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Let $F$ be the real sequence which satisfies $ \sup _{n} {|nx_n|}<\infty$ in $\ell^2$. Show that

  1. $F$ are linear subspace of $\ell^2$.
  2. $F$ is closed in $\ell^2$.

My idea:

  1. is trivial.
  2. As is known to all, $\ell^2$ is a Banach space. The subspace of a Banach space is closed if and only if it is complete (Is this right?).

So we only to prove $F$ is complete subspace. But how can I show that every Cauchy sequence converge in $F$? Or we need to show that the closure of $F$ is itself? Could someone give me some details. Thank you!

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  • $\begingroup$ To refer to $F$ as a sequence is a mistake given that the subspace is uncountable. Also the simplest method to show that $F$ is closed, is probably by taking sequences from $F$ and showing that their limits are in $F$. $\endgroup$ Jan 29, 2021 at 14:21
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    $\begingroup$ Hint: prove it is sequentially closed. $\endgroup$
    – fcz
    Jan 29, 2021 at 14:21
  • $\begingroup$ Thanks four edit! $\endgroup$ Jan 29, 2021 at 14:21
  • $\begingroup$ This step puzzles me. $x$ is a limit point of $F$. Then there exists a sequence ${x^k}$ in $F$ and converges to ${x}$ in the total space. And $\sup_n |nx_n^k|<a_k$. We can not get $a_k$ bounded. I have no idea how to prove it. $\endgroup$ Jan 29, 2021 at 14:38

1 Answer 1

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The set $F$ is not closed. To see it consider a sequence.

$$x^k =\left(\frac{1}{1} , \frac{1}{2^{\frac{3}{4} }}, \frac{1}{3^{\frac{3}{4} }} , ...,\frac{1}{k^{\frac{3}{4} }} , 0,0,0,....\right)$$

Clearly $$\sup_n |nx^k_n |<\infty$$ for all $k\in\mathbb{N} .$ Hence $x^k\in F.$ Moreover $x^k \to x$ in $\ell^2 $ where $$x= (n^{-\frac{3}{4}})_{n\in\mathbb{N}}\in\ell^2.$$ But $x\notin F.$

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