2 Clarified Question

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?

1

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.