Let $X,Y$ be Banach spaces, and $T:X\longrightarrow Y$ be a linear operator. I want to understand the following fact.
The graph $\Gamma(T) = \{(x,Tx)\;|\;x\in X\}$ is closed in $X\oplus Y$ if and only if for every sequence $x_n\longrightarrow x$ such that $Tx_n$ converges, $Tx_n\longrightarrow Tx$.
I take the first part of this statement as the definition of a closed operator, i.e. $T$ is closed if $\Gamma(T)$ is closed. All the sources I have tried so far (Yosida's book, Lax's book, Wikipedia…) simply mention this as a self-evident fact, so I assumed it would be easy to see, but I just was't able to warp my head around it.
Does anyone have a proof for this? Please refrain from invoking the closed graph theorem, as it is a further development. Thanks!
Best Answer
If the condition holds and $(x_n,Tx_n)$ is a sequence in the graph converging to some point $(u,v)$ then $x_n \to u$ and $Tx_n \to v$. By the given condition this implies $Tx_n \to Tu$. But $Tx_n \to v$ also. This gives $v=Tu$ so $(u,v)=(u,Tu) $ belongs to the graph. Hence,, the graph is closed.
Conversely. if the graph is closed and $x_n \to x, Tx_n \to z$ then $(x_n,Tx_n) \to (x,z)$ so $(x,z) $ belongs to the graph of $T$. Hence, $z=Tx$.