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

Solution 1:

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 $X^\perp\cap X=\{0\}$

Solution 2:

$(\rightarrow)$ 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)