Orthogonal complement of closure

Question

In a Hilbert space, is orthogonal complement of the span of a sequence equal to the orthogoal complement of its closure? i.e. is $\overline{(\operatorname{span}}\left\{x_{n}\right\})^{\perp}=(\operatorname{span}\left\{x_{n}\right\})^{\perp}$?

Answer 1

Yes . $\overline{(\operatorname{span}}\left\{x_{n}\right\})^{\perp}=(\operatorname{span}\left\{x_{n}\right\})^{\perp}$ Because if $x\in (\operatorname{span}\left\{x_{n}\right\})^{\perp}$ , $y\in \overline{(\operatorname{span}}\left\{x_{n}\right\}) $, there is a sequence $a_n \in (\operatorname{span}\left\{x_{n}\right\})$ such that $a_n\to y$. Therefore $x.a_n\to x.y$ , So $0\to x.y$, and $x.y=0$

Answer 2

Yes. In general , if $A\subset B$ than $B^\perp \subset A^\perp$ so , $\overline{span\{a_n\}}^\perp\subset span\{a_n\} ^\perp$.

For the other inclusion, Let, $x\in span\{a_n\}^\perp$.

This means that $<x,y>=0 $ for every $y\in span\{a_n\}$.

We want to show $x\in \overline{span\{a_n\}}^\perp $.

So Let $y\in \overline{span\{a_n\}}$. than there are $y_n\to y$ where $y_n\in span\{a_n\}$.

So, $<x,y> = <x,lim_n y_n> = lim_n <x,y_n> = 0$ where I used the continuity of the inner product. This shows $x\in \overline{span\{a_n\}}^\perp$.