Forward direction $ (\Longrightarrow) $
Suppose that $ T \in \mathcal{L}(X,Y) $ and that $ x_{n} \stackrel{\text{wk}}{\longrightarrow} x $. Then for any $ \varphi \in Y^{*} $, we have
\begin{align}
\lim_{n \to \infty} \varphi(T(x_{n}) - T(x))
&= \lim_{n \to \infty} \varphi(T(x_{n} - x)) \\
&= \lim_{n \to \infty} (\varphi \circ T)(x_{n} - x) \\
&= 0. \quad (\text{As $ \varphi \circ T \in X^{*} $.})
\end{align}
Therefore, $ T(x_{n}) \stackrel{\text{wk}}{\longrightarrow} T(x) $.
Backward direction $ (\Longleftarrow) $
Suppose that $ T: X \to Y $ is a linear operator (not assumed to be bounded) and that $ x_{n} \stackrel{\text{wk}}{\longrightarrow} x $ implies $ T(x_{n}) \stackrel{\text{wk}}{\longrightarrow} T(x) $.
For the sake of contradiction, assume that $ T $ is not bounded. Then we can easily construct a sequence $ (x_{n})_{n \in \mathbb{N}} $ in $ X $ that converges in norm to $ 0_{X} $ but for which $ \| T(x_{n}) \|_{Y} \geq n^{2} $ for all $ n \in \mathbb{N} $. As $ \dfrac{1}{n} \cdot x_{n} \stackrel{\| \cdot \|_{X}}{\longrightarrow} 0_{X} $, we obviously have $ \dfrac{1}{n} \cdot x_{n} \stackrel{\text{wk}}{\longrightarrow} 0_{X} $. From our initial hypothesis, it follows that $ T \left( \dfrac{1}{n} \cdot x_{n} \right) \stackrel{\text{wk}}{\longrightarrow} T(0_{X}) = 0_{Y} $.
Now, for each $ n \in \mathbb{N} $, we can view $ T \left( \dfrac{1}{n} \cdot x_{n} \right) $ as an element $ \Psi_{n} $ of $ Y^{**} $, via the canonical embedding $ J: Y \to Y^{**} $. As
\begin{align}
\forall \varphi \in Y^{*}: \quad
\lim_{n \to \infty} {\Psi_{n}}(\varphi)
&= \lim_{n \to \infty} \varphi \left( T \left( \frac{1}{n} \cdot x_{n} \right) \right) \\
&= \varphi(0_{Y}) \\
&= 0,
\end{align}
we see that $ \displaystyle \sup_{n \in \mathbb{N}} |{\Psi_{n}}(\varphi)| < \infty $ for all $ \varphi \in Y^{*} $. Applying the Uniform Boundedness Principle (also known as the Banach-Steinhaus Theorem) to $ Y^{**} $, we obtain $ \displaystyle \sup_{n \in \mathbb{N}} \| \Psi_{n} \|_{Y^{**}} < \infty $. This yields $ \displaystyle \sup_{n \in \mathbb{N}} \left\| T \left( \dfrac{1}{n} \cdot x_{n} \right) \right\|_{Y} < \infty $ as the canonical embedding $ J $ is an isometry. We have thus arrived at a contradiction as our construction of the sequence $ (x_{n})_{n \in \mathbb{N}} $ forces us to have $ \left\| T \left( \dfrac{1}{n} \cdot x_{n} \right) \right\|_{Y} \geq n $ for all $ n \in \mathbb{N} $ instead.
The assumption is therefore false, so we conclude that $ T $ is indeed bounded.
Note: There is no need to assume that $ X $ is reflexive.
Best Answer
If $X$ is reflexsive $X''=X$ that means $\forall x \in X \Rightarrow x\in X''$.Also $f_n \rightarrow^{w*} f$ then $\forall x \in X$ ; $f_n(x)\rightarrow f(x)$.$X$ reflexsive then $x:X' \rightarrow \Bbb{F}$ $\forall h\in X'$ ; $h(x)=x(h)$ is a linear functional$X''$. Hence $\forall x\in X''$; $x(f_n) \rightarrow x(f) $ because we known $x(f_n)=f_n(x)\rightarrow x(f)=f(x)$ so that says $(f_n)$weak convargence $f$