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Let $\{x_n\}_{n=1}^\infty$ be a sequence of vectors in a Hilbert space $$l^2_{k^{-2}}:=\{z=\{z(k)\}_{k=1}^\infty:\sum\limits_{k=1}^\infty z(k)^2k^{-2}<\infty\}.$$ It is known that for some $x\in l^2_{k^{-2}}$ $$x_n(k)=k(x_n(1)-x(1))+\sum_{r=1}^n\left[\frac{k}{r}\right]\sum_{d|r}\mu\left(\frac{r}{d}\right)y(d),$$where $y\in l^2_{k^{-2}}$ such that $y(n)=x(n-1)-x(n); \ x(0)=0.$ One can deduce from here, $$x_n(k)-x(k)=k(x_n(1)-x(1)), \ \ \forall k\leq n-1. \ \ \ \dots(A),$$ and, $$x_n(1)-x(1)=\sum_{r\leq n}\frac{(\mu*y)(r)}r=\sum_{r=1}^n\frac{1}{r}\sum_{d|r}\mu\left(\frac{r}{d}\right)y(d)=\sum_{ab\leq n}\frac{\mu(a)}{a}\frac{y(b)}{b}.\dots(B)$$ where $\mu$ is the Mobius inversion function and $*$ is the Dirichlet convolution. From above its not difficult to see that $\lim\limits_{n\to\infty}x_n(1)=x(1)$ and hence for a fixed $k$; $\lim\limits_{n\to\infty}x_n(k)=x(k),$ from $(A)$ and $(B)$.

My question is from here how one can show that ,$$\|x_n-x\|_{l^2_{k^{-2}}}=\sum_{k=1}^\infty\frac{(x_n(k)-x(k))^2}{k^2}\xrightarrow{n\to\infty} 0?$$Using the information above it can be seen that the convergence highly depends on a good bound of $\sum\limits_{n\leq x}\frac{\mu(n)}{n}$. Even if assuming Riemann Hypothesis I am being unable to prove this.

ANY help is highly appreciated.

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    $\begingroup$ From your definition of $l^2_{k^{-2}}$ it looks as though every element of $l^2$ belongs to this space. And therefore $l^2_{k^{-2}} = l^2$. Is that really what you meant? $\endgroup$ – Nik Weaver Apr 19 '14 at 23:31
  • $\begingroup$ That was a pretty serious typo. I corrected it. Thanks! $\endgroup$ – Subhajit Jana Apr 20 '14 at 0:01
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I don't think so, because the conditions you've given don't really give us any control over $x_n(k)$ for $k > n$. For example, if we have some $\{x_n\}$ and $x$ satisfying your hypotheses and such that $\|x_n - x\|_{l^2_{k^{-2}}} \to 0$, we can define a new sequence $\{x_n'\}$ by setting $$x_n'(k) = \begin{cases}x_n(k)&k \neq n+1\cr x(n+1) + n+1&k=n+1\end{cases}$$ and then your conditions will still be satisfied but you certainly won't have $\|x_n - x\|_{l^2_{k^{-2}}} \to 0$.

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  • $\begingroup$ Thanks! I haven't realized it before. But the problem still is remaining as I noted that we can actually determine all $x_n(k)$'s. Kindly check the edited question. $\endgroup$ – Subhajit Jana Apr 20 '14 at 6:14

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