# 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 $T:D(T)→Y$ 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 $(x_{n},Tx_{n})_{n}$ be a sequence in $G(T)$ that converges to some element $(x,y)∈X×Y$. This means that $x_{n}→x$ and $Tx_{n}→y$ for $n→∞$. The continuity of $T$ implies $Tx_{n}→Tx$. 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 $X×Y$. Note that $X×Y$ is a Banach space with norm $∥(x,y)∥=∥x∥+∥y∥$. Therefore, $G(T)$ is itself as Banach space in the restricted norm $∥(x,Tx)∥=∥x∥+∥Tx∥$. The canonical projections $π_{X}:G(T)→X$ and $π_{Y}:G(T)→Y$ are bounded. Clearly, $π_{X}$ is bijective, so its inverse $π_{X}:X→G(T)$ is a bounded operator by the Bounded Inverse Theorem. Consequently, the composition, $π_{Y}∘π_{X}:X→Y$ is bounded. To complete the proof, observe that $π_{Y}∘π_{X}=T$. $□$