# Orthogonal complement of the orthogonal complement

Consider the infinite dimensional vector space of functions $$M$$ over $$\mathbb{C}$$.

The inner product defined as in square integrable functions we use in quantum mechanics.

How can we show that the orthogonal complement of the orthogonal complement gives the topological closure of the vector space and not the vector space itself?

If we already know that the orthogonal complement is itself closed.

$$M^{{\perp}{\perp}}=\overline M$$

• Orthogonal complement with respect to which inner product? Commented Mar 12, 2020 at 10:40
• Whatever the inner product, though, a good place to start is probably showing that $M^{\perp}$ is always closed. Commented Mar 12, 2020 at 10:43
• The inner product used in quantum mechanics defined as square integrable functions $L^2(\mathbb{R})$
– VVM
Commented Mar 12, 2020 at 10:44
• See corollary 21.6. (iii) on Schilling's "Measures integrals and martingales" Commented Mar 12, 2020 at 11:21

Let $$M \subseteq \mathcal{H}$$ be a linear subspace of the Hilbert space.

(i) Consider $$\{\varphi_n\}$$ as a Cauchy sequence in $$M$$

Because $$\forall \varphi_n \in M \implies \forall \mu \in M^\perp: \langle\mu, \varphi_n\rangle=0\implies \varphi_n\in (M^\perp)^\perp$$

Therefore, $$M \subseteq M^{\perp\perp}$$

(ii) $$M^{\perp\perp}$$ is an orthogonal complement and hence it is a closed linear subspace of $$\mathcal{H}$$.

Therefore, it is a sub Hilbert space. Hence complete. So $$\{\varphi_n\}$$ is a Cauchy sequence in $$M^{\perp\perp}$$ and hence it converges in $$M^{\perp\perp}$$:

$$\lim_{n \to \infty} \varphi_n= \varphi \in M^{\perp\perp}$$

We have shown that $$\{\varphi_n\}$$ as a Cauchy sequence in $$M$$ converges to $$\varphi$$ in $$M^{\perp\perp}$$. That is the topological closure of $$M$$ is $$M^{\perp\perp}$$:

$$\overline{M}=M^{\perp\perp}$$

Here is an answer based on the fact that for in a Hilbert space $$H$$, given any closed linear subspace $$L\subset H$$, $$H=L \oplus L^\perp$$ This follows from the existence end uniqueness of the orthogonal projection $$P:H\rightarrow L$$, with maps any $$x\in H$$ to a point $$v\in L$$ such that $$\|x-v\|\leq \|x-\ell\|$$ for all $$\ell\in L$$.

Let $$M$$ be a linear subspace of $$H$$. Then $$H=\overline{M} \oplus (\overline{M})^{\perp}$$ Since $$M\subset \overline{M}$$, $$M^\perp\supset (\overline{M})^\perp$$. If $$y_n\in M^\perp$$ and $$y=\lim_ny_n=y$$, then for any $$u\in M^\perp$$ $$|\langle u,y\rangle|\leq \|u\|\|y-y_n\|+|\langle u,y_n\rangle|\xrightarrow{n\rightarrow\infty}0$$ Hence $$M^\perp =(\overline{M})^\perp$$ and $$H=\overline{M} \oplus M^{\perp}$$ It is clear that $$M\subset (M^\perp)^\perp$$. Since $$A^\perp$$ is closed for any $$A\subset H$$, we have that $$\overline{M}\subset (M^\perp)^\perp$$ Now, since $$M^\perp$$ is a closed linear subspace, $$H=(M^\perp)^\perp \oplus M^\perp$$

Let $$u\in (M^\perp)^\perp$$. Then there is are unique vectors $$u_1\in \overline{M}$$ and $$u_2\in M^\perp$$ such that $$u=u_1+u_2$$. Then $$u-u_1=u_2\in (M^\perp)^\perp \cap M^\perp$$ since $$u-u_1\in(M^\perp)^\perp$$ and $$u_2\in M\perp$$. Hence $$u-u_1=u_2=0$$, ans so, $$u=u_1\in \overline{M}$$. This prove that $$(M^\perp)^\perp\subset \overline{M}$$ and we are done.