Closed Graph Theorem

Closed Graph Theorem

An (everywhere-defined) linear operator between Banach spaces is bounded iff its graph is closed.

We prove a slightly more general version:

Closed Graph Theorem (Variant)

Let $X$ and $Y$ be Banach spaces and $T:D(T)\to Y$ a linear operator with domain $D(T)$ closed in $X$. Then $T$ is bounded if and only if its graph $G(T)$ is closed.

Proof   Let us assume first that $T$ is bounded. Let $(x_n,T{x}_n{)}_n$ be a sequence in $G(T)$ that converges to some element $(x,y)\in X\times Y$. This means that $x_n\to x$ and $T{x}_n\to y$ for $n\to \infty$. The continuity of $T$ implies $T{x}_n\to Tx$. Since a convergent series in a Hausdorff space has a unique limit, it follows that $Tx=y$; hence $(x,y)$ lies in $G(T)$. This shows that $G(T)$ is closed.

Conversely, suppose that $G(T)$ is a closed subspace of $X\times Y$. Note that $X\times Y$ is a Banach space with norm $\Vert (x,y)\Vert =\Vert x\Vert +\Vert y\Vert$. Therefore, $G(T)$ is itself as Banach space in the restricted norm $\Vert (x,Tx)\Vert =\Vert x\Vert +\Vert Tx\Vert$. The canonical projections ${\pi }_X:G(T)\to X$ and ${\pi }_Y:G(T)\to Y$ are bounded. Clearly, ${\pi }_X$ is bijective, so its inverse ${\pi }_X^{-1}:X\to G(T)$ is a bounded operator by the Bounded Inverse Theorem. Consequently, the composition, ${\pi }_Y\circ {\pi }_X^{-1}:X\to Y$ is bounded. To complete the proof, observe that ${\pi }_Y\circ {\pi }_X^{-1}=T$. $\square$