Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 4 / 12G (Analysis II)

Two norms $N_1,N_2$ on the same vector space are equivalent norms, or Lipschitz equivalent, if there are constants $0<c\le C<\infty$ such that $c{N}_1(v)\le N_2(v)\le C{N}_1(v)$ for every $v$. The constants are uniform over the whole space.

Let $N$ be any norm on ${\mathbb{R}}^n$. The triangle inequality and the Cauchy-Schwarz inequality giveThe reverse triangle inequality then gives $|N(x)-N(y)|\le C\Vert x-y{\Vert }_2$, so $N$ is Euclidean-continuous; this continuity was derived, rather than assumed. On the compact Euclidean unit sphere, $N$ attains a strictly positive minimum $c$, because it vanishes only at zero. Homogeneity yieldsFinally a linear map $A:{\mathbb{R}}^n\to {\mathbb{R}}^m$ satisfies $\Vert Ax{\Vert }_2\le ({\sum }_{i,j}|a_{ij}{|}^2{)}^{1/2}\Vert x{\Vert }_2$, again by Cauchy-Schwarz inequality. The equivalence just proved transfers this bound to any norms on the source and target. Hence every such linear map is continuous.

If $f$ is a bounded linear operator, linearity gives $\Vert f(u)-f(v)\Vert \le C\Vert u-v\Vert$. It is therefore Lipschitz and continuous everywhere.

Conversely, suppose $f$ is continuous at zero. There is $\delta >0$ such that $\Vert u\Vert <\delta$ implies $\Vert f(u)\Vert <1$. For $u\ne 0$, apply this to $v=\delta u/(2\Vert u\Vert )$ and use homogeneity:The value at $u=0$ is zero by linearity. ConsequentlyThus boundedness and continuity are equivalent for linear maps between normed spaces. No finite dimensionality or completeness is needed for this equivalence.

First $w\in {\ell }^2$, since ${\sum }_{n\ge 1}{4}^{-n}=1/3$. Any finite linear dependence among $w$ and the $e_m$ has the form $aw+{\sum }_{m\in F}{b}_m{e}_m=0$ for a finite index set $F$. At an index outside $F$, its coordinate equals $a{2}^{-n}$, so $a=0$. Each remaining coordinate then gives $b_m=0$. Therefore the set is linearly independent, and the prescribed values define a unique linear map on its algebraic span.

Take the tailsThey belong to $V$, satisfy $f(v_N)=1$, and haveThus $v_N\to 0$ while $f(v_N)$ does not approach $f(0)=0$. The map is not continuous, and part (b) also shows it is unbounded. This is a discontinuous functional detected by vanishing sequence tails, illustrating why the finite-dimensional conclusion in part (a) cannot be extended to arbitrary normed spaces.