Let $X$ and $Y$ be Banach spaces. Let $\{f_i \} \subset Y^∗$ separates points in $Y$ . Suppose that $f_i T$ is continuous for each $f_i$ , then prove that $T$ is continuous.
I know that if I show that if $T: X \to Y$ is a closed map then the problem is solved. So, I need to show that Graph($T$) is closed i.e. for a sequence $\{x_n\}\in X$, if $x_n\to x$ and $T(x_n)\to y$ then $T(x)=y$.
But I am not sure how to use the concept of Dual of $Y$ to solve the problem. Any help will be greatly appreciated. Thanks.
Assume $x_n \to x$ and $Tx_n \to y$.
Since $f_i$ is continuous and $Tx_n \to y$, we have $f_i(Tx_n) \to f_i(y)$.
Since $f_i T$ is continuous and $x_n \to x$, we have $f_i(Tx_n) \to f_i(Tx)$.
Therefore $f_i(Tx) = f_i(y), \forall i$. Since $\{f_i\}$ separates points, we conclude $y = Tx$.