Graduate Orthogonal complement of the orthogonal complement

[victorvmotti]

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.

If we already know that the orthogonal complement is itself closed, 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?

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

 

[jambaugh]

Is this a homework question or are you merely curious? I would say you should look carefully at the definition of the orthogonal complement (keeping in mind the infinity of the dimension).

 

[member 587159]

Clearly $\overline{M}$ is contained in the double orthogonal complement as limits of sequences preserve orthogonality. So you can't expect that it is equal to $M$ in general.

 

[victorvmotti]

Is this proof a good one:

Let $M \subseteq \mathcal{H}$ be a linear subspace.

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

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 $M^{\perp\perp}$ or $M$ is densely defined in $M^{\perp\perp}$. That is the topological closure of $M$ is $M^{\perp\perp}$:

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

 

[Infrared]

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

This is true in any inner product space (or indeed in any space with a symmetric bilinear form). You don't need to talk about Cauchy sequences. Here is an argument: let $v\in M$. Then, by the definition of orthogonal complement, $\langle v,w\rangle=0$ for all $w\in M^{\perp}$. This equation also means $v\in (M^{\perp})^{\perp}.$

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

Right, although you might want to check this fact (that orthogonal complements are topologically closed) if you haven't thought about the proof. Using this, taking closures of $M\subset (M^{\perp})^{\perp}$ shows that $\overline{M}\subset (M^{\perp})^{\perp}.$

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 $M^{\perp\perp}$ or $M$ is densely defined in $M^{\perp\perp}$.

What you wrote doesn't imply that $M$ is dense in $(M^{\perp})^{\perp}$. e.g. consider $\mathbb{R}^1\subset\mathbb{R}^2$ included in the obvious way. Any Cauchy sequence in $\mathbb{R}^1$ converges in $\mathbb{R}^2$, but it's definitely not dense.