Given $A_n : X\rightarrow B$, a linear and continuous operator between two Banach spaces, for each $x\in X$ $A_n(x) \rightharpoonup A(x)$ in $Y$, is $A$ linear and continuous?
My attempt:
$A$ is linear, given $A(x + y)$, then there exists a sequence $A_n(x+y)$ which converges weakly to $A(x+y)$ and $A_n(x+y) = A_n x + A_n y $ which converges weakly to $Ax + Ay$.
To show $A$is bounded, since $A_n (x) \rightharpoonup A(x)$, we have $$\|A(x)\|_Y \leq \liminf \|A_n (x) \|_Y$$ and that $\|A_n (x) \|_Y$ is bounded.
Using uniform boundedness principle, we get that $\|A_n\|_{op}$ is bounded.
Thus, $$\|A(x)\|_Y \leq \liminf \|A_n (x) \|_Y \leq \liminf \|A_n\|_{op} \|x\|_X .$$
Question: Is this correct? I am not so sure since the Banach-Saks-Steinhaus theorem has the same result with the assumption that $A_n(x)$ is strongly convergent (Cauchy) in $Y$ (instead of weak convergence like in this problem).