Examples reflexive spaces Let $(E, \| \cdot \|_E)$ be a normed vector space over a field $\mathbb{K}$ and $(E', \| \cdot \|_{\mathrm{op}})$ its dual.


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*Theorem 1). If $(E, \| \cdot \|_E)$ is reflexive, then each bounded sequence in $E$ (bounded in the topology induced by $\| \cdot \|_E$) has a weakly convergent subsequence.

*Theorem 2). Is $(E, \| \cdot \|_E)$ reflexive, then is $(E', \| \cdot \|_{op})$ reflexive as well.


Question: Find counterexamples on the conditions of the theorems where:


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*For 1): If $(E, \| \cdot \|_E)$ is not reflexive.

*For 2): $(E', \| \cdot \|_{\mathrm{op}})$ is reflexive but $(E, \| \cdot \|_E)$ not.
Thanks.
 A: 1) Let $(E,\|\cdot\|_E)=(\ell^1(\Bbb N),\|\cdot\|_1)$ and $(e_n)_{n\ge 1}$ be its standard basis. For every $n\ge 1$, $\|e_n\|_{1}$ is $1$, hence $(e_n)_{n\ge 1}$ is a bounded sequence. Suppose it admits a weakly convergent subsequence $(e_{n(k)})_{k\ge 1}$ and let $y\in (\ell^1)'$ be defined by $y(\sum_{n}x_n e_n)=\sum_{i=1}^\infty (-1)^ix_{n(i)}$. Then, $$
y(e_{n(k)})=(-1)^k
$$ does not converge, and this contradicts $(e_{n(k)})_{k\ge 1}$ is weakly convergent. So, $(e_n)_{n\ge 1}$ does not have a convergent subsequence.
2) If the dual space $E'$ is reflexive, then its predual $E$ is also reflexive provided that $E$ is Banach. For a counterexample, take
$(E,\|\cdot\|)=(c_{00},\|\cdot\|_2)$ where $c_{00}$ is the space of all finite sequences $x=(x_n)_{n\ge 1}$ and $\|x\|_2=(\sum_n |x_n|^2)^{\frac12}$ is the 2-norm. Since $c_{00}$ is dense in $\ell^2$, any bounded linear functional $y:c_{00}\to\Bbb C$ can be uniquely and continuously extended to $\bar y:\ell^2\to\Bbb C$. The correspondence $y\mapsto \bar y$ ,$\bar y|_E\leftarrow \bar y$gives an isomorphism between $E'$ and $(\ell^2)'=\ell^2$, hence $E'$ is reflexive. But note that $E$ is not even a Banach space, so cannot be a reflexive space.
