Closed Graph Theorem

Closed Graph Theorem

Let $X$ and $Y$ be Banach spaces and $T$ a linear map of $X\to Y$.Then $X\times Y$ is also normed vector space with norm $\Vert (x,y){\Vert }_{X\times Y}:=\max \{\Vert x{\Vert }_X,\Vert y{\Vert }_Y\}$ and $\text{Graph}(T):=\{(x,y)\in X\times Y:y=Tx\}$ is a vector subspace of $X\times Y$. If $\text{Graph}(T)$ is closed inside $X\times Y$, then $T$ is continuous(bounded).

$T$ is continuous, then $\text{Graph}(T)$ is a closed subspace of $X\times Y$. Proof : Suppose $(x_n,y_n)\in \text{Graph}(T)$ and suppose $(x_n,y_n)\to (x,y)$ s.t. $\Vert (x_n,y_n)-(x,y){\Vert }_{X\times Y}\to 0$. Then $x_n\to x$ in $X$, $y_n\to y$ $y_n=T{x}_n⟹\Vert T{x}_n-y\Vert \to 0$. By continuity, $y=Tx$. in $Y$. So the theorem can be rephrased as $T$ is bounded if and only if the graph of $T$ is closed.

If

Proof Since $X,Y$ are Banach spaces, $X\times Y$ is also Banach (complete). Since $\text{Graph}(T)$ is closed inside $X\times Y$, we have $\text{Graph}(T)$ also Banach space. Define $\pi :\text{Graph}(T)\to X$ by $\pi (x,y)=x$. $x|(x,y)\in \text{Graph}(T)$. Then $\pi$ is bounded linear bijection: $\Vert \pi (x,y)\Vert =\Vert x\Vert \le \max \{\Vert x\Vert ,\Vert y\Vert \}=\Vert (x,y){\Vert }_{X\times Y}\forall (x,y)\in \text{Graph}(T)$. So by the bounded inverse theorem , ${\pi }^{-1}:X\to \text{Graph}(T)$ is bounded. $\exists c>0\text{ s.t. }\Vert {\pi }^{-1}(x){\Vert }_{X\times Y}\le c\Vert x{\Vert }_X\forall x\in X$ $\Vert {\pi }^{-1}(x){\Vert }_{X\times Y}=\Vert (x,Tx){\Vert }_{X\times Y}=\max \{\Vert x{\Vert }_X,\Vert Tx{\Vert }_Y\}$ $⟹\Vert Tx{\Vert }_Y\le c\Vert x{\Vert }_X\forall x\in X$ Hence $T$ is bounded (continuous).

Remark

If we want to check whether $T:X\to Y$ is continuous, by definition we need to check whether $x_n\to x⟹T{x}_n\to Tx$. But if $X,Y$ happen to be both Banach spaces, then according to the closed graph theorem, we need only check whether $\text{Graph}(T)$ is closed in $X\times Y$ i.e. whether $(x_n\to x\text{ implies }y=Tx)$ $\underset{\text{you can assume for free the existence of limit of }T{x}_n.}{\underbrace{T{x}_n\to y}}\text{ implies }y=Tx.$