subnet in reflexive spaces Let $X$ be a reflexive banach space. Let $(a_i)$ be a net in $X$ such that 
$$          \liminf_{i} \| a_i \| = \alpha \in R$$
Then can we conclude that there exist a subnet of $(a_i)$ which is weakly convergent ?
If the answer is no in general, what if $X$ be a separable Hilbert space? 
 A: Let $X$ be a reflexive Banach space. Let $a:D\rightarrow X$ be a
net in $X$, where $(D,\preceq)$ is a directed system. Suppose that
$\liminf_{i}||a_{i}||=\alpha\in[0,\infty)$.
Firstly, we show that there exists a bounded subnet of $a$. For each
$i\in D$, define $\beta_{i}=\inf\{||a_{j}||\mid j\in D\mbox{ and }i\preceq j\}\in[0,\infty)$.
Then we obtain a net of non-negative real numbers $\beta:D\rightarrow[0,\infty)$,
$i\mapsto\beta_{i}$. Clearly, the net $\beta$ is monotonic increasing.
By assumption, $\lim_{i}\beta_{i}=\alpha<\infty$. Hence $\beta(i)\leq\alpha<\infty$
for all $i\in D$. For each $i\in D$, let $A_{i}=\{j\in D\mid||a_{j}||<\alpha+1\mbox{ and }i\preceq j\}$,
which is non-empty because $\beta_{i}<\alpha+1\Rightarrow\alpha+1$
is not a lower bound of the set $\{||a_{j}||\mid j\in D\mbox{ and }i\preceq j\}$.
By the Axiom of Choice, there exists a function $\theta:D\rightarrow\bigcup_{i\in D}A_{i}\subseteq D$
such that $\theta(i)\in A_{i}$. Let $b=a\circ\theta:D\rightarrow X$.
Note that $\theta$ is a co-final map from $D$ into $D$: For, let
$i_{0}\in D$ be arbitrary. Let $j_{0}=i_{0}$. For any $j\in D$
with $j_{0}\preceq j$, we have $\theta(j)\in A_{j}$, which implies
that $j\preceq\theta(j)$ by the very definition of $A_{j}$. That
is, $i_{0}\preceq j$ and $j\preceq\theta(j)$, which further implies
that $i_{0}\preceq\theta(j)$ by the transitive property of direction
$\preceq$. This shows that $\theta$ is a co-final map and hence
$b$ is a subnet of $a$. By the way we construct $\theta$, for each
$i\in D$, $||b(i)||=||a(\theta(i))||<\alpha+1$. Therefore, $b$
is a bounded subnet of $a$.
By assumption, $X^{\ast\ast}=X$ (more accurately, $X^{\ast\ast}$
is isometric isomorphic to the image of $X$ in $X^{\ast\ast}$ via
the canonical embedding $x\mapsto\hat{x}$, where $\hat{x}(f)=f(x),$
$f\in X^{\ast}$). Regard $b$ as a bounded net in $(X^{\ast})^{\ast}$
and recall that every bounded ball of $(X^{\ast})^{\ast}$ (bounded
ball means the set $\{\xi\in X^{\ast\ast}\mid||\xi||\leq r\}$, for
some $r\in[0,\infty)$) is $\sigma(X^{\ast\ast},X^{\ast})$-compact
(the Banach-Alaoglu Theorem), so $b$ further has a subnet $c$ which
is convergent with respect to the $\sigma(X^{\ast\ast},X^{\ast})$-topology
on $X^{\ast\ast}$. Clearly $c$ is a subnet of $a$. Moreover, since
$X^{\ast\ast}=X$, the $\sigma(X^{\ast\ast},X^{\ast})$-topology is
actually the $\sigma(X,X^{\ast})$-topology on $X$. In short, $a$
has a subnet $c$ which is convergent with respect to the weak topology.
A: Hint: There is a bounded subnet...
