Example on $X^*$ is reflexive but $\sigma(X^*,X)\neq\sigma(X^*,X^{**})$ for a normed space $X$. I just learned this theorem:
Theorem: Let $X$ be a Banach space. Then the following statements are equivalent.
(a) $X$ is reflexive.
(b) $X^*$ is reflexive.
(c) $\sigma(X^*,X)=\sigma(X^*,X^{**})$.
(d) ball $X$ is weakly compact.
And I know the condition $X$ is a Banach space is necessary. But I am wondering can I find an example that satisfies (b) but not (c) for a non-Banach space. 
Thanks in advance!
 A: If $X$ is a vector space then I will denote by $X'$ the algebraic dual space of $X$. For a subspace $Y$ of $X'$, $\sigma(X,Y)$ is the coarsest topology on $X$ such that all elements of $X'$ are continuous. We have the following result:

Lemma: Let $X$ be a vector space and suppose that Y is a subspace
  of $X'$. Then $f \in X'$ is $\sigma(X,Y)$-continuous if and only if $f \in Y$.

In particular, it follows that $\sigma(X^*,X^{**}) = \sigma(X^*,X)$ if and only if $X$ is reflexive. So $(a)$ and $(c)$ are equivalent without the assumption that $X$ is a Banach space.
However we cannot drop this assumption in $(a) \iff (b)$. $X$ is reflexive iff $X^*$ is reflexive and $X$ is complete. So to find an example of a space such that $X^*$ is reflexive and $\sigma(X^*,X^{**}) \neq \sigma(X^*,X)$ we simply need to take a non-reflexive normed space $X$ such that $X^*$ is reflexive. 
An easy way to do this is to take your favourite example of a reflexive Banach space $X$ with proper dense subspace $Y$. Then $Y$ is not complete and hence isn't reflexive. However, since $Y$ is dense in $X$, $Y^* = X^*$ and $X^*$ is reflexive. So $Y$ is a suitable example. 
Explicitly, you can take e.g. $Y=C(\mathbb{R}) \subseteq L^2(\mathbb{R})$ to be the space of continuous functions equipped with the $L^2$-norm.
