Orthonormal system/basis

[Солунац]

Is it true that an orthonormal system $\{e_i|i \in \mathbb{N}\}$ in separable Hilbert's space H is an orthonormal basis (full orthonormal system)? If true, how to prove it?

[ewert]

Неверно. Если верно, то доказать невозможно, не используя специфику системы.

Верно другое: сепарабельность равносильна существованию ортонормированного базиса.

[Солунац]

Спасибо ewert.

[zoo]

Солунац писал(а):

Is it true that an orthonormal system $\{e_i|i \in \mathbb{N}\}$ in separable Hilbert's space H is an orthonormal basis (full orthonormal system)? If true, how to prove it?

take a basis of a Hilbert space and then exclude from this basis the elements $e_i$ with odd $i$:)

[ewert]

Зачем все нечётные? -- вполне достаточно выкинуть первый

[zoo]

ewert писал(а):

Зачем все нечётные? --

штоп дольше думал

[Солунац]

My apologies to moderator, but I think it would be better to go on in this topic in my next question.

It doesn't have anything with previous question, but I hope that's OK It goes like this:
If $Y$ is proper Hilbert's subspace of Hilbert's space $H$, then there exists non-zero vector who is orthogonal on $Y$. Prove.

If we suppose otherwise that doesn't exists such $x$ ($x\in H$ and $x\neq 0$ and $x$ is normal to $Y$), we have that $Y^\perp =0$. From $0^\perp =H$ and $H^\perp =0$ we have that $Y^\perp =H^\perp$.
But what then? $Y^\perp^\perp = H$, so $\overline{LinY} =H$ and ... ?

[Narn]

Солунац писал(а):

If $Y$ is proper Hilbert's subspace of Hilbert's space $H$, then there exists non-zero vector who is orthogonal on $Y$. Prove.

If we suppose otherwise that doesn't exists such $x$ ($x\in H$ and $x\neq 0$ and $x$ is normal to $Y$), we have that $Y^\perp =0$. From $0^\perp =H$ and $H^\perp =0$ we have that $Y^\perp =H^\perp$.
But what then? $Y^\perp^\perp = H$, so $\overline{LinY} =H$ and ... ?

...$Y$ - не собственое подпространство $H$. $LinY=Y$ и $\overline{Y}=Y$.

[Солунац]

а да, ...

Спасибо Narn, спасибо.