Annihilator of a subset in the dual space. Im reading Peter Lax book and he says:
For any subset $S \subset X'$, we define $S^\perp$ as the subset of those vectors in $X$ that are annihilated by every vector in S.
This confuses me a bit, shoudent it be every functional in X'' that vanishes on S?
Or is this the same thing by identifying those vectors in X? 
 A: Yes, it should be the functionals. I'll try to give you a more accurate description:
Let V be a vector space over a field F, and $$V^{*}$$ be V's dual space (meaning the space of functions from V to F).
Let $$ S \subseteq V^{*}$$ be a subset of the dual space, meaning, it's a set of functionals.
Let's define $$S^{\perp}$$ to be the set of vectors v in V, such that for all the functionals f in $$S \subseteq V^{*}$$, f(v)=0.
Or in formal language:
$$ S^{\perp}=\left \{ v \in V: \forall f \in S, f(v)=0  \right \}  $$
A: There are two choices for definition of annihilators in $V^*$, the dual of a vector space $V$. The first is the one you give, $\{v \in V : \alpha(v) = 0 \space \forall \alpha \in S\}$. The second, using the same definition for annihilators in $V^*$ as we do in $V$, $\{\theta \in V^{**}: \theta(\alpha) = 0 \space \forall \alpha \in S\}$.
If the space is finite dimensional, then $V$ is naturally isomorphic to $V^{**}$ by the map $v \mapsto\hat {\hat v}$ where $\hat{\hat v}(\alpha) = \alpha(v)$. Under this isomorphism, the two different annihilators are isomorphic.
If $V$ is not finite dimensional then $V$ and $V^{**}$ are not generally isomorphic and so the different definitions give genuinely different sets.
