TITLE: Map from a normed space to its double dual.
QUESTION [5 upvotes]: Let $X$ be a normed space. Let the function $J:X \rightarrow X'' $ be defined by 
$$J(x)(x')=x'(x)\ \forall\ x'\in X' $$
where $X''=\{f:X'\rightarrow\mathbb{C} \mid \hbox{$f$ is bounded linear}  \}$ and $X'$ is the dual of $X$.
Is $J$ injective?
I was trying this,
Suppose we have $x,y\in X $such that 
$$J(x)=J(y)$$
$$\Rightarrow  J(x)(x')=J(y)(x') $$
$$\Rightarrow x'(x)=x'(y)$$
$$\Rightarrow x'(x-y)=0$$
But to get $x=y $ we must have $x'$ to be injective.
I am stuck here. Is this the correct way to go about it?

REPLY [2 votes]: The mapping $J$ defined as above is always injective. And it is onto only if the space $X$ is reflexive (this is the definition for reflexive Banach space actually). That from $x'(x-y)=0\Rightarrow x=y$ is because the dual space is a total set and it follows from Hahn-Banach Theorem.
The corollary is:
If $x\in X\,\,$ and $x'(x)=0,\quad\forall x'\in X'$ then $x=0$