Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 22I (Linear Analysis)

The bounded inverse theorem states that a bounded bijective linear map between Banach spaces has a bounded inverse.

The closed graph theorem states that if $X$ and $Y$ are Banach spaces and $T:X\to Y$ is linear, then $T$ is continuous if and only if its graphis closed in the Banach space $X\times Y$.

If $T$ is continuous and $(x_n,T{x}_n)\to (x,y)$, then $T{x}_n\to Tx$, so uniqueness of limits gives $y=Tx$. Hence the graph is closed.

Conversely, suppose $\Gamma (T)$ is closed. It is then a Banach space. The coordinate projectionis bounded and bijective. By the bounded inverse theorem, $P_X^{-1}:x\mapsto (x,Tx)$ is bounded. Composing it with the bounded second-coordinate projection givesso $T$ is continuous.

Take $X=Y={\ell }^2$ and defineThis is bounded and injective. Its inverse on $\mathrm{im}S$ is not continuous: for the standard unit vectors $e_n$,

Set $R=S^{-1}\circ T:X\to X$, which is well-defined and linear because $\mathrm{im}T\subseteq \mathrm{im}S$. We show that its graph is closed.

Suppose $x_n\to x$ and $R{x}_n\to z$ in $X$. Continuity of $S$ and $T$ givesSince $S$ is injective, $z=Rx$. Thus $\Gamma (R)$ is closed. The domain and codomain $X$ are Banach spaces, so the closed graph theorem implies

Let $X=c_{00}$, the finitely supported sequences, with the ${\ell }^2$ norm. This normed space is not complete. Let $Y={\ell }^2$ and defineBoth maps $X\to Y$ are bounded, and $S$ is injective. Moreoverbecause multiplication of a finitely supported sequence by $(1,2,3,\dots )$ preserves finite support. Butso $\Vert (S^{-1}T)e_n{\Vert }_2=n$ while $\Vert e_n{\Vert }_2=1$. Hence $S^{-1}T$ is not continuous, showing that completeness of $X$ is essential.