Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 3 / 1A (Analysis II)

A norm on a real vector space $V$ is a map $p:V\to [0,\infty )$ satisfying $p(x)=0$ exactly when $x=0$, $p(tx)=|t|p(x)$ for real $t$, and $p(x+y)\le p(x)+p(y)$. Its induced metric is $d_p(x,y)=p(x-y)$. The three axioms respectively give definiteness, compatibility with scaling, and the triangle inequality.

Here equivalent metrics means that the two metrics induce the same topology, and Lipschitz equivalent norms means that there are constants $c,C>0$ such that $cp(x)\le q(x)\le Cp(x)$ for every $x$.

Suppose first that the inequalities hold. Since they also hold for $x-y$, the identity maps between the two metric spaces are Lipschitz continuous. In particular, each is continuous, so the topologies agree.

Conversely, equality of the topologies makes the identity from the $p$-norm to the $q$-norm continuous at zero. There is therefore $\delta >0$ such that $p(x)<\delta$ implies $q(x)<1$. For nonzero $x$, apply this to $\delta x/(2p(x))$ and use the absolute homogeneity of both norms:The reverse identity is also continuous, so there is $\epsilon >0$ with $q(x)<\epsilon \Rightarrow p(x)<1$. Rescaling again gives $p(x)<2q(x)/\epsilon$. ThusThe inequalities also hold at zero. This is topology determines norm equivalence; it works in arbitrary dimension because the rescaling argument needs neither a basis nor compactness of a unit sphere.

Expand using the bilinearity and symmetry of the real inner product:Addition proves the parallelogram law,For the maximum norm on ${\mathbb{R}}^2$, take $x=(1,0)$ and $y=(0,1)$. All four vectors $x,y,x+y,x-y$ have maximum norm one. The left side of the parallelogram law is then $2$, whereas the right side is $4$. The maximum norm cannot come from an inner product.