Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 4 / 22I (Linear Analysis)

$X^{\ast }$ is the Banach space of bounded linear functionals. For $T:X\to Y$, define $T^{\ast }:Y^{\ast }\to X^{\ast }$ by $T^{\ast }f=f\circ T$. It is linear and $\Vert T^{\ast }f\Vert \le \Vert T\Vert \Vert f\Vert$, so bounded (indeed $\Vert T^{\ast }\Vert =\Vert T\Vert$).

If $S:X\to Y$ is an isomorphism, then $S^{\ast }:Y^{\ast }\to X^{\ast }$ is an isomorphism with inverse $(S^{-1}{)}^{\ast }$. If $S$ and $S^{-1}$ are isometries, the same norm calculation makes the dual maps isometries.

The canonical embedding puts $X$ isometrically, hence with closed range, inside $X^{\ast \ast }\simeq {\ell }_2$. Every closed subspace of a Hilbert space is Hilbert and, in the infinite-dimensional separable case here, is isomorphic to ${\ell }_2$. Thus $X\simeq {\ell }_2$.

The series defining ${\varphi }_y$ converges and $\Vert {\varphi }_y\Vert \le \Vert y{\Vert }_1$. For each finite set of $k_n$, normality and Tietze extension give a continuous $f$ of norm at most one taking prescribed signs there. Letting the finite set grow proves $\Vert {\varphi }_y\Vert =\Vert y{\Vert }_1$. Thus $y\mapsto {\varphi }_y$ is an isometric embedding of ${\ell }_1$ into $C(K{)}^{\ast }$.

No. The point evaluations ${\delta }_x$ satisfy $\Vert {\delta }_x-{\delta }_y\Vert =2$ for $x\ne y$ (choose a continuous function taking values $1,-1$ at the two points). Thus $C[0,1{]}^{\ast }$ is nonseparable, whereas ${\ell }_1$ is separable.