let $V$ is inner product space.and let $B$ is closed subspace of $V.$ then is it true $(B^\bot)^\bot=B$

Question

let $V$ is inner product space.and let $B$ is closed subspace of $V.$ then is it true $(B^\bot)^\bot=B$

i know $B\subset(B^\bot)^\bot $.is the other way around is also true??

Answer 1

Let $\mathcal{H}$ be the inner product space consisting of all $x\in \ell^2$ for which only finitely many elements of $x=\{x_n\}_{n=0}^{\infty}$ are non-zero. $\mathcal{H}$ is a linear subspace of $\ell^2$ that is an inner product space under the $\ell^2$ inner product. Clearly $\mathcal{H}$ is not a Hilbert space. Define $B$ be the subspace of $\mathcal{H}$ for which $x\in B$ satisfies $$ Lx = \sum_{n=0}^{\infty}\frac{1}{n+1}x_n = 0. $$ This is a closed subspace of $\mathcal{H}$ because $L : \mathcal{H}\rightarrow\mathbb{C}$ is continuous, as it satisfies $$ \|Lx\|_{\mathcal{H}}\le \left(\sum_{n=0}^{\infty}\frac{1}{(n+1)^2}\right)^{1/2}\|x\|_{\mathcal{H}},\;\;\; x\in \mathcal{H}. $$

Then $B^{\perp_{\mathcal{H}}} = \{ y \in \mathcal{H} : y \perp B \} = \{ 0 \}$ because the only $y\in \ell^2$ which is orthogonal to $B$ is $\{1,\frac{1}{2},\frac{1}{3},\cdots\}$, and this element is not in $\mathcal{H}$. Therefore, $B^{\perp_{\mathcal{H}}}=\{0\}$. Hence, $$ (B^{\perp_{\mathcal{H}}})^{\perp_{\mathcal{H}}}=\mathcal{H}\ne B. $$

Answer 2

Let $y\in (B^\bot)^\bot$ then $$\langle y,z\rangle = 0 $$ for all $z\in B^\bot$. Let $P_By\in \overline{B}=B $ the orthogonal projection of y on $B$. Then, is well known that $y-P_By\in B^\bot.$ therefore, \begin{split} \|y-P_By\|^2&=&\langle y,y-P_By\rangle+\langle P_By,y-P_By\rangle\\ &=& \underbrace{\langle y,y-P_By\rangle}_{0}+\underbrace{\langle y,P_By-P_By\rangle}_{\text{since $P_B $ is self-adl $P_B=P_B\circ P_B$}}\\ &=& 0 \end{split} hence $y=P_By\in B$