A sufficient and necessary condition for the boundedness of linear operator Let $X$ and $Y$ be two normed linear spaces and $T: X \rightarrow Y$ is a linear operator(mapping). Prove that $T$ is bounded if and only if it maps weakly convergent sequences to weakly convergent sequences.
To my best knowledge, I think it's easy to prove that bounded linear operator maps weakly convergent sequences to weakly convergent sequences. However, I wonder how the reverse direction can be proved.
Thank you for your help.
 A: To prove that the first proposition implies the second one you can apply the definition of weakly convergent sequence.
Our hypothesis now is that $T$ is a bounded linear operator from $(X,||\;||_x)$ to $(Y,||\;||_y)$, so there exists a constant $c>0$ such that $||Tx||_y\leq c||x||_x,\;\forall x\in X$. 
Moreover, let's take an arbitrary sequence $\{x_k\}\subset X$ which converges weakly to $\bar{x}\in X$, hence we have that $\forall T\in X',\;T(x_k)\rightarrow T(\bar{x})$ as $k\rightarrow \infty$, where with the notation $X'$ I just mean the dual space of $X$, so the space of linear and continuous (hence bounded) functionals from $X$ to $\mathbb{R}$.
Hence we are now required to prove that the sequence $T(x_k)=y_k$ converges weakly in the space $Y$, which means that it should happen the following fact: $\forall L\in Y',\;L(y_k)\rightarrow L(\bar{y})$ as $k\rightarrow \infty$.
But this is quite an immediate consequence of the linearity of $L$:
$|L(y_k)-L(\bar{y})|=|L(y_k-\bar{y})|\leq ||L||_{y'}\,|T(x_k)-T(\bar{x})|\leq ||L||_{y'}||T||_{x'}||x_k-\bar{x}||_x \leq \bar{c}||x_k-\bar{x}||\rightarrow 0$ as $k\rightarrow 0$.
Hence we get the desired result.
