# A subspace $X$ is closed iff $X =( X^\perp)^\perp$

Let $X$ be a subspace of a Hilbert space $H$.

prove:

$X$ is closed iff $X =( X^\perp)^\perp$

I do not know how to proceed. any hints would be appreciated.

thanks.

$X^\perp =${ $y \in H$:$\langle x,y\rangle = 0 \forall x \in X\}$.

Assume $$(X^\perp)^\perp=X$$

Let $$x_n$$ be a sequence in $$X$$ such that $$x_n \rightarrow x$$ where $$x\in H$$.To show $$x \in X$$.

Let $$y\in X^\perp$$ be arbitrary $$\implies \langle x_n,y\rangle=0$$ $$\forall n\in \mathbb N$$.Since $$\langle ,\rangle$$ is continuous

So $$\langle \lim_{n\to \infty}x,y\rangle=\lim_{n\to \infty}\langle x_n,y\rangle=0$$

Thus $$x\in (X^\perp)^\perp=X$$ .Hence $$X$$ is closed

For Converse,

$$X\subseteq(X^\perp)^\perp$$ is trivial

to show $$(X^\perp)^\perp\subseteq X$$ use the fact that $$X$$ is closed $$\implies H= X^\perp+X$$ and $$X^\perp\cap X=\{0\}$$

$(\rightarrow)$ If $X$ is closed linear subspace of a Hilbert space then $X=X^{\perp\perp}$

Proof: Let $x\in X$. Then for all $y\in X^\perp$, $(x,y)=\overline{(y,x)}=0$, so $x\in X^{\perp\perp}$. Thus, $X\subset X^{\perp\perp}$.

Now suppose that $x\in X^{\perp\perp}$. By the orthogonal decomposition theorem; $x=y+z$ where $y\in X$ and $z\in X^\perp$. Since $y\in X$ and $x\in X^{\perp\perp}$, $(x,z)=0=(y,z)$. Thus, $$0=(x,z)=(y+z,z)=(y,z)+(z,z)=||z||^2.$$ So, $z=0$ and $x=y\in X$. Therefore $X^{\perp\perp}\subset X$. Which gives $X=X^{\perp\perp}$, as required.

$(\leftarrow)$ If $X=X^{\perp\perp}$ then $X$ is a closed linear subspace of a Hilbert space.

You could try the proof of the converse for yourself. Hint: use a sequence in $X$ together with the definition of the orthogonal complement, show that the limit point is in $X$ or $X^{\perp\perp}$. (Edit: Alternatively, see @learnmore's answer)