Closed graph theorem

Remarks

If $T$ is continuous, its graph is closed because $Y$ is Hausdorff. Conversely, if $\Gamma(T)$ is closed, it is a Banach space, and the projection $(x, Tx) \mapsto x$ is a continuous linear bijection from $\Gamma(T)$ onto $X$. The bounded inverse theorem makes its inverse $x \mapsto (x, Tx)$ continuous, and so $T$ is continuous.

In practice the closed graph condition is easier to check: it says that if $x_n \to x$ and $T x_n \to y$, then $y = Tx$. One does not have to show that $T x_n$ converges. Completeness is necessary. Let $X = C^1[0, 1]$ and $Y = C[0, 1]$, both with the sup norm, and let $T f = f'$. If $f_n \to f$ and $f_n' \to g$ uniformly, then $f$ is differentiable with $f' = g$, so the graph is closed. But $T$ is not continuous, since $f_n(t) = t^n$ has $\norm{f_n} = 1$ and $\norm{f_n'} = n$; here $X$ is not complete.

Going deeper

For a student

The theorem is used to prove automatic continuity: for example, a linear map $T$ defined on a whole Hilbert space with $\ip{Tx}{y} = \ip{x}{Ty}$ for all $x, y$ is bounded (Hellinger–Toeplitz theorem). It is also used to show that a projection onto a closed subspace along a closed complement is bounded, see complemented subspace. Unbounded operators in quantum mechanics are typically closed operators defined only on dense subspaces, which is consistent with the theorem.