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Is the set \begin{align} A=\left\{a=(a_1,a_2,\dots)\in\ell^2 \ \ \lvert \ \ \sum_{k=1}^\infty \frac{a_n}{n}=0 \right\}\subset\ell^2 \end{align} dense in $\ell^2$

Is the following argument correct? Let $x=(1,0,0,\dots)$ and assume $\forall \epsilon>0\ \ \exists a\in A$ such that $\lVert x-a\lVert_2<\epsilon $. This implies $\lvert\sum_{n\geq1}\frac{x_n-a_n}{n}\lvert\leq \frac{\pi}{\sqrt{6}}\epsilon$, i.e. $\lvert\sum_{n\geq 1}\frac{a_n}{n}\lvert\geq1-\frac{\pi}{\sqrt{6}}\epsilon$. Choosing $\epsilon$ sufficiently small therefore implies $a\not\in A$

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You argument seems fine. –  Davide Giraudo Jul 12 '12 at 21:33

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up vote 7 down vote accepted

Actually, $A$ is closed. Indeed, let $L\colon \ell^2\to \Bbb C$, $La=\sum_{k=1}^{+\infty}\frac{a_k}k$. It's a linear map, and by Cauchy-Schwarz inequality it's continuous. We have $A=\ker L$. Since $A$ is strictly contained in $\ell^2$ ($(1,1,0,\dots)\in\ell^2\setminus A$), it cannot be dense.

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What do you mean by "$A$ is strict"? –  Mercy Jul 12 '12 at 21:52
    
Nice argument Davide –  Adam Rubinson Jul 13 '12 at 11:29

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