Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 22F (Linear Analysis)

The open mapping theorem states that every surjective bounded linear map $S:X\to Y$ between Banach spaces maps open sets to open sets. Equivalently, there is $c>0$ such that

The closed graph theorem states that a linear map $T:X\to Y$ between Banach spaces is continuous if and only if its graphis closed in $X\times Y$.

Assume the open mapping theorem and suppose $\Gamma (T)$ is closed. As a closed subspace of the Banach space $X\times Y$, the graph is itself Banach. The coordinate projectionis a bounded linear bijection. By the open mapping theorem, $P^{-1}$ is bounded. The other coordinate projection $Q:\Gamma (T)\to Y$ is bounded, andThus $T$ is bounded and hence continuous. This is the closed graph theorem from the bounded inverse theorem argument.

The continuous dual space of a normed vector space $V$ iswith the operator norm

Let $(f_n)$ be a Cauchy sequence in this norm. For each $v\in V$,so $(f_n(v))$ is Cauchy in the scalar field. Define $f(v)={\lim }_n{f}_n(v)$. Passing to limits in the linearity identities shows that $f$ is linear.

A norm-Cauchy sequence is bounded, say $\Vert f_n\Vert \le M$. Hence $|f(v)|\le M\Vert v\Vert$, so $f\in V^{\ast }$. Given $\epsilon >0$, choose $N$ such that $\Vert f_n-f_m\Vert <\epsilon$ for $m,n\ge N$. Letting $m\to \infty$ givesfor every $v$, and therefore $\Vert f_n-f\Vert \le \epsilon$. Thus $f_n\to f$ in operator norm, proving completeness of the dual space: $V^{\ast }$ is Banach.

Write $T_v=T(v)$. We prove that $T$ has a closed graph. SupposePut $u_n=v_n-v$ and $g=f-T_v$. Then $u_n\to 0$ and $T_{u_n}\to g$ in $V^{\ast }$.

Fix $w\in V$ and $t\in \mathbb{R}$. Positivity and linearity giveBecause $(T_{u_n})$ converges in the continuous dual space, its norms are bounded, soAlso $T_{u_n}(w)\to g(w)$, while $T_w(u_n)\to 0$ because the fixed functional $T_w$ is continuous. Passing to the limit yieldsfor every real $t$. If $g(w)\ne 0$, a sufficiently small $t$ of the opposite sign makes the right-hand side negative. Hence $g(w)=0$ for every $w$, so $g=0$ and $f=T_v$.

The graph is therefore closed. The closed graph theorem now proves that $T:V\to V^{\ast }$ is continuous. This is precisely continuity of a positive linear map into a dual space.