Let $T: X \to Y$ be a linear operator where $X$ and $Y$ are Banach spaces and $\mathcal F$ be a subset of $Y^*$ that separates the points of $Y$. Given that $f(T(x_n))\to 0$ whenever $f$ is in $\mathcal F$ and $||x_n|| \to 0$. Show that $T$ is bounded.
My work:
I was trying to solve this using the definition of a operator, $$i.e.,\, ||T|| = sup\{ ||Tx|| : ||x||=1 \}$$ but I failed.Need some help.
As suggested in the comment, we show that the graph of $T$ is closed. Assume that $(x_n , T(x_n) ) \to (x, y)$. Then $x_n - x \to 0$. From the assumption, for all $f\in \mathcal F$, $$f(T(x_n -x)) \to 0 \Rightarrow f(T(x_n)) \to f(T(x)).$$ On the other hand, for all $f\in Y^*$ we have $$ T(x_n) \to y \Rightarrow f(T(x_n)) \to f(y).$$ That is, $$f(T(x)) = f(y)$$ for all $f\in \mathcal F$. By assumption of $\mathcal F$, $T(x) =y$, so the graph of $T$ is closed and so $T$ is bounded by the closed graph theorem.