Closed graph theorem

A linear map between Banach spaces is continuous if and only if its graph is closed.

For an everywhere-defined linear map $T:E\to F$ between Banach spaces, its separating space measures the possible limiting defect in continuity at zero. It is a closed linear subspace: linear combinations combine witnessing sequences, and diagonal choices with $\Vert x_j\Vert <1/j$ and $\Vert T{x}_j-y_j\Vert <1/j$ handle a convergent sequence $y_j$ of defects. The closed graph theorem gives $T$ continuous exactly when $S(T)=\{0\}$, since a limit $(x_n,T{x}_n)\to (x,y)$ has $y-Tx\in S(T)$.

Let $S:E\to Y$ be linear between Banach spaces, let $T_n:E\to E$ be bounded, and let $U_n:Y\to Z_n$ be bounded. If $U_{n}S{P}_m$ is continuous whenever $m>n$, then $U_{n}S{P}_n$ is continuous for all sufficiently large $n$. After rescaling the bounded operators, assume their norms are at most one. If the conclusion failed, discontinuity would allow successively tiny vectors $x_j$ whose diagonal images dominate all previous contributions. Choose their norms also to control the continuous maps $U_{n_i}S{P}_{n_i+1}$ on subsequent tails. The convergent sum $x={\sum }_j{P}_{n_j}{x}_j$ would then satisfy $\Vert U_{n_j}Sx\Vert >j$ despite $\Vert U_{n_j}Sx\Vert \le \Vert Sx\Vert$, a contradiction. The tail is factored through $P_{n_i+1}$ before applying $S$, so no continuity of $S$ is assumed.

A linear map $T:E\to F$ between Fréchet spaces is continuous if its graph of a linear operator is closed in $E\times F$. The closed graph is itself a Fréchet space. Its first-coordinate projection is a continuous linear bijection onto $E$, so the open mapping theorem for Fréchet spaces makes the inverse continuous. Compose that inverse with the second-coordinate projection to obtain $T$. Completeness of both original spaces is part of the theorem.

For a closed graph, the projection $\Gamma (T)\to X$ is a bounded bijection between Banach spaces. Its bounded inverse followed by projection to $Y$ is $T$.

If bounded $S:X\to Y$ is injective and $\mathrm{im}T\subseteq \mathrm{im}S$, then $R=S^{-1}T$ has closed graph: $x_n\to x$ and $R{x}_n\to z$ imply $Sz=Tx=S(Rx)$ and hence $z=Rx$.

On incomplete $c_{00}$ with the ${\ell }^2$ norm, let $S(x_n)=(x_n/n)$ and let $T$ be inclusion into ${\ell }^2$. Their images agree, but $S^{-1}T(e_n)=n{e}_n$ is unbounded.

Let $V$ be a real Banach space and $T:V\to V^{\ast }$ be linear with $T_v(v)\ge 0$ for every $v$. If $v_n\to 0$ and $T_{v_n}\to f$ in $V^{\ast }$, positivity of $T_{v_n+tw}(v_n+tw)$ and passage to the limit givefor every real $t$ and every $w$. Taking small $t$ of either sign forces $f(w)=0$. Thus the graph of $T$ is closed, and the closed graph theorem makes $T$ continuous.