Closed graph theorem

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

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.