I came up with a property of bounded linear operators and i am not sure if it is correct,because the proof seems easy.
Let $T:X \to Y$ be a linear operator with the property that: if $x_n \in X$ a sequence such that $x_n \to 0$ then $||Tx_n||$ is bounded (as a sequence of real numbers).Then $T$ is a bounded operator.
I proved it twice,with different proofs.
$\textbf{Proof 1}$ Assume that $T$ is not bounded.Then exists a sequence $x_n \in X$ such that $||x_n||=1,\forall n \in \Bbb{N}$ and $||Tx_n|| \geq n^2$.
Then $y_n:=\frac{x_n}{n} \to 0$ thus $||Ty_n||$ is bounded,by hypothesis.
But $||Ty_n|| \geq n,\forall n \in \Bbb{N}$,which is a contradiction.
$\textbf{Proof 2}$ Assume that is not continuous at some point $x \in X$.
Thus exists $c>0$ with the property: exists a sequence $x_n$ such that $||x_n-x||<\frac{1}{n}$ and $||T(x_n-x)|| \geq c,\forall n \in \Bbb{N}$
Define $y_n=\frac{||x_n-x||}{\sqrt{||x_n-x||}}=\sqrt{||x_n-x||} <\frac{1}{\sqrt{n}} \to 0$
But $||Ty_n|| \geq c\sqrt{n} \to +\infty$ which is a contradiction.
Are these proofs correct?
Thank you in advance.