Bidual of Banach Space

Question

I am trying to prove that given a reflective Banach space $X$, i.e $X^{**}=X$, and a complet subspace $E \subseteq X$ then $E$ is reflective to. I had som problem proving it so i looked at a solution from a functional analysis class. Here the lecture states that it is obvious that $E \subseteq E^{**}$. I do not see why this should be true and have tryed to prove this without sucess. I would be very happy if someone could give an explenation or a proof of why this is true.

Answer 1

Any normed linear space $X$ is regarded as a subset of $X^{**}$ by identifying $x \in X$ with $F_x \in X^{**}$ defined by $F_x(f)=f(x)$ for all $f \in X^{*}$. In this sense $E \subseteq E^{**}$. The identification $x \to F_x$ is a nice map: it is linear, injective and preserves distances.

Answer 2

We think of every element $x\in E$ as a functional $x:E^*\to\mathbb{C}$ defined by $x(f)=f(x)$. You can easily check that $x$ is indeed a bounded linear functional, i.e an element of $E^{**}$. So that way we have a map $E\to E^{**}$. It is linear, injective and preserves the norm. (the last two parts are not absolutely trivial, they follow from Hahn-Banach theorem).

So the "inclusion" $E\subseteq E^{**}$ is not a real inclusion as defined in set theory, it is just a very natural embedding.