# The second dual space

There is something basic I don't understand about the second dual space. Assume $$X$$ is a normed linear space. Denote by $$X^{*}$$ the dual space of $$X$$. If we fix an $$x \in X$$, $$X^{*}(x)$$ is a bounded linear functional and their space is called the second dual.

Here comes the part I don't understand. Isn't the second dual just $$X$$ again? What am I missing?

• If $X^*$ is a space, what is $X*(x)$. Do you mean $f \in X^*$ and then $f(x)$? Because then $f(x) \in \mathbb R$ or $\mathbb C$. Apr 20 '20 at 8:24

Spaces $$X$$ for which $$X^{**}=X$$ are called reflexive spaces. Not every Banach space is reflexive. (An incomplete normed linear space can never be reflexive). $$C[0,1]$$ is an example of a non-reflexive space. All finite dimensional spaces are reflexive.
I think your problem is that you mixed the "continuous" dual and the "algebraic" dual. For a given normed vector space $$X$$ over $$\mathbb{R}$$ or $$\mathbb{C}$$, we have the dual $$X^{\ast }$$ consisting of all continuous linear functionals on $$X$$, and the "dual space" $$X'$$ consisting of all linear functionals on $$X$$. For the second one, we naturally have $$X''\cong X$$ algebraically as vector spaces(this is a wrong argument. In fact, we have $$\dim X'\geq 2^{\dim X}$$. See page 207 of Hungerford's Algebra.), but for the first one, there is no reason that the second dual is isomorphic to $$X$$. The answer of Murthy has given a counter-example.
• In general, we only have $X\hookrightarrow X''$ with the algebraic dual as well, they are not isomorphic if $\dim X\ge\omega$, are they? Apr 20 '20 at 11:05
• @Berci You are right. Hungerford's Algebra gives the result that a dual of a vector space with dimension infinity would have $\dim X'>\dim X$ as cardinal numbers.(Exactly, exercise 12 in page 207.) Apr 21 '20 at 6:25