# $M$ and $N$ are subspaces of a Hilbert space. If $M\subset N$, show that $N^{\perp}\subset M^{\perp}$. Show also that $(M^{\perp})^{\perp}=M$.

$M$ and $N$ are subspaces of a Hilbert space. If $M\subset N$, show that $N^{\perp}\subset M^{\perp}$. Show also that $(M^{\perp})^{\perp}=M$.

I know that the orthogonal complement of $X$ is the set $X^{\perp}=\{x\in H : x{\perp}X\}$ where $X$ is any subset of a Hilbert space $H$. I'm not sure how to proceed. Any hints or solutions are greatly appreciated.

• You need closures somewhere to make any of these statements true. Given that, the inclusion follows directly from the definition. One direction of the equality is straightforward, the other benefits from the Hahn Banach theorem. – copper.hat Jun 14 '15 at 21:41
• When you write $M\subset N$, do you allow $M=N$? – egreg Jun 14 '15 at 22:11
• I think so, the book I'm using doesn't specify. – 1233dfv Jun 14 '15 at 22:15

If $x$ $\in$ $N^{\perp}$ then $x$ $\perp$ $N$ and since $M \subset N$ then $x \perp M$ therefore $x \in M^{\perp}$
Now let $x \in M$, then $x \perp M^{\perp}$ and therefore $x \in M^{\perp \perp}$. That means $M \subset M^{\perp \perp}$. Since $M^{\perp \perp}$ is a closed and linear subspace containing M that means $\overline{\lt M \gt} \subset M^{\perp \perp}$
Suppose now that there is a $x \in M^{\perp \perp} \setminus \overline{\lt M \gt}$. There exists a $y \in \overline{\lt M \gt}$ so that $x - y \perp \overline{\lt M \gt}$. Then $x - y \perp M$ and therefore $x - y \in M^{\perp}$. Also $x - y \in M^{\perp \perp}$ since it is a linear combination of elements of $M^{\perp \perp}$. This means that $x - y = 0$ and therefore $x = y$ and $x \in \overline{\lt M \gt}$ which is false.