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Given a Banach space $X$.

Denote the annihilator: $$A\subseteq X:\quad A^\perp:=\{x'\in X':x'A=0\}$$

Then does it hold: $$Z_\lambda=\overline{Z_\lambda}\leq X:\quad\overline{\left\langle\bigcup_{\lambda\in\Lambda}Z_\lambda^\perp\right\rangle}=\left(\bigcap_{\lambda\in\Lambda}Z_\lambda\right)^\perp$$ How can I check this?

Similar thread: Hilbert vs. De Morgan

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  • $\begingroup$ @Crostul: So far I have checked that the LHS is included in the RHS. Unfortunately, I don't have a computer with me right now. So it would be very inconvenient to write down a formal check. I will add a check to the question later, though. Is that fine? $\endgroup$ – C-Star-W-Star Sep 16 '16 at 11:50
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Here is a counterexample, consider $C([0,1])$. Note that $C([0,1])^*$ is given by bounded measures on $[0,1]$. Now look at

$$\Lambda=\left\{\text{finite subsets of }[0,1]\right\}\\ Z_\lambda = \{f\in C([0,1]) \mid f\lvert_\lambda = 0\}$$ Then you get the following: $$\bigcap_{\lambda} Z_\lambda = Z_{\,\bigcup \lambda}\\ Z_\lambda^\perp=\left\{\mu\in C([0,1])^*\mid \mu\left(\overline{\lambda}\right)=0\right\}$$ Since $\lambda\in\Lambda$ consists of isolated points $Z_\lambda^\perp$ always consists of linear combinations of dirac measures on $\lambda$. So for example the standard Lebesgue measure $\mu = dx$ does not lie in $\overline{\left\langle\bigcup_{\lambda\in\Lambda}Z_\lambda^\perp\right\rangle}$, but clearly it does in $\left(\bigcap_{\lambda}Z_\lambda\right)^\perp=Z_{[0,1]}^\perp=\{0\}^\perp$.

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  • $\begingroup$ Which topology do you consider on $\mathcal{C}^\infty([0,1])$? Wouldn't work too the Banach space $\mathcal{C}([0,1])$? $\endgroup$ – C-Star-W-Star Oct 8 '16 at 16:40
  • $\begingroup$ @AlexanderFrei sorry, I meant $C([0,1])$. I don't know why I put the $^\infty$ there, this way it is wrong! $\endgroup$ – s.harp Oct 8 '16 at 17:44

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