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Let $(x_n)_{n \in \mathbb{N}}$ be a sequence in a Hilbert space $H$. Show that the following are equivalent:

  1. the zero vector is the only vector orthogonal to all $x_n$, and

  2. the subspace spanned by the $x_n$ is dense in $H$.

Thanks for the help!

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1 Answer 1

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Let $M$ be the closure of the span of the $x_n$.

For $1 \Rightarrow 2$, think about the projection operator $P$ onto that closed linear subspace. If $M \neq H$, then there is a vector $h \notin M$. What do you know about $h - Ph$?

For $2 \Rightarrow 1$, think about $M^\perp$

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You might need that for any $U\subseteq H$, the closure of span$(U)$ is just $U^{\perp\perp}$. –  Berci Sep 30 '12 at 12:39
    
@Zach L. Thanks, the problem is done! –  P. M. O. Sep 30 '12 at 16:42

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