Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 2 / 12E (Analysis II)

Two norms $\Vert \cdot \Vert$ and $\Vert \cdot {\Vert }^{\prime }$ on one vector space are Lipschitz equivalent when constants $c,C>0$ exist such that

Choose a basis of the finite-dimensional vector space and let $|x{|}_{\infty }$ be the maximum absolute coordinate. For any norm $N$, the triangle inequality givesso $N$ is continuous in the coordinate topology. On the compact set $\{x:|x{|}_{\infty }=1\}$, the positive continuous function $N$ attains a positive minimum $m$ and a finite maximum $M$. By absolute homogeneity,This proves the equivalence of norms in finite dimensions: every norm is equivalent to the maximum norm, and transitivity of these inequalities proves that any two norms on a finite-dimensional vector space are Lipschitz equivalent.

SupposeIf $(v_n)$ is a Cauchy sequence in $\Vert \cdot \Vert$, thenConversely, $\Vert v_n-v_m\Vert \le c^{-1}\Vert v_n-v_m{\Vert }^{\prime }\to 0$. Hence the sequence is Cauchy for one norm exactly when it is Cauchy for the other.

For each positive integer $N$, let $x^{(N)}$ have its first $N$ coordinates equal to one and all later coordinates zero. This is an element of the l-p sequence space for every finite $p$, andNo constant can bound $\Vert x{\Vert }_p$ by a fixed multiple of $\Vert x{\Vert }_{\infty }$ for all $N$. Thus $\Vert \cdot {\Vert }_p$ and $\Vert \cdot {\Vert }_{\infty }$ are not Lipschitz equivalent.

For $1\le p<q<\infty$, the same finitely supported sequence givesOne of the two inequalities required for equivalent norms therefore fails. Hence