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?



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


