Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 22I (Linear Analysis)

The dual $X^{\ast }$ is the vector space of bounded linear maps $f:X\to \mathbb{F}$, with normIf $(f_n)$ is Cauchy in this norm, then $(f_n(x))$ is Cauchy in the complete scalar field for every $x$. Define $f(x)={\lim }_n{f}_n(x)$. Pointwise passage to the limit preserves linearity. Given $\epsilon >0$, choose $N$ such that $\Vert f_n-f_m\Vert <\epsilon$ for $m,n\ge N$; sending $m\to \infty$ givesThus $f$ is bounded and $\Vert f_n-f\Vert \le \epsilon$. Hence $X^{\ast }$ is Banach, even if $X$ was only normed.

Let $1<p<\infty$ and $q=p/(p-1)$. Every $y\in {\ell }^q$ definesHölder's inequality gives $\Vert f_y\Vert \le \Vert y{\Vert }_q$, and equality follows by using a normalized sequence proportional to $\mathrm{sgn}(y_n)|y_n{|}^{q-1}$.

Conversely, for $f\in ({\ell }^p{)}^{\ast }$ put $y_n=f(e_n)$. On the first $N$ coordinates, choose the same norming vector to obtainLetting $N\to \infty$ shows $y\in {\ell }^q$. On finitely supported sequences, linearity gives $f=f_y$; density of those sequences in ${\ell }^p$ extends this identity to all $x$. Thereforeisometrically.

The endpoint duals areunder the same coordinate pairing $⟨x,y⟩={\sum }_n{x}_n{y}_n$.

Yes. For $x=(x_n)\in c$, let $L={\lim }_n{x}_n$ and defineThe tail tends to zero, so $T(x)\in c_0$, and $\Vert T(x){\Vert }_{\infty }\le 2\Vert x{\Vert }_{\infty }$. Conversely, for $y=(y_n)\in c_0$, defineThis sequence converges to $y_1$ and has norm at most $2\Vert y{\Vert }_{\infty }$. The formulas are inverse bounded linear maps, so $c$ and $c_0$ are isomorphic Banach spaces.