closed graph theorem

[[concept]]

Closed Graph Theorem

If $B_{1}, B_{2}$ are Banach spaces and $T : B_{1} \to B_{2}$ is a linear operator, then

T \in B (B_{1}, B_{2}) ⟺ Γ (T) := {(u, T u) : u \in B_{1}} \subset B_{1} \times B_{2} is closed

Note

this may be easier to prove than proving something is a bounded linear operator. Normally, we need to show that sequences $u_{n} \to u$ have $T u_{n} \to T u$

Proof

(⟹)

Suppose $T \in B (B_{1}, B_{2})$ . Let ${(u_{n}, T u_{n})}_{n}$ be a sequence in $Γ (T)$ such that $u_{n} \to u$ and $T u_{n} \to v$ . Then

v = lim_{n \to \infty} T u_{n} = T (lim_{u \to \infty} u_{n}) = T u

since $T$ is continuous. Thus

(u, v) = (u, T u) \in Γ (T) $ $ i e $ Γ (T) $ i s c l o s e d .

Proof

(⟸)

Define

$π_{1} : Γ (T) \to B_{1}, π_{1} (u, T u) = u$
$π_{2} : Γ (T) \to B_{2}, π_{2} (u, T u) = T u$

Note that $Γ (T)$ is a Banach space since $Γ (T) \subset B_{1} \times B_{2}$ is a closed subspace of $B_{1} \times B_{2}$ (which is Banach by the corrolary above)

Further, we have $π_{1} \in B (Γ (T), B_{1})$ and $π_{2} \in B (Γ (T), B_{2})$

| | π_{2} (u, v) | | = | | v | | \leq | | u | | + | | v | | = | | (u, v) | |

(we can see it is bounded by the norm of $B_{1} \times B_{2}$ as we defined above, which must also be bounded)

$π_{1} : Γ (T) \to B_{1}$ is one-to-one and onto (bijective). Thus $S := π_{1}^{- 1} : B_{1} \to Γ (T)$ is also a bounded linear operator

And so $T = π_{2} \circ S$ is the composition of two bounded linear operators, and is thus itself a bounded linear operator.

ie $T \in B (B_{1}, B_{2})$

Note

closed graph theorem $⟺$ open mapping theorem

Mentions

File	Last Modified
Functional Analysis Lecture 4	2025-06-05
open mapping theorem	2025-06-05

Created 2025-06-05 Last Modified 2025-06-05