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I am trying to solve the following problem:
Let $\bar{H}$ be a Hilbert space, $H$ - its dense linear subspace, $z_0\in \bar{H}-H$. Consider the following subspace $M\subset H$ defined as $\{x\in H:<x,z_0>=0\}$. Is is true that $Cl(M)=Span(z_0)^{\perp}$ ($Cl(M)$ is the closure of $M$ in the topology of $\bar{H}$, $<,>$ is the given inner product).

It is easy to verify that $Cl(M)\subset Span(z_0)^{\perp}$ (really, if $x\in Cl(M)$ then there exists a sequence $\{x_n\}\subset M$ such that $x=\lim x_n$; we see that an inner product is continuous, hence $<x,z_0>=<\lim x_n,z_0>=\lim <x_n,z_0>=0$). But what about the converse $Span(z_0)^{\perp}\subset Cl(M)$?

I will be grateful for any help!

PS
$Span(z_0)^{\perp}$ is the orthogonal completion for one-dimensional $Span(z_0)$

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    $\begingroup$ Use \langle and \rangle for the angle-brackets of the inner product. That looks nicer, and gives proper spacing. $\endgroup$ – Daniel Fischer Oct 24 '13 at 20:40
  • $\begingroup$ Regarding the question, I'm not sure. Thinking about it. $\endgroup$ – Daniel Fischer Oct 24 '13 at 20:42
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Yes.

Suppose $z \in \operatorname{Span}(z_0)^\perp$, i.e. $\langle z, z_0 \rangle = 0$. There is a sequence $x_n \in H$ with $x_n \to z$. Let $u$ be any vector in $H$ with $\langle u, z_0 \rangle \ne 0$ (this $u$ must exist because $\operatorname{Span}(z_0)^\perp$ is not dense, so it cannot contain $H$). By rescaling, assume $\langle u, z_0 \rangle = 1$. Now set $y_n = x_n - \langle x_n, z_0 \rangle u$; clearly $y_n \in H$. By continuity, $\langle x_n, z_0 \rangle \to \langle z, z_0 \rangle = 0$, so $y_n \to z$. And $\langle y_n, z_0 \rangle = \langle x_n, z_0 \rangle - \langle x_n, z_0 \rangle \langle u, z_0 \rangle = 0$, so $y_n \in \operatorname{Span}(z_0)^\perp$.

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  • $\begingroup$ Exellent, thanks! $\endgroup$ – user74574 Oct 24 '13 at 22:10

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