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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!

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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.

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  • $\begingroup$ I am sorry that I am new to analysis. Can you explain your example why $Y$ is not complete and how to get is not reflexive and why $Y$ is dense in $X$ we have $Y^*=X^*$? Thank you! $\endgroup$ – Answer Lee Jun 13 '18 at 23:24
  • $\begingroup$ Well if $Y$ were complete it would be closed in $X$. But $Y$ is a dense subspace of $X$ and $Y \neq X$ so it cannot be closed. It follows that $Y$ is not reflexive since all reflexive spaces are complete since they are isometrically isomorphic to their double dual which is a complete space. The last point is really an abuse of notation on my part. In fact, $Y^*$ is isometrically isomorphic to $X^*$ which is good enough for our purposes. The isometric isomorphism is given by sending $f \in Y^*$ to its unique continuous extension to a functional on $X$. $\endgroup$ – Rhys Steele Jun 13 '18 at 23:30
  • $\begingroup$ Thanks for you answer. Can you be more specific about why $Y^*\cong X^*$? $\endgroup$ – Answer Lee Jun 13 '18 at 23:36
  • $\begingroup$ I can certainly try, but maybe the comments aren't the best place to go into detail on this. The point is that since $Y$ is dense in $X$, for every $g \in Y^*$ there is a unique $\Phi(g) \in X^*$ such that $\Phi(g)(y) = g(y)$ for all $y \in Y$ and furthermore we have $\| \Phi(g) \| = \|g\|$. It's then not to hard to check that $\Phi$ is linear and surjective so that $\Phi: Y^* \to X^*$ is an isometric isomorphism. $\endgroup$ – Rhys Steele Jun 13 '18 at 23:42
  • $\begingroup$ I am sorry if I have too many questions. I basically get the idea but wht $Y$ is dense can imply that there is a there is a unique $\Phi(g)\in X^*$ such that.... Thank you so much!! $\endgroup$ – Answer Lee Jun 13 '18 at 23:49

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