Let $X, Y$ be a Banach spaces, $T: X \rightarrow Y$ be a bounded linear operator. What I want to show is if $T^{**}(X^{**}) \subseteq Y$, then $T$ is weakly-compact operator.($T$ is weakly compact operator :<=> cl$[T(ballX)]$) is weakly compact, where $ball X$ is closed unit ball in $X$).
However, I actually show that $every$ bounded operator is weakly-compact, which is nonsense. But I can't find what's wrong in my proof.
Here is my proof
(1) First, note that if $\sigma(Y^{**}, Y^*)$ (weak star topology on $Y^{**}$) is restricted to a subspace $Y$, then it coincides $\sigma(Y, Y^*)$ (weak topology on $Y$). This is because, if $\hat{y_i}$ (resp.,$\hat{y}$) is isomorphic embedding of $y_i (\text{resp.,}y) \in Y$ into $Y^{**}$, then $$ <\hat{y_i}, y^*> \longrightarrow <\hat{y}, y^*> \text{in } \sigma(Y^{**}, Y^*) \text{ if and only if } <y^*, y_i> \longrightarrow <y^*, y> \text{in } \sigma(Y, Y^*).$$
(2) By an Alaoglu theorem, $ballX^{**}$ (unit ball in $X^{**}$) $\sigma(X^{**}, X^*)$-compact. Since $T$ is continuous, $T^{**}: (X^{**}, wk^*) \longrightarrow (Y^{**}, wk^*)$ is continuous. Therefore, its image $T^{**}(ball(X^{**}))$ is $\sigma(Y^{**}, Y^*)$ - compact. Now, by (1), we obtain $T^{**}(ball(X^{**})) \cap Y$ is $\sigma(Y, Y^*)$ - compact. Finally, $T(ball X) = T^{**}(ballX) \subseteq T^{**}(ball(X^{**})) \cap Y$ implies cl[${T(ballX)}$] is $\sigma(Y, Y^*)$ - compact. Thus, $T$ is weakly compact map.
=== Edit ===
$T: X \rightarrow Y$ is continuous then $T^*: (Y^*, wk^*) \rightarrow (X^*, wk^*)$ is continuous.
proof) We have to show $y_i^* \rightarrow y^* (\text{in }\sigma(Y^*, Y))$ implies $T^*(y_i^*) \rightarrow T^*(y^*) (\text{in } \sigma(X^*, X))$. To this, let $x \in X$. Then, $<x, T^*(y_i^*)> = <Tx, y_i^*> \rightarrow <Tx, y^*> = <x, T^*(y^*)>$. This shows our claim.