Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
Which of his books? $S^\bot=\bigcap_{f\in S} Ker(f)$ makes sense to me. –  AD. Dec 25 '12 at 10:34
1  
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
add comment

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

share|improve this answer
    
+1 Welcome to Math.SE! Please do keep coming back. –  user53153 Jan 5 '13 at 8:18
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.