Bidual space of a normed space I have trouble understanding the bidual space of the normed space $(E, \|\cdot \|)$. Could someone help me understand it?
 A: *

*Every normed space $(E,\|\cdot\|)$ has a dual space, denoted by $E^*$, defined as the vector space of all
continuous linear functionals on $E$, equipped with the norm
$$\|x^*\|:=\sup\{|x^*(x)|: x\in E,\ \ \|x\|\leq 1\}\quad (x^*\in E^*)$$

*As $(E^*,\|\cdot\|)$ is also a normed space, with the norm defined in 1., it also has a dual space, denoted $E^{**}$, equipped with the norm
$$\|x^{**}(x)\|:=\sup\{|x^{**}(x^*)|: x^*\in E^*,\ \ \|x^*\|\leq 1\}\quad (x^{**}\in E^{**})$$
The normed space $E^{**}$ (equipped with its norm) is called the bidual of $E$.


Some basic properties duality are the following:
A.  A dual space is a Banach space. That is, even if $E$ is not a complete normed space, its dual $E^*$ is always complete, hence a Banach space. In particular, the bidual of a normed space is also a Banach space, whether $E$ is complete or not.
B. There is a natural embedding of $E$ into its bidual $E^{**}$. In fact, every $x\in E$ defines a continuous linear functional $i(x)$ on $E^*$ by the formula
$$(i(x))(x^*):=x^*(x)\quad x^*\in E^{*}$$
It is not hard to check that the map $x\to i(x)$ is a linear isometry from $E$ into $E^{**}$. It is called the cannonical embedding of $E$ in its bidual.
C. If the cannonical embedding of part B. is also onto, then $E$ is said to be reflexive. A reflexive space is therefore a space $E$ which coincides with its bidual. Note that an incomplete normed space (i.e. not Banach) cannot be reflexive, because the bidual is complete, and isometries preserve completeness. Thus, an incomplete normed space is always "strictly smaller" than its bidual.
