Let $X$ be a normed space and $F\subseteq X$. How to prove ${^\perp}(F^\perp)$ is the closed linear span of $F$? Here is some definitions: If $A\subseteq X$, then we denote $$A^{\perp}=\{x^*\in X^*:\langle a,x^*\rangle=0 \textrm{ for all }a\in A\}.$$ Similarly, if $B\subseteq X^*$, then we denote $$^{\perp}B=\{x\in X:\langle x,b^*\rangle=0 \textrm{ for all }b^*\in B\}.$$

Thank you!

  • $\begingroup$ $A^{\perp}$ is closed as it equals $\bigcap_{a \in A} (e_a)^{-1}[\{0\}]$, where $e_a(x^\ast) = x^\ast(a) (= <a,x^\ast>)$ for $x^\ast \in X^\ast$. All $e_a$ are continuous on $X^\ast$ and $\{0\}$ is closed, so we have an intersection of closed sets (hence closed). Similarly for ${^\perp}B$. $\endgroup$ – Henno Brandsma Aug 27 '17 at 16:35

Clearly $A^{\perp}$ and $^{\perp}B$ are linear subspaces of the respective space. Also both of them are closed as preimage of a closed set. So it is only left to show that $F\subseteq {}^{\perp}(F^{\perp})$. $\renewcommand{\Re}{\operatorname{Re}} \newcommand{\cls}{\operatorname{cls}}$

take $f\in F$ then by Definition of $F^{\perp}$ we have $$\langle f,g \rangle = 0 \quad\text{for all}\quad g\in F^{\perp}.$$ Well and this justifies that $f$ is in ${}^{\perp}(F^{\perp})$.

From this we have $\cls F \subseteq {}^{\perp}(F^{\perp})$, where $\cls F = \overline{\operatorname{span} F}$.

For the other inclusion we need Hahn-Banach

Theorem (part of Hahn Banach) Let $X$ be a local convex vector space and $A$, $B$ disjoint, convex and non empty and $A$ is compact and $B$ is closed. Then there is a $f\in X'$ and $\gamma_1,\gamma_2\in \mathbb{R}$ such that $$ \Re f (A) \leq \gamma_1 <\gamma_2 \leq \Re f(B) $$

Note that for $f\in X'$ we have that $f(\operatorname{cls} F)$ is a subspace of $\mathbb{C}$, which means it is either $\mathbb{C}$ or $\{0\}$. Using this and the theorem for $A = \{x\}$ where $x\notin \operatorname{cls}F$ and $B = \operatorname{cls}F$ we get an $f\in X'$ and $\gamma_1,\gamma_2 \in \mathbb{R}$ such that \begin{align*} \Re f (\{x\}) \leq \gamma_1 <\gamma_2 \leq \Re f(\cls F) \end{align*} since the right-hand-side is a subspace of $\mathbb{R}$ is must be $\{0\}$ and consequently $f(\cls F)=\{0\}$. In particular $f(F)= 0$. So this $f\in X'$ has the property $$ f\in F^\perp \quad \text{and}\quad \langle x,f\rangle \neq 0. $$ Hence $x\notin {}^{\perp}(F^{\perp})$

  • 1
    $\begingroup$ The hardest part is missing. Why is it the smallest closed subspace containing $F$ ? $\endgroup$ – Gribouillis Aug 26 '17 at 20:05
  • $\begingroup$ you are right I will edit it $\endgroup$ – Nathanael Skrepek Aug 26 '17 at 20:06
  • $\begingroup$ It's a consequence of the Hahn Banach extension theorem. $\endgroup$ – Gribouillis Aug 26 '17 at 20:18
  • $\begingroup$ @Gribouillis done $\endgroup$ – Nathanael Skrepek Aug 26 '17 at 20:43
  • $\begingroup$ Correct. You can also avoid using complexes in the case the scalars are in $\mathbb{R}$. This is not said in the original question. Another way to view Hahn Banach is to define a continuous linear form on $[x] + cls(F)$ by $f(\lambda x + y) = \lambda$, and to use that this linear form can be extended to $X$. $\endgroup$ – Gribouillis Aug 26 '17 at 20:48

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