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?
Annihilator of a subset in the dual space.
-
0Some books, like *Functional Analysis* by Conway, use different notation for the annihilator $S^{\perp}$ and the pre-annihilator ${}^{\perp}\!S$ to avoid such confusion. – 2013-01-01
2 Answers
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 \} $
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.