$T$ is closed $\iff$ for arbitrary $\{ x_n \}\in D(T)$ such that $x_n\to x,$ and $Tx_n\to y,$ we have $x\in D(T)$ and $Tx=y.$

Question

Let $X$ and $Y$ be normed linear spaces and $T:X\to Y$ be any map. Then, $T$ is closed if and only if for arbitrary $\{ x_n \}\in D(T)$, domain of $T$, with $x_n\to x,$ and $Tx_n\to y,$ we have $x\in D(T)$ and $Tx=y.$

My Trial

Assume that $T$ is closed and $\{ x_n \}\in D(T)$ be arbitrary such that $x_n\to x,$ as $n\to\infty.$ Since $T$ is closed, $$x\in D(T).$$

If $T$ were continuous, then $x_n\to x,$ implies $Tx_n\to Tx,$ and by uniqueness of limits, $Tx=y.$ However, we don't have this. Hence, I am stuck here! Could any one help out? Various proofs are welcome!

Answer 1

Since $T$ is closed by definition $G(T)$ is a closed subspace of $X \times Y$. Let $x_n \to x$ and $Tx_n \to y$. Then $(x_n, Tx_n) \in G(T)$ and therefore by closedness the limit $(x, y) \in G(T)$. Therefore we can deduce $x \in D(T)$ and $Tx = y$.
For the converse direction we have to show that $G(T)$ is a closed subspace of $X \times Y$. Let $(x_n, Tx_n)_n$ be a sequence in $G(T)$ which converges to $(x,y)$ in $X \times Y$. Therefore $x_n \to x$ in $X$ and $Tx_n \to y$ in $Y$. By assumption this yields $x \in D(T)$ and $Tx =y$. But this means $(x,y) \in G(T)$.