$T$ is bounded if and only if for $f_n, f \in X$ and $f_n \to f$ then $T(f_n) \to T(f)$

Question

Let $T$ be a linear functional on a normed linear space $X$. Show that $T$ is bounded if and only if for $f_n, f \in X$ and $f_n \to f$ then $T(f_n) \to T(f)$

I proved one direction but I am need some help on the other direction:

$"\Rightarrow"$

Consider $T: X \to \mathbb{R}$ bounded on $X$ $\Leftrightarrow \exists M \geq 0$ such that $|T(f)| \leq M ||f|| $ for all $f \in X$ and the norm $||T||_* = M$

Let $f_n, f \in X$ such that $f_n \to f$, by linearity of $T$ $$|T(f_n) - T(f)| \leq ||T||_* ||f_n - f||$$ Since $f_n \to f$ we conclude that $|T(f_n) - T(f)| \to 0$ or $T(f_n) \to T(f)$

$"\Leftarrow"$

Consider $f_n, f \in X, f_n \to f$ and $T(f_n) \to T(f)$

Answer 1

Assume $T$ is unbounded, then there exists a sequence $(f_n)\subseteq X$ with $\|f_n\|=1 \quad\forall n$ and $$\|Tf_n\|\geq n$$ Then define $$g_n=\frac{1}{n}f_n$$ then clearly $g_n\to 0$,but $$\|Tg_n\|\geq 1,\forall n\in\mathbb{N}$$ which clearly does not converge to $0$

Answer 2

Suppose there exist $f_n$ with $||Tf_n||\ge n||f_n||$. Replacing $f_n$ by a scalar multiple of $f_n$ we can assume that $||f_n||=\sqrt n$. So $f_n\to0$ but $Tf_n\not\to T0$.