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.
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.
