# Orthogonal and closedness in normed spaces with inner product

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 ?

• $S=\{(x,y)|x, y \in R^2, x>0\}$ is not a subspace of $\mathbb R^2$ ! – Fred Feb 2 '18 at 9:27
• You must go infinite-dimensional to get your examples. – max_zorn Feb 2 '18 at 9:29
• see math.stackexchange.com/questions/1980777/… the answer there shows a subset $S$ with $\bar S\oplus S^\perp\ne X$. – daw Feb 2 '18 at 9:35
• – daw Feb 2 '18 at 9:35

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