Prove that result that a linear transformation is continuous iff it transforms every sequence converge to zero into a bounded sequence. Proof.
(=>) Let $T$ be a continuous linear transformation.
Then we get $T$ is continuous at $0$.
Then $T$ is bounded.
Define $T$ transform sequence $x_{n}$ into $T(x_{n})$ ,then we will show $T(x_{n})$ is bounded sequence.
Since $x_{n}\rightarrow  0$ and $T$ is continuous, $T(x_{n})$ converges to $0$.
That is, $T(x_{n})$ is bounded sequence.

Is this proof correct? ####

And how to prove (<=) .
Thanks a lot! :-)
 A: Your proof of $\implies$ is a bit excessive.  By the definition of continuity (in normed spaces), if $x_n \to 0$, then $T(x_n) \to T(0)$.  Since $\{T(x_n)\}$ is a convergent sequence, it is bounded.
For the other direction ($\Longleftarrow$): I prefer a proof by contrapositive.  Suppose that $T$ is not continuous.  Then, the image of the unit ball under $T$ is unbounded.  We can then take a sequence $y_n$ in the unit ball such that $\|T(y_n)\| \to \infty$.  Now, define
$$
x_n = \frac{1}{\sqrt{\|T(y_n)\|}} y_n
$$
we then have $x_n \to 0$, but $\{T(x_n)\}$ is an unbounded sequence.
A: suppose $T$ isnot countinous then 
Then there exists a sequence $\{x_n \}_{n\geq 1}$ such that $\left\| T(x_n)\right\|> n\| x_n\| \ $, and so
$$
1<\frac{1}{n\|x_n\|}\|T(x_n)\|=\left\| T\left(\frac{x_n}{n\|x_n\|}\right)\right\|\ \ .
$$
But $\left\|\frac{x_n}{n\|x_n\|}\right\|\rightarrow 0\ $, and so by continuity at $0$
$$
\left\| T\left(\frac{x_n}{n\|x_n\|}-0\right)\right\|\rightarrow\left\| T\left(0\right)\right\|=\|0\|=0,
$$
which is a contradiciton.  
