Suppose $X$ is a normed vector space, $X'$ is its dual space. My question is is there a $X$ satisfied the following condition.
There is a element $x$ in $X$ which is not $0$ and it satisfied that every functional of the dual space $X'$ mapping $x$ to $0$.
The reason that I ask this question is because that I read the mapping $J$ whose value $Jx$ at $x\in X$ is given by $Jx(x')=x'(x) \quad x'\in X'$ is a injection in the book. So why $J$ is a injection?
If $x \neq 0$, then you can show via the Hahn-Banach Theorem that there is a linear functional $f: X \to \mathbb{C}$ such that $f(x) = \left|\left|x\right|\right|$. As for your second question, you're concerned with the map: $$J: X \to X'' \ , \ x \mapsto J(x)$$ where we have that: $$\forall x \in X: \forall f \in X': J(x)(f) := f(x)$$ This can be shown to be injective by proving that it is an isometry (this is usually how it is done when this inclusion is first introduced) but a direct argument is also possible. Indeed, let $J(x) = 0$. That means that for every $f \in X'$, we have that $J(x)(f) = f(x) = 0$. But this implies that $x = 0$ by the statement I mentioned above (the one involving the Hahn-Banach Theorem).