# How to show $\hat{x}\in X^{**}$ without using $(X^*,wk^*)^*\subseteq X^{**}$?

Definition: Let $X$ be a topological vector space and let $x\in X$. Then $x$ defines a linear functional $\hat{x}$ on $X^*$ via $\hat{x}(f)=f(x)$ $(f\in X^*)$.

From this definition we can easily get $\hat{x}$ is weak* continuous on $X^*$. Therefore, we have $\hat{x}\in (X^*,wk^*)^*$.

Now let $X$ be a normed space. My question is from $\hat{x}\in (X^*,wk^*)^*$ how can we get $\hat{x}\in X^{**}$ without using $(X^*,wk^*)^*\subseteq X^{**}$?

If you mean how to show that $\hat x$ is bounded, it is simply $$|\hat x(f)|=|f(x)|\leq\|f\|\,\|x\|$$
• Thanks for your answer. Can we use the fact $\hat{x}\in (X^*,wk^*)^*$ to get $\hat{x}\in X^{**}$? – Answer Lee Mar 10 '18 at 21:11
• I'm not entirely sure what you expect: it is well-known that $(X^*,wk^*)^*=X$. – Martin Argerami Mar 10 '18 at 21:21
• I am sorry if I make you confused. I saw a conclusion on a paper and it is: $\hat{x}\in (X^*, wk^*)^*$, and hence $\hat{x}\in X^{**}$. I just don't know how he gets it. Is it because some property about weak star topology? Thank you! – Answer Lee Mar 10 '18 at 21:31
• As I said, if $\hat x\in (X^*,wk^*)^*$, then $\hat x\in X$ (so, in particular, it is in $X^{**}$. It is not entirely obvious, but it is not a very complicated proof either. I've seen the result in Conway's Functional Analysis book, but it should also be elsewhere. – Martin Argerami Mar 10 '18 at 21:38
• How to get $\hat{x}\in X^{**}$ from $\hat{x}\in X$? – Answer Lee Mar 10 '18 at 22:05