Consider a sequence $(x_n)_n$ in Hilbert space $H$ such that $\langle x_m,x_n\rangle=\delta_{mn}$ where $\delta_{mn}$ equals one if $m = n$ and $C$ otherwise. Prove that $(x_n)_n$ is a weakly convergent sequence.
Weakly convergent sequence
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0I see that your consequences $\langle x_m,x_1\rangle =C$ for $m\geq 2$ is the assumption of problem. – 2012-12-09
2 Answers
Let $F:=\overline{\operatorname{Span}\{x_n,n\geqslant 1\}}$: it's a closed subspace of $H$, so we have $H=F\oplus F^\perp$.
We know, as $\{x_n\}$ is bounded, that it has a weakly converging subsequence. What we have to prove is that the limit doesn't depend on the choice of the subsequence.
Let $u_0$ a weak limit of $\{x_{k'}\}$. We can write $u_0=u'_0+u''_0$, where $u'_0\in F$ and $u''_0\in F^\perp$.
Testing the definition of weak convergence $\langle x_{k'},v\rangle\to \langle u,v\rangle$ with $v=u''$, we get that $u''=0$, and for each $j$, $\langle u',x_j\rangle=C$.
This determines $u'$, because if $u_1$ and $u_2$ are weak limits of a subsequence of $\{x_n\}$, using the preceding argument we get that $\langle u_1'-u_2',x_j\rangle=C-C=0$ for all $j$, so $\langle u'_1-u'_2,w\rangle=0$ for all $w$ in the linear span of the $x_k$, and by density for all $w\in F$, and $u''_1-u''_2=0$.
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0We _show_ that they are the same. If I take the sequence $(v_0,e_1,v_0,e_2,v_0,e_3,\dots,v_0,e_n,\dots)$ where $v_0\neq 0$, if I take the subsequence $y_{2k}$ I get $v_0$ as a limit, but if I take $y_{2k+1}$ I get $0$. The limit is unique once the subsequence is fixed. – 2012-12-10
We have $y_n=\frac{1}{n}\sum\limits_{i=1}^{n}x_i$ converges to $y$ in $H$. Then $e_i=x_i-y$, $1\leq i\leq n$ is an orthonormal system. Hence, $(e_i)$, $1\leq i\leq n$ are linearly independent.
Let $F_n=\operatorname{Span}\{x_1,\ldots ,x_n\}$, it is a closed subspace of $H$. For all $x$ in $H$, $x=u_n+v_n$, $u_n\in F_n$, $v_n\in F_n^\perp$. Then $u_n=\sum\limits_{k=1}^{n}\lambda_ke_k$ and $\langle x,e_k\rangle=\lambda_k$. So $u_n=\sum\limits_{k=1}^{n}\langle x,e_k\rangle e_k$.
By Pythagorean Theorem, $||x||=||\sum\limits_{k=1}^{n}\langle x,e_k\rangle e_k+v_n||=\sum\limits_{k=1}^{n}|\langle x,e_k\rangle|^2+||v_n||^2$. Therefore, $\sum\limits_{k=1}^{n}|\langle x,e_k\rangle|^2\leq||x||$. In other words, the series $\sum\limits_{k=1}^{\infty}|\langle x,e_k\rangle|^2$ converges. So $\lim\limits_{n\to\infty}\langle x,e_k\rangle=0$. We can derive that $\lim\limits_{n\to\infty}\langle x,x_k\rangle=\lim\limits_{n\to\infty}\langle x,y\rangle$. Q.E.D