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I am very confused whether or not the set

$G=\{(x_n)_{n\in\mathbb{N}} \in \ell_2: \sum_{n=1}^{\infty}\frac{x_n}{\sqrt{n}}=0\}$

is closed in $\ell_2$. Any hint is appreciated!

Daniel
  • 5,872

1 Answers1

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Note that $(\frac{1}{\sqrt{2}},-1,0,\dots), (\frac{1}{\sqrt{3}},0,-1,0,\dots), \ldots $ belong to $G$. Moreover, note that $G\neq\ell_2$.

Let us assume that $G$ is closed. Since $G\neq\ell_2$, there is a non-null vector $\left(x_n\right)_{n\in\mathbb{N}}$ orthogonal to $G$. Therefore $(x_n)$ is orthogonal to each sequence above.

Thus, $\frac{x_1}{\sqrt{2}}-x_2=0$, $\ \frac{x_1}{\sqrt{3}}-x_3=0$, $\ \ldots.\ $ So $x_n=\frac{x_1}{\sqrt{n}}$.

Note that $(x_n)_{n\in\mathbb{N}}=\left(\frac{x_1}{\sqrt{n}}\right)_{n\in\mathbb{N}}$ belongs to $\ell_2$ if only if $x_1=0$. Absurd!

Daniel
  • 5,872