Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 1 / 21F (Linear Analysis)

The continuous dual space $X^{\ast }$ consists of all bounded real linear functionals on $X$, with operator norm $\Vert f\Vert ={\sup }_{\Vert x\Vert \le 1}|f(x)|$. Let $(f_n)$ be Cauchy in this norm. For each $x$, $|f_n(x)-f_m(x)|\le \Vert f_n-f_m\Vert \Vert x\Vert$, so $f_n(x)$ converges in $\mathbb{R}$; define its limit to be $f(x)$.

Pointwise limits of the linear identities show that $f$ is linear. The sequence of norms is bounded, say by $M$, so $|f(x)|\le M\Vert x\Vert$ and $f\in X^{\ast }$. Given $\epsilon >0$, choose $N$ so that $\Vert f_n-f_m\Vert \le \epsilon$ for all $m,n\ge N$. Holding $n$ fixed and taking the pointwise limit in $m$ gives $|f_n(x)-f(x)|\le \epsilon \Vert x\Vert$ for all $x$. Thus $\Vert f_n-f\Vert \le \epsilon$, provingCompleteness of $X$ is not needed; completeness of the scalar field supplies the pointwise limit.

On the one-dimensional linear subspace $Y=\mathbb{R}{x}_0$, define $g(t{x}_0)=t\Vert x_0\Vert$. It is linear and obeys $g(y)\le \Vert y\Vert$ even for negative $t$. The norm on $X$ is a sublinear functional, so the real Hahn-Banach theorem extends $g$ to a linear functional $f$ with $f(x)\le \Vert x\Vert$ for every $x$.

Apply this also to $-x$ to get $-f(x)\le \Vert x\Vert$. Hence $|f(x)|\le \Vert x\Vert$, proving continuity and $\Vert f\Vert \le 1$. At $x_0\ne 0$ the extension satisfies $f(x_0)=\Vert x_0\Vert$, forcing the reverse norm inequality. ThereforeThe exclusion of $x_0=0$ is what ensures that this particular construction attains norm one.

Take $Y=X^{\ast \ast }$, which is a Banach space by the completeness of the dual just proved. The canonical embedding into the bidual isFor fixed $x$ this is a bounded linear functional on $X^{\ast }$, since $|f(x)|\le \Vert f\Vert \Vert x\Vert$, and $\Phi$ is linear in $x$. This gives $\Vert \Phi (x)\Vert \le \Vert x\Vert$. For $x\ne 0$, the norming functional from the Hahn-Banach theorem has $\Vert f\Vert =1$ and $f(x)=\Vert x\Vert$, giving the reverse inequality. At zero equality is immediate. ThusIn particular $\Phi$ is injective and isometric. If a dense isometric embedding is wanted, rather than merely the stated embedding, take the norm closure of $\Phi (X)$ in $X^{\ast \ast }$; a closed subspace of a Banach space is complete and supplies a completion of a normed space.