Let $\mathrm T$ be a linear operator from a Banach space $\mathrm X $ into a Banach space $\mathrm Y $. Suppose that $\mathrm F $ is a subset of $\mathrm Y^* $that separates points of $\mathrm Y $. Assume that f (T ($x_n$)) $\to$ $\mathrm0$ whenever $\mathrm f $ $\in$ $\mathrm F$ and {$x_n$} $\subset$ $\mathrm X$ is such that $\Vert$$x_n$$\Vert$$\to$$\mathrm 0$. Show that $\mathrm T$ is a bounded operator.
I figured that I need to use the Closed Graph theorem, but can't get to the application. If anyone would come in help I would be grateful.
Following the closed graph theorem approach, we need to show that if $x_n\to x$ and $Tx_n\to y$, then $y=Tx$.
If $x_n\to x$ then $\|x_n-x\|\to 0$, so $f(T(x_n-x))=f(Tx_n)-f(Tx)\to 0$ for all $f\in F$.
On the other hand $f\in Y^*$, so $f(Tx_n)\to f(y)$. Hence $f(Tx)=f(y)$ for all $f\in F$. Since $F$ separates points, it follows that $Tx=y$.