Let $V$ be a real vector space. What is a norm on $V$ ? Show that if $\|-\|$ is a norm on $V$, then the maps $T_{v}: x \mapsto x+v\left(\right.$ for $v \in V$ ) and $m_{a}: x \mapsto a x$ (for $a \in \mathbb{R}$ ) are continuous with respect to the norm.

Let $B \subset V$ be a subset containing 0 . Show that there exists at most one norm on $V$ for which $B$ is the open unit ball.

Suppose that $B$ satisfies the following two properties:

• if $v \in V$ is a nonzero vector, then the line $\mathbb{R} v \subset V$ meets $B$ in a set of the form $\{t v:-\lambda<t<\lambda\}$ for some $\lambda>0$;

• if $x, y \in B$ and $s, t>0$ then $(s+t)^{-1}(s x+t y) \in B$.

Show that there exists a norm $\|-\|_{B}$ for which $B$ is the open unit ball.

Identify $\|-\|_{B}$ in the following two cases:

(i) $V=\mathbb{R}^{n}, B=\left\{\left(x_{1}, \ldots, x_{n}\right) \in \mathbb{R}^{n}:-1<x_{i}<1\right.$ for all $\left.i\right\}$.

(ii) $V=\mathbb{R}^{2}, B$ the interior of the square with vertices $(\pm 1,0),(0, \pm 1)$.

Let $C \subset \mathbb{R}^{2}$ be the set

$$C=\left\{\left(x_{1}, x_{2}\right) \in \mathbb{R}^{2}:\left|x_{1}\right|<1,\left|x_{2}\right|<1, \text { and }\left(\left|x_{1}\right|-1\right)^{2}+\left(\left|x_{2}\right|-1\right)^{2}>1\right\}$$

Is there a norm on $\mathbb{R}^{2}$ for which $C$ is the open unit ball? Justify your answer.