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I was trying to prove the following fact:

Suppose that $X$ is a Banach space and $N\subset X^*$ a linear subspace. Then $(^\perp N)^\perp = \overline{N}^{w*}$, where the closure is in the weak* topology.

I found this statement here

$G$ is dense in $X^*$ in weak* sense if and only if $G$ is total set

where the author of the post references to an exercise written (without solution) in the book of Dunford and Schwartz.

Tommaso Seneci
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1 Answers1

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$(^{\perp}N)^{\perp}$ is $w^{\ast}$-closed, so $\overline{N}^{w^{\ast}}\subseteq(^{\perp}N)^{\perp}$.

For the other direction, if $x^{\ast}\in X^{\ast}-\overline{N}^{w^{\ast}}$, then an application of Hahn-Banach Theorem gives an $x\in X$ such that $\left<x,x^{\ast}\right>=1$ and $\overline{N}^{w^{\ast}}\subseteq\ker(Qx)$, where $Qx$ is the canonical induced element in $X^{\ast\ast}$. Since $x\in{^{\perp}N}$, then $x^{\ast}\notin(^{\perp}N)^{\perp}$.

Edit:

If we endow $X^{\ast}$ with $w^{\ast}$ topology $(X^{\ast},w^{\ast})$, this is not the norm topology, and the continuous maps on $(X^{\ast},w^{\ast})$ are those $Q(X)$, see here.

user284331
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  • Actually, could you please be more precise of what you mean by "an application of the Hahn-Banach theorem"? Because I can get $x^{}\in X^{}$ separating the two sets, but not $x^{**}\in X$. – Tommaso Seneci Dec 07 '19 at 01:52
  • Actually you can re-examine the proof of the Hahn-Banach separation theorem in $X^{\ast}$ to get the assertion, I will try to refer you a reference of that exact statement. – user284331 Dec 07 '19 at 02:17
  • If you were to give me a better reference that would be great, because a simple application of Hahn-Banach only gives an element into the bi-dual space. – Tommaso Seneci Dec 07 '19 at 02:25
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    See the edited post, something is very important. – user284331 Dec 07 '19 at 02:38
  • Now it is all clear, thanks a lot!! – Tommaso Seneci Dec 07 '19 at 02:43