Closed graph theorem between Banach spaces: sufficiency of null sequence

Question

For for a linear operator between Banach spaces $T:E \rightarrow F$, why does the seemingly weaker implication $$x_n \rightarrow 0,\quad Tx_n \rightarrow y \implies y = 0$$ yield $$x_n \rightarrow x,\quad Tx_n \rightarrow y \implies y = Tx.$$ I'm new to operator theory and would like to understand the Closed graph theorem better.

Answer 1

It's a consequence of linearity. Suppose you have the formally weaker condition, and you have $x_n \to x$, $Tx_n \to y$. Then put $\xi_n := x_n - x$. It follows that $\xi_n \to 0$, and $T\xi_n = T(x_n - x) = Tx_n - Tx \to \eta := y - Tx$.

By the assumed condition $\eta = 0$, i.e. $y = Tx$. Thus the more general condition follows.

Answer 2

Assume the first statement you wrote is true. Now suppose we have $x_n\longrightarrow x$ and $Tx_n\longrightarrow y$. This means $x_n - x \longrightarrow 0$ and $Tx_n - Tx = T(x_n - x)\longrightarrow y - Tx$ (due to linearity of $T$) and it follows that $y - Tx = 0$ or $y = Tx$.