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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?

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Which of his books? $S^\bot=\bigcap_{f\in S} Ker(f)$ makes sense to me. – AD. Dec 25 '12 at 10:34
You have two options: the annihilator in $X$ Lax defines or the one in $X''$ which you propose. If $X$ is not reflexive those are distinct in general. Consider $S = \{0\}$ for an easy example. – Martin Dec 25 '12 at 10:38
Oki good! thanks! – Johan Dec 25 '12 at 10:56
Some books, like Functional Analysis by Conway, use different notation for the annihilator $S^{\perp}$ and the pre-annihilator ${}^{\perp}\!S$ to avoid such confusion. – user53153 Jan 1 '13 at 17:40

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: $$ S^{\perp}=\left \{ v \in V: \forall f \in S, f(v)=0 \right \} $$

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+1 Welcome to Math.SE! Please do keep coming back. – user53153 Jan 5 '13 at 8:18

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.

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