# Can you tell whether a space is Banach from the unit ball?

Let $V$ be a real vector space. It is well known that a subset $B\subset V$ is the unit ball for some norm on $V$ if and only if $B$ satisfies the following conditions:

1. $B$ is convex, i.e. if $v,w\in B$ and $\lambda\in[0,1]$ then $\lambda v+(1-\lambda)w \in B$.

2. $B$ is balanced, i.e. $\lambda B \subset B$ for all $\lambda \in [-1,1]$.

3. $\displaystyle\bigcup_{\lambda > 0} \lambda B = V$ and $\displaystyle\bigcap_{\lambda>0} \lambda B = \{0\}$.

My question is: is there some simple way to determine from $B$ whether the resulting norm on $V$ will be complete? Keep in mind that $V$ does not yet have a topology.

Edit: I guess the word "simple" is a bit misleading. What I'm looking for is some geometric insight into how the shape of $B$ affects whether the result is a Banach space. When $V$ is finite dimensional, all sets $B$ satisfying conditions (1) - (3) give equivalent norms, so all $B$'s are somehow roughly the same shape. In what way do the shapes vary when $V$ is infinite-dimensional, and how does this affect the completeness of the resulting norm?

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If the dimension is finite all norms induce the same topology, so that's something. – Adam Hughes Feb 28 '11 at 17:46
More directly to the point, if the dimension is finite all norms give $V$ a complete metric structure. An easy necessary condition for completeness is that $B$ satisfies a version of the nested interval property: any nested sequence of translates of dilates of $B$ has a nonempty intersection. – Mark Meckes Feb 28 '11 at 18:46
Translating the notion of 'converging sequence' and of 'Cauchy sequence' in terms of $B$ in place of the norm is quite immediate. The resulting formulation of completeness in terms of $B$ is not that different form the usual one. Do you expect a simpler way than that? – Pietro Majer Feb 28 '11 at 19:48
Another silly condition is that B must be an algebra for the monad "$\ell^1$" (that sends a set to the unit ball of the $\ell^1$-space on it). Aka, $B$ must be totally convex. (silly for the same reason) – Loop Space Mar 1 '11 at 14:14
Your ambient space, Jim, with the product topology is a complete locally convex space. If $B$ is closed in the product topology, the norm induced by $B$ is complete. – Bill Johnson Mar 2 '11 at 16:57

A sufficient (additional) condition is that $B$ be compact for some Hausdorff vector topology for $V$. The proof goes as follows.

Letting $\langle x_i\rangle$ be a Cauchy sequence, it is contained in some $nB$, and so it has some cluster point $y$ there. Given $\varepsilon>0$, there is $i_0$ such that $x_i-x_j\in\frac 12\varepsilon B$ for $i,j\ge i_0$ , and it remains to show that $x_{i_0}-y\in\frac 12\varepsilon B$ . Indeed, if this does not hold, we have $y\not\in x_{i_0}-\frac 12\varepsilon B$ . Since $B$ is compact, it is closed in the Hausdorff case, and so is $x_{i_0}-\frac 12\varepsilon B$ as we have a vector topology. By the cluster point property, there must be some $i\ge i_0$ with $x_i\not\in x_{i_0}-\frac 12\varepsilon B$ , which is impossible.

Edit. Actually, I originally had in mind a more general condition, but I couldn't correctly recall it when writing the answer. Namely, a sufficient condition is that there be a Hausdorff topological vector space $E$ and there an absolutely convex compact set $C$ and a linear map $\ell:V\to E$ with $B=\ell^{-1}[C]$ , and such that we have $y\in{\rm rng\ }\ell$ whenever $y\in E$ is such that $(y+\varepsilon C)\cap({\rm rng\ }\ell)\not=\emptyset$ for all $\varepsilon>0$ . The proof is essentially the same as the one given above with $C$ in place of $B$ . This more general condition applies for example in the case where $B$ is the closed unit ball of $C^k([0,1])$ since we can take $E=({\mathbb R}^{[0,1]})^{k+1}$ and $C=([-1,1]^{[0,1]})^{k+1}$ and $\ell$ given by $y\mapsto\langle y,y',y'',\ldots y^{(k)}\rangle$ .

II Edit. The sufficient condition I gave above is of "extrinsic nature", and as such probably not in the spirit requested in the original question. An "intrinsic" condition, which is (probably) "simple", and in the line already suggested above in the first answer and in the comments, is that for any sequence $\langle x_i:i\in\mathbb N_0\rangle$ in $V$ satisfying $\lbrace 2^{i+2}(x_i-x_{i+1}):i\in\mathbb N_0\rbrace\subseteq B$ , there be $x\in V$ with $\lbrace 2^i(x_i-x):i\in\mathbb N_0\rbrace\subseteq B$ .

However, obviously this is not very practical to be verified in concrete situations. Basing on my experience and intuition, I would generally say that "extrinsic" conditions probably are more convenient than "intrinsic" ones. So, I think the question is good, but the restriction put there on the direction for searching for the answer is wrong. In practice, when one constructs (prospective new) Banach spaces, there is often some surrounding "larger" topological vector space where the new spaces will be continuously injected. In view of this, it is natural to look for extrinsic conditions.

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Actually this condition is necessary and sufficient for $B$ to be the unit ball of a space that is isometrically isomorphic to a dual Banach space. This is another exercise in books, IIRC. – Bill Johnson Mar 1 '11 at 17:20
It is very reasonable to argue that the restrictive nature of the question is wrong. Indeed, the point of view that I'm taking is very different from the usual perspective on such things. This is intentional -- I'm curious what can be said about completeness in this context, without any of the usual tools available. It's certainly possible that nothing very interesting can be said, in which case the question is not successful. – Jim Belk Mar 2 '11 at 21:14
Then I have no idea what you are looking for, Jim. You should give some motivation for your question and explain what type of condition you want. From my perspective, from what you have written, what you have asked has already been over answered. – Bill Johnson Mar 3 '11 at 0:53
That's fair. I'll accept this answer and move on. Thanks everybody! – Jim Belk Mar 3 '11 at 4:44

I don't know if this counts as "simple". But $V$ is Banach if and only if, whenever $(x_n)$ is a sequence in $B$, and $\sum_n \|x_n\|<1$, then $\sum_n x_n$ converges in $V$ (and necessarily to something in $B$). Now, you can phrase this convergence purely in terms of $B$. You need that there is $x\in B$ such that, for all $\epsilon>0$, there exists $N$ such that $\epsilon^{-1}(x - \sum_{n=1}^N x_n) \in B$.

That doesn't seem super-simple to me.

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Edit: While I typed this, Pietro make a comment. Of course, all I've done is actually carry out Pietro's comment more explicitly... – Matthew Daws Feb 28 '11 at 19:50
Thank you, I was wondering if I had to expand the comment, with the same remark ;-) Also note that for the completeness of a normed space it it sufficient the convergence of all geometric series, that is with $|x_n|\leq 2^{-n}$ – Pietro Majer Feb 28 '11 at 20:00
Take $B$ to be the closed unit ball of the Minkowski functional for $B$. The usual way of checking that the normed space that has $B$ as the unit ball is complete is to verify that $B$ is closed in some Banach space that contains $B$ s.t the unit ball of the Banach space contains $B$. You can find this as an exercise in some books (not that I recall which ones). – Bill Johnson Feb 28 '11 at 21:00
Bill, I did wonder a bit whether the question was just a homework problem. – Deane Yang Mar 1 '11 at 4:19

Unit balls with precisely the property that you are looking for have been studied under the rather awkward name of completant (presumably directly from the French) in the book on applications of bornologies to functional analysis by Hogbe-Nlend. I think that the only result of any substance that you will find is a variant of Grothendieck's completeness theorem which can be found there. One assumes that the ball is a closed bounded set in an ambient topological vector space which is complete. This, amongst others, provides what is probably the simplest and most transparent proof of the completeness of the $\ell^p$ and $L^p$-spaces.

By the way the class of spaces of Bill Johnson's answer has also been investigated. They were introduced by Waelbroeck and called Waelbroeck spaces by Buchwalter. They form a concrete representation of the category opposite to that of Banach spaces---see Cigler-Losert-Michor on functors on categories of Banach spaces (available onine). A good example of their use is in the characterisation of von Neumann algebas as $C^*$ algebras which, as Banach spaces, are Waelbroeck. This givea a useful pointer on how to form limits in the category of von Neumann algebras.

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Call me grumpy, but I can't help feeling Sakai's name should be mentioned in the last para, even if the words "dual Banach space" are for some reason being avoided. (And yes, I have read that section and others of Cigler-Losert-Michor) – Yemon Choi Oct 23 '12 at 13:39
tried to remove this answer since it seems to have caused offence but apparently don't have the power. perhaps somebody who does could do me the favour. – jbc Oct 23 '12 at 18:14
I wouldn't say offence; just a certain irritation with the style of the last para. BTW, I am not sure it makes to say that some object is "Waelbroeck, as a Banach space". It is a rather subtle consequence of Sakai's theorem/proof that in the isometric sense, there is only one way for a von Neumann algebra to be Waelbroeck: that is, the underlying Banach space of a $W^\ast$-algebra has a unique isometric predual. – Yemon Choi Oct 29 '12 at 21:50