Given a normed vector space $X$ such that the norm can be derived from an inner product ( pre Hilbert space ) and $S\subset X$ a subspace of $X$, then generally $\overline{S}\neq S ^{\bot\bot}.$ Also in general, $\overline{S}\oplus S ^{\bot}\neq X. $

I am wondering if the case $S=\{(x,y)|x, y \in R^2, x>0\}\subset R^2$ is a good example of the first situation ? Can you give some other examples ? How about examples for the second situation ?

Many thanks in advance.


As an example of the second situation, take $X$ as the set of all sequences $(a_n)_{n\in\mathbb N}$ of real numbers such that $n\gg1\implies a_n=0$. Consider the inner product$$\bigl\langle(a_n)_{n\in\mathbb N},(b_n)_{n\in\mathbb N}\bigr\rangle=\sum_{n=1}^\infty a_nb_n.$$Let$$S=\left\{(a_n)_{n\in\mathbb N}\in X\,\middle|\,\sum_{n=1}^\infty\frac{a_n}{n^2}=0\right\}.$$and note that $S\neq X$. Then $S^\perp=\{0\}$ and therefore $S\oplus S^\perp\neq X$.

  • $\begingroup$ Thanks. What is meant by $n\gg1 $. $\endgroup$ – user249018 Feb 2 '18 at 11:35
  • $\begingroup$ @user249018 It means “if $n$ is much bigger than $1$”. $\endgroup$ – José Carlos Santos Feb 2 '18 at 11:35
  • $\begingroup$ What do you mean by $H$ ? $\endgroup$ – user249018 Feb 2 '18 at 11:40
  • $\begingroup$ @user249018 It was a typo. I meant $X$. $\endgroup$ – José Carlos Santos Feb 2 '18 at 11:41
  • $\begingroup$ Can we resume the condition $n\gg1\implies a_n=0 $ by saying that for only finitely many $n, a_n\neq 0 ?$ $\endgroup$ – user249018 Feb 2 '18 at 11:46

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