Let $H$ be an Hilbert space and define $M^{\perp}$ to be:

$$M^{\perp}=\{x\in H\vert(x,m)=0\,\forall\, m\in M\}$$

where $M\subset H$ is any subset of $H$. It is easily seen that $M,\overline{M}\subset (M^{\perp})^{\perp}=:M^{\perp\perp}$. I don't manage to prove that $M^{\perp\perp}\subset M$ if $M$ closed.

I know I have to use completeness since there are counterexamples in non-complete spaces, but I am kind of stuck. I tried to use the orthogonal decomposition theorem twice or to use the projection onto a closed convex in order to get a contradiction by supposing there exists $y\in M^{\perp\perp}\setminus M$, but it leads me nowhere.

Any hint is appreciated. Thank you.


2 Answers 2


If you're allowed to use the orthogonal decomposition theorem, you can argue as follows: Let $v \in M^{\perp \perp}$. Then we can write $v = v_1 + v_2$, where $v_1 \in M$ and $v_2 \in M^{\perp}$. Hence we know that $\langle v_1, v_2 \rangle = 0$, and also $\langle v, v_2 \rangle = 0$. This implies $\langle v_2, v_2 \rangle = \langle v - v_1, v_2 \rangle = 0$, so $v_2 = 0$, and we get $v \in M$.

  • $\begingroup$ Very smart answer! Is it possibile to prove the other direction by similar means? $\endgroup$
    – sound wave
    Jan 29, 2019 at 13:03
  • $\begingroup$ hope I don't sound stupid, but why does this work only for closed subspace ? $\endgroup$
    – vib_29
    Sep 14 at 14:49

Ok, I finally found the answer, using the orthogonal decomposition theorem.

Suppose $x\in M^{\perp\perp}\setminus M$. By the orthogonal decomposition theorem, there exists $m\in M$ and $y\in M^{\perp}$ such that $x=m+y$ since $M$ is closed. Thus, $x-m=y$. But as $M^{\perp\perp}$ is a subspace of $H$ containing $M$, we have that $x-m\in M^{\perp\perp}$. But as $y\in M^{\perp}$, this implies that $x-m\in M^{\perp}$ and thus $x-m\in M^{\perp}\cap M^{\perp\perp}=\{0\}$ and we deduce that $m=x\in M$, which is a contradiction.


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