Closed Graph Theorem in the weak topology

Question

Let $T$ be a bounded linear operator mapping from $E$ to itself. Assume further $T$ is closed.

My question is as follows: is the graph of $T$ closed in the weak topology of $E\times E$ ?

My attempt: Intuitively, the range is in E forever as $T$ acts on $E$. So if it is closed in the strong sense, then it should be closed in the weak sense. But the most important thing is how to derive the above thought rigorously? I would be very happy if you could help me. Thanks!

Answer 1

If $(x_i,Tx_i)$ is a net in the graph of $T$ converging weakly to $(x,y)$ then $x_i \to x$ weakly and $Tx_i \to y$ weakly. But $T$ is weak-weak continuous, so $Tx_i \to Tx$ weakly. Hence, $y=Tx$. This proves that the graph is weakly closed.

Answer 2

Assume $(x_0,y_0)\notin \Gamma(T).$ Hence $y_0\neq Tx_0.$ There is a linear functional $\varphi\in E^*$ such that $\varphi(Tx_0-y_0)=\|Tx_0-y_0\|$ and $\|\varphi\|_{E^*}=1.$ Let $$ U=\{(x,y)\in E\times E\,:\, {\rm Re}\,\varphi(Tx-y)>0\}$$ The set $U$ is an open neighborhood of $(x_0,y_0)$ in the weak topology on $E\times E$ and $U$ is disjoint with $\Gamma(T).$ Therefore $\Gamma(T)$ is weakly closed in $E\times E.$