We know that if $X$ is a normed linear spaces, the strong convergence of a sequence implies the weak convergence.
Suppose that $X$ is a normed linear spaces. An element in $X$ is denoted by $x$ and in $X^{\star\star}$ is denoted by $x^{\star\star}$. Then we can easily get:
\begin{align*} \left \| x_{n}-x \right \|& =\left \| x^{\star\star}_{n}- x^{\star\star} \right \|\\ &=\underset{\left \| f \right \|\leq 1}{\sup}\left \{\left | \left \langle f,x_{n}-x \right \rangle \right |: f\in X^{\star}\right \} \end{align*}
That is, the condition that $\left \langle f, x_{n} \right \rangle$ converges to $\left \langle f, x \right \rangle$ uniformly for arbitrary $f\in B^{\star} \left ( B^{\star}:= \left \{ f\in X^{\star}: \left \| f \right \|\leq 1 \right \} \right )$ is equivalent to the strong convergence of $\left \{x_{n} \right \}$ to $x$.
However, even if $X$ is reflexive and $x_{n}\overset{weak}{\rightarrow }x$, we may still not make a conclusion that $\left \| x_{n}-x \right \|\rightarrow 0$, where $n\rightarrow \infty$. On the other hand, the weak convergence in $X^{\star}$ is equivalent to the weak-$\star$ convergence when $X$ is reflexive.
My questions are the following two somewhat appeared to be relevant problems:
Firstly, suppose that $X$ is reflexive, can we weaken the condition "$\left \langle f, x_{n} \right \rangle$ converges to $\left \langle f, x \right \rangle$ uniformly for arbitrary $f\in B^{\star}$" in the above such that $\left \{ x_{n}\right \}$ still converges strongly to $x$ ?
Secondly, if the weak convergence in $X^{\star}$ is equivalent to the weak-$\star$ convergence, can we get a conclusion that $X$ is reflexive ?
