete.

An example where the source space is not complete is given by first taking a non-continuous linear map $S\,:\,Y\rightarrow W$

where Y and W are Banach spaces (note that the existence of such a map is essentially dependent on the Axiom of Choice!), and then defining $V=\text{graph of } S=\{(v,w)\in Y\times W\,\mid\, w=S(v)\}$

with the norm induced from the product norm on Y x W. The non-continuity ensures that this space is not complete, but the map $T\,:\, V\rightarrow W,\quad\quad (v,S(v))\mapsto v$

is continuous, surjective, but not open.