Unit balls in normed spaces.

Question

Assume we talk about the $n$ dimensional vector space over the reals.

It is easy to see that for any norm the unit ball is a convex symmetric set. And here is my question :

Let $A$ be a bounded , symmetric, closed, convex set. Assume that $A$ is not contained in any $k$ dimensional affine subspace for $k\leq n-1$. Show (disprove ?) that $A$ is a closed unit ball for some norm.

Answer 1

Okay, let us assume without loss of generality, our set $A$ is symmetric to $0$. Then we can just look at the closed unit ball with centre $0$. Define a function $$N\colon \mathbb{R}^n \to \mathbb{R}_+, v \mapsto \inf \left\{ r > 0 : \frac1r \cdot v \in A \right\}.$$ Now, we have to check that $N$ is a norm:

1. For all $a \in \mathbb{R}$ and $v \in \mathbb{R}^n$: $N(av) = |a|N(v)$: \begin{align} N(av) &= \inf \left\{ r > 0 : \frac{a}r \cdot v \in A \right\} \\ &= |a| \cdot \inf \left\{ r > 0 : \frac1r \cdot v \in A \right\} &\text{Since $A$ is symmetric}\\ &= |a| \cdot N(v) \end{align}
2. If $N(v) = 0 \Longrightarrow v = 0:$ Assume $N(v) = 0$ and $v \neq 0$. Then $\frac1r \cdot v \in A$ for all $r \in \mathbb{R}$ which contradicts our assumption that $A$ bounded. Thus, $v = 0$.
3. For all $v, w \in \mathbb{R}^n, N(v + w) \leq N(v) + N(w):$ I think this is a bit harder to show. However, it is also the most interesting part since you need the convexity of $A$ to show subadditivity. How do we show this? Consider $v, w \in \mathbb{R}^n$ and assume that $v, w \neq 0$. By definition, we find that $v/N(v), w/N(w) \in A$. Since $A$ is convex, we know that $$\frac{v + w}{N(v)+N(w)} \in A$$ and thus $$\frac{N(v + w)}{N(v)+N(w)} \leq 1.$$ which implies that $N(v + w) \leq N(v)+N(w)$.

Last but not least we have to show that $A$ is the unit ball of $N$. Since all norms in $\mathbb{R}^n$ are continuous, $N$ is continuous (See other posts). $A \subseteq \{v \in \mathbb{R}^n: N(v) \leq 1\}$ is trivial by definition. The other inclusion $A \supseteq \{v \in \mathbb{R}^n: N(v) \leq 1\}$ follows from continuity of $N$.