---
title: Closed Graph Theorem
parent: The Fundamental Four
grand_parent: Functional Analysis Basics
nav_order: 4
---

# {{ page.title }}

{% theorem * Closed Graph Theorem %}
An (everywhere-defined) linear operator between Banach spaces is bounded
iff its graph is closed.
{% endtheorem %}

We prove a slightly more general version:

{% theorem * Closed Graph Theorem (Variant) %}
Let $X$ and $Y$ be Banach spaces
and $T : \dom{T} \to Y$ a linear operator
with domain $\dom{T}$ closed in $X$.
Then $T$ is bounded if and only if
its graph $\graph{T}$ is closed.
{% endtheorem %}

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

Conversely, suppose that $\graph{T}$ is a closed subspace of $X \times Y$.
Note that $X \times Y$ is a Banach space with norm $\norm{(x,y)} = \norm{x} + \norm{y}$.
Therefore $\graph{T}$ is itself as Banach space in the restricted norm $\norm{(x,Tx)} = \norm{x} + \norm{Tx}$.
The canonical projections $\pi_X : \graph{T} \to X$ and $\pi_Y : \graph{T} \to Y$ are bounded.
Clearly, $\pi_X$ is bijective, so its inverse $\pi_X^{-1} : X \to \graph{T}$ is a bounded operator by the
[Bounded Inverse Theorem](/pages/functional-analysis-basics/the-fundamental-four/open-mapping-theorem.html#bounded-inverse-theorem).
Consequently the composition, $\pi_Y \circ \pi_X^{-1} : X \to Y$ is bounded.
To complete the proof, observe that $\pi_Y \circ \pi_X^{-1} = T$.
{% endproof %}