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 $T\in \mathcal{B}(B_1,B_2)⟺\Gamma (T):=\{(u,Tu):u\in B_1\}\subset B_1\times B_2\text{ is closed}$

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 $T{u}_n\to Tu$

Proof

Proof $(⟹)$ Suppose $T\in \mathcal{B}(B_1,B_2)$. Let $\{(u_n,T{u}_n){\}}_n$ be a sequence in $\Gamma (T)$ such that $u_n\to u$ and $T{u}_n\to v$. Then $v={\lim }_{n\to \infty }T{u}_n=T({\lim }_{u\to \infty }{u}_n)=Tu$ since $T$ is continuous. Thus $(u,v)=(u,Tu)\in \Gamma (T)$ ie $\Gamma (T)$ is closed.

Proof $(⟸)$ Define

• ${\pi }_1:\Gamma (T)\to B_1,{\pi }_1(u,Tu)=u$
• ${\pi }_2:\Gamma (T)\to B_2,{\pi }_2(u,Tu)=Tu$

Note that $\Gamma (T)$ is a Banach space since $\Gamma (T)\subset B_1\times B_2$ is a closed subspace of $B_1\times B_2$ (which is Banach by the corrolary above)

Further, we have ${\pi }_1\in \mathcal{B}(\Gamma (T),B_1)$ and ${\pi }_2\in \mathcal{B}(\Gamma (T),B_2)$ $||{\pi }_2(u,v)||=||v||\le ||u||+||v||=||(u,v)||$ (we can see it is bounded by the norm of $B_1\times B_2$ as we defined above, which must also be bounded)

${\pi }_1:\Gamma (T)\to B_1$ is one-to-one and onto (bijective). Thus $S:={\pi }_1^{-1}:B_1\to \Gamma (T)$ is also a bounded linear operator

And so $T={\pi }_2\circ S$ is the composition of two bounded linear operators, and is thus itself a bounded linear operator.

ie $T\in \mathcal{B}(B_1,B_2)$

NOTE

closed graph theorem $⟺$ open mapping theorem

Mentions

Mentions

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