# Closed Graph Theorem

## Theorem

Let $\struct {X, \norm {\,\cdot\,}_X}$ and $\struct {Y, \norm {\,\cdot\,}_Y}$ be Banach spaces.

Let $T : X \to Y$ be a linear transformation.

Let $G_T \subseteq X \times Y$ be the graph of $T$.

Suppose that:

- $G_T$ is closed in the direct product $X \times Y$ equipped with the direct product norm $\norm \cdot_{X \times Y}$.

Then:

- $T$ is bounded.

## Proof

Let $\norm {\,\cdot\,}_{G_T}$ be the restriction of $\norm \cdot_{X \times Y}$ to $G_T$.

By Closed Subspace of Banach Space forms Banach Space, $\struct {G_T, \norm \cdot_{G_T} }$ is a Banach space.

Define $\pi_X : G_T \to X$ by:

- $\map {\pi_X} {x, y} = x$ for each $\tuple {x, y} \in X \times Y$.

Note that by the definition of $G_T$, we have $\tuple {x, y} \in G_T$ if and only if $y = T x$.

So, we have that $\pi_X$ is a bijection with inverse $\pi_X^{-1} : X \to G_T$ with:

- $\map {\pi_X^{-1} } x = \tuple {x, T x}$ for each $x \in X$.

From Projections on Direct Product of Normed Vector Spaces define Bounded Linear Transformations, $\pi_X$ is a bounded linear transformation.

By the Banach Isomorphism Theorem:

- since $\pi_X$ is a bijective bounded linear transformation between Banach spaces, its inverse $\pi_X^{-1}$ is a bounded linear transformation.

So there exists $C > 0$ such that:

- $\norm {\pi_X^{-1} x}_{G_T} \le C \norm x_X$

for each $x \in X$.

That is:

- $\norm {\tuple {x, T x} }_{X \times Y} \le C \norm x_X$ for each $x \in X$.

From the definition of the direct product norm, we have:

- $\max \set {\norm x_X, \norm {T x}_Y} \le C \norm x_X$ for each $x \in X$.

From the definition of the max operation, we have:

- $\norm {T x}_Y \le \max \set {\norm x_X, \norm {T x}_Y}$

So:

- $\norm {T x}_Y \le C \norm x_X$ for all $x \in X$.

So:

- $T$ is bounded.

$\blacksquare$