Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 2 / 12G (Analysis II)

The unheaded preliminary requests are addressed before the first example. A norm is a function $p:V\to [0,\infty )$ satisfying $p(x)=0$ exactly for $x=0$, $p(ax)=|a|p(x)$, and $p(x+y)\le p(x)+p(y)$. Its induced metric is $d(x,y)=p(x-y)$. Translation $T_v$ preserves this distance, and scalar multiplication satisfies $d(ax,ay)=|a|d(x,y)$, so both are continuous; multiplication by zero is constant.

The open unit ball determines the norm uniquely through its Minkowski functional:because $x/t\in B$ is equivalent to $p(x)<t$. This also works for $x=0$.

Under the given radial and convexity hypotheses, set $p_B(0)=0$ and $p_B(v)=1/\lambda (v)$ for $v\ne 0$. The finite symmetric segment along each line gives positivity, absolute homogeneity and $v\in B⟺p_B(v)<1$. If $s>p_B(x)$ and $t>p_B(y)$, then $x/s,y/t\in B$, and the stated convexity condition yields $(x+y)/(s+t)\in B$. Hence $p_B(x+y)<s+t$; taking $s\downarrow p_B(x)$ and $t\downarrow p_B(y)$ proves the triangle inequality. Thus this is a norm determined by a convex radial unit ball. The case $V=\{0\}$ is immediate.

For the coordinate box, the Minkowski functional is the supremum norm:Indeed $x/t\in B$ exactly when $t>{\max }_i|x_i|$.

The interior of the diamond is $B=\{(x_1,x_2):|x_1|+|x_2|<1\}$. Its Minkowski functional, or the uniqueness argument already proved, gives the L1 norm:The final unheaded set $C$ cannot be an open unit ball of a norm, because every norm ball is convex by homogeneity and the triangle inequality. Both $p=(9/10,0)$ and $q=(0,9/10)$ lie in $C$: their squared-distance expression is $101/100>1$. Their midpoint has that expressionso $(p+q)/2\notin C$. Thus $\boxed{\text{no norm has }C\text{ as its open unit ball}}$. The strict inequalities in the definition are respected by this explicit counterexample.