# How to show a subspace is closed in $\ell^2$

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!

• 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 $F$. Jan 29, 2021 at 14:21
• Hint: prove it is sequentially closed.
– fcz
Jan 29, 2021 at 14:21
• Thanks four edit! Jan 29, 2021 at 14:21
• 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. Jan 29, 2021 at 14:38

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 $$x\notin F.$$