# A direct example on $X^*$ is reflexive but the weak topology and weak* topology on $X^*$ are not equal for a normed space $X$.

I asked this question Example on $X^*$ is reflexive but $\sigma(X^*,X)\neq\sigma(X^*,X^{**})$ for a normed space $X$. before. And the answer is to use $X$ is reflexive iff $\sigma(X^*,X)=\sigma(X^*,X^{**})$ to find a normed space X is not reflexive but $X^*$ is reflexive.

But my professor told me that there is an easy example that show $X^*$ is reflexive but $\sigma(X^*,X)\neq\sigma(X^*,X^{**})$ without using the equivalence. Can anyone give me a hint? Thank you in advance!

• Do you know that a Banach space is reflexive if and only if its dual is reflexive? – Daniel Fischer Jun 16 '18 at 15:59
• @DanielFischer♦ Yes. But X is not a Banach space. – Answer Lee Jun 16 '18 at 16:05
• The answer by Rhys Steele in the linked answer not only provides ways of finding easy examples, but also provides a proof that they work (which does not depend on the equivalence). I would presume this is what the professor had in mind; perhaps you should ask him and check? – ktoi Jun 16 '18 at 17:03
• That's the point, $X$ mustn't be complete, and its completion must be reflexive. And the answer contains an explicit example. Another example is $c_{00} \subset \ell^2(\mathbb{N})$, but really, all examples are in a way the same. – Daniel Fischer Jun 16 '18 at 17:33