# If $M=M^{\perp\perp}$ for every closed subspace $M$ of a pre-Hilbert space then $H$ is complete

It is well-known that if $$H$$ is a Hilbert space then for every closed subspace $$M$$ we have $$M=M^{\perp\perp}$$ see here

I would like to prove the converse by showing that: if $$M=M^{\perp\perp}$$ for every closed subspace $$M$$ of a pre-Hilbert space then $$H$$ is complete

• I find this question very interesting and I upvoted it. However I see someone has voted to close this. I guess this is because you have not commented anything on what you tried; apparently the rule of "show your effort" does not apply only to low rep users. Jan 13, 2022 at 12:31
• Five seconds of googling gives math.stackexchange.com/questions/631769/… Is this answer of use to you? Jan 17, 2022 at 8:28
• @QuantumSpace: are you ok with that answer? If so, you'll have to explain it to me. Jan 17, 2022 at 19:49
• Yeah, I see what you mean. The problem is precisely that $H_0$ is not closed. What amazes me about this whole thing is the following: if the OP's assertion is true, as it likely is, it means that there is a subspace $M$ of $\mathbb C[x]$, considered as a subspace of $L^2[0,1]$, such that it's double orthogonal contains a polynomial not in $M$. It is really hard for me to imagine this situation. Jan 17, 2022 at 22:49
• @Martin Argerami My intuition is probably garbage but why do you expect this to be true? Jan 17, 2022 at 22:56

We will show that the embedding $$i:H\hookrightarrow \hat H$$ of $$H$$ into its completion is onto. Given $$z\in \hat H\setminus \{0\}$$, the subspace $$M=\{x\in H:\langle i(x),z\rangle =0\}$$ is closed in $$H$$ and its orthogonal complement $$M^\perp$$ is different from $$\{0\}$$ since otherwise $$M=M^{\perp\perp}=\{0\}^\perp=H$$ which implies $$z=0$$. For fixed $$y\in M^\perp\setminus \{0\}$$ and the linear functionals $$\varphi(x)= \langle x,y\rangle$$ and $$\psi(x)=\langle i(x),z\rangle$$ we have kern$$(\psi)=M\subseteq$$ kern$$(\varphi)$$ which implies $$\varphi=a\psi$$ for some scalar $$a$$ (indeed, if $$e\in H$$ satisfies $$\psi(e)=1$$ we get $$x-\psi(x)e\in$$ kern$$(\psi)$$ so that $$0=\varphi(x-\psi(x)e)=\varphi(x)-a\psi(x)$$ for all $$x\in H$$ with $$a=\varphi(e)$$). Now, the functionals on $$\hat H$$ given by $$\langle\cdot,i(y)\rangle$$ and $$\langle\cdot, az\rangle$$ coincide on the dense subspace $$i(H)$$ and hence on $$\hat H$$ which implies $$i(y)=az$$. Since $$a\neq 0$$ this gives $$z\in i(H)$$.

• As this proof is essentially the same as the one of @P.Luis it is his answer which deserves the bounty. Jan 18, 2022 at 11:45

I hope there aren't any gaps, I tried to write my previous proof in a less convoluted way. Feel free to ask for any clarifications as I was really a novice when I wrote this and right now I am not being able to feel that there are any gaps in my intuition about this proof, but it might have gone terribly wrong.

The proof:

Let $$X$$ be a pre Hilbert Space in which for every closed subspace $$M$$ we have $$M = M^{\perp \perp}$$.

We want to prove that $$X$$ is Hilbert. We will do it by embedding $$X$$ in it's completion $$\overline{X}$$ and proving that they are equal.

Let's call $$\overset{\sim}{X}$$ the space of continuous linear functionals of $$X$$ and $$\overset{\sim}{\overline{X}}$$ the one of continuous linear functionals over $$\overline{X}$$

Let's observe that given $$f \in \overset{\sim}{\overline{X}}$$, $$f_{|X} \in \overset{\sim}{X}$$. Also, given $$f \in \overset{\sim}{X}$$ there exists a unique extension $$\overline{f} : \overline{X} \to \overset{\sim}{\overline{X}}$$. So we have a canonical correspondence between $$\overset{\sim}{X}$$ and $$\overset{\sim}{\overline{X}}$$, we will treat both spaces as if they were the same.

Let define the following conjugate linear operators $$T: \overline{X} \rightarrow \overset{\sim}{ \overline{X} } \qquad T (x)(y) = (x,y)$$ If $$T(x_{1})=T(x_{2})$$ then $$(x_{1}-x_{2},y)=0$$ for all $$y \in \overline{X}$$ which implies $$\| x_{1}-x_{2} \|^{2} = (x_{1}-x_{2},x_{1}-x_{2})=0$$ so $$x_{1}=x_{2}$$, so $$T$$ is injective.

Let's assume that $$x_{0} \in \overline{X} - X$$ and arrive a contradiction, so we will know that $$X = \overline{X}$$.

Let's call $$N_{f} = f^{-1} (0) \cap X$$. This is clearly a closed subspace of $$X$$, so by hyphotesis $$N_{f} = N_{f}^{\perp \perp}$$. If $$N_{f} = \{ 0 \}^{\perp}$$, then

$$N_{f} = N_{f}^{\perp \perp} = (N_{f}^{\perp} )^{\perp} = \{ 0 \}^{\perp} = X$$ Which would imply $$f=0$$, and as $$T$$ is injective, $$T(x_{0}) = f = 0$$ would imply $$x_{0} =0$$. But $$x_{0} \in \overline{X} - X$$, and as $$0 \in X$$ it cannot be $$x_{0} = 0$$.

Let's see that $$N_{f}^{\perp} = \{ 0 \}$$ and arrive the desired contradiction. Take an arbitrary $$x \in N_{f}^{\perp}$$, let's see that $$x = 0$$.

As $$x \in N_{f}^{\perp}$$, $$(x,y) =0$$ holds for every $$y \in N_{f}$$. So $$T(x)(y) = 0$$ for all $$y \in N_{f}$$, then $$N_{f} \subset Ker ( T(x) )$$. This implies that either $$T(x)=0$$ or $$T(x) = \lambda f = \lambda T(x_{0})$$ with $$\lambda \neq 0$$ as the span of a linear functional is characterized by the kernel of it's generator, and this kernel is either the entire space or has codimension $$1$$. If $$T(x) = 0$$ then $$x=0$$, as $$T$$ is injective, and we are done.

On the other case $$T(x) = T( \lambda x_{0} )$$ so $$x = \lambda x_{0}$$. As $$\lambda \neq 0$$ and $$x_{0} \neq 0$$, we have $$x \neq 0$$. So as $$x \in N_{f} \subset X$$, we have $$X \cap span (x_{0}) \neq \{ 0 \}$$. So there is a $$\mu \neq 0$$ such that $$\mu x_{0} \in X$$. But as $$X$$ is a vector space $$x_{0} = \mu^{-1} \mu x_{0} \in X$$, which contradicts the fact that $$x_{0} \not\in X$$.

One way or another, we have reached a contradiction.

• I think that your proof is correct but a bit legthy. This is the reason that I'll write another answer. Jan 18, 2022 at 10:06
• Thank you! I have to continue working and improving my presentation, I am really messy and that's no good. Jan 18, 2022 at 14:49