Updated: 25/2/26

Ref:

• Functional Analysis lectures by Casey Rodriguez. Lecture on the Closed Graph Theorem. Link to the lecture: Link.

Obs [Product of Banach spaces]

Let ${ B _1 , }$ ${ B _2 }$ be Banach spaces. Then ${ B _1 \times B _2 }$ under

\[{ \lVert (b _1, b _2) \rVert = \lVert b _1 \rVert + \lVert b _2 \rVert }\]

is a Banach space.

Thm [Closed Graph Theorem]

Let ${ B _1 , }$ ${ B _2 }$ be Banach spaces. Let ${ T : B _1 \to B _2 }$ be a linear operator.

Then ${ T \in \mathcal{B}(B _1, B _2) }$ if and only if the graph ${ \Gamma(T) = \lbrace (u, T(u)) : u \in B _1 \rbrace }$ is closed in ${ B _1 \times B _2 . }$

Hence in terms of sequential criteria,

${( \forall u \in B _1 , \quad u _n \to u \implies T(u _n) \to T(u)) }$

is equivalent to

${ (\forall u \in B _1, \quad u _n \to u, T(u _n) \to v \implies v = T(u)) . }$

Pf: ${ \underline{\implies} }$ Clear from the above sequential version.

${ \underline{\impliedby} }$ Let ${ \Gamma(T) }$ be closed in ${ B _1 \times B _2 . }$

Consider the commutative diagram

Here

\[{ {\begin{aligned} &\, \pi _1 : \Gamma(T) \longrightarrow B _1, \\ &\, \pi _1 (u,T(u)) = u \end{aligned}} }\]

and

\[{ {\begin{aligned} &\, \pi _2 : \Gamma(T) \longrightarrow B _2, \\ &\, \pi _2 (u,T(u)) = T(u) \end{aligned}} }\]

Note that ${ \Gamma(T) }$ is a closed subspace of a Banach space, and hence is a Banach space.

Note that ${ \pi _1, }$ ${ \pi _2 }$ are bounded linear operators. For example, ${ \lVert \pi _2 (u, Tu) \rVert }$ ${ = \lVert Tu \rVert }$ ${ \leq \lVert u \rVert + \lVert Tu \rVert }$ ${ = \lVert (u, Tu) \rVert . }$

Note that ${ \pi _1 }$ is a bijection. By the Open Mapping Theorem, ${ \pi _1 ^{-1} }$ ${ = s }$ is a bounded linear operator.

Hence ${ T = \pi _2 \circ s }$ is a bounded linear operator, as needed.