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I have to show that $G_p=\left\{\{x_n\}_{n\inℕ} \in l^p(R) :\displaystyle \sum_{k=1}^\infty x_k=0 \right\}$ is not closed for $p\in (1,\infty)$.

My idea was to take a sequence:

$x_n(k)=-1 \text{ for } k=1$

$x_n(k)=\frac{1}{n} \text{ for } 2\le k<n+1$

$x_n(k)=0 \text{ for } k>n+1$

which is element of $G_p$ for all $n $

but the pointwise limit:

$x(k)=-1 \text{ for } k=1$

$x(k)=0 \text{ for rest}$

is not element of $G_p$

...

Let's try to show that $x_n \to x$ in $l^p$

$$||x_n-x||_p^p={\sum_{k=2}^{n+1} \frac{1}{n^p}}= n \frac{1}{n^p} \to_{n\to\infty} 0$$

am I missing something here?

hbghlyj
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Dave
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  • If you work in $l^p$ then you need to show that the $l^p$-limit is not in the set, not the pointwise limit. Prove that $x_n\to x$ in $l^p$. (which wouldn't be true by the way if we had $p=1$) – Mark Nov 18 '23 at 18:59
  • The set $G_p$ is dense in $\ell^p,$ so it cannot be closed, as $G_p\subsetneq\ell^p.$ – Ryszard Szwarc Nov 18 '23 at 19:02
  • From general results about functionals on a normed vector space, it follows that hyperplanes are either closed or dense. – hbghlyj Nov 18 '23 at 19:03
  • How could I show that $G_p$ is dense in $l^p$ ? – Dave Nov 18 '23 at 19:15
  • @Dave Just show $x_n\to x$ in $l^p$, this is easy and solves the problem. Unless you want to see other methods. – Mark Nov 18 '23 at 19:18

2 Answers2

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The denseness of $G_p$ can be shown straightforward. It suffices to show that the elements $\delta_n$ of the standard basis can be approximated by elements in $G_p,$ with respect to $\|\cdot \|_p$ norm. We have $$f_k:=\delta_n-{1\over k}\sum_{j=1}^k\delta_{n+j}\in G_p$$ and $$\|f_k-\delta_n\|_p^p={1\over k^{p-1}}\to 0$$ Therefore $G_p$ cannot be closed as $G_p\subsetneq \ell^p.$

Ryszard Szwarc
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Consider linear functional $f(x)=\displaystyle\sum\limits_{k=1}^\infty x_k$, defined over $c_{00}$ -- set of finite sequences (i.e. sequences, all elements of which are zero, starting with some number) with induced norm of $l^p$. By taking $x_n=\left(\underbrace{\dfrac{1}{n},...,\dfrac{1}{n}}_n,0,0,...\right)\in c_{00}$, we have $\|f\|\ge\dfrac{|f(x_n)|}{\|x_n\|}=n^{\frac{p-1}{p}}\to\infty$ , i.e. $f$ not bounded over $c_{00}$. This means, that $\overline{\ker f}=c_{00}$. But $c_{00}$ is dense in $l_p$, so $\overline{\ker f}=l^p$. Since $\ker f\subset G_p$, we have $\overline{G}_p=l^p$. Its clear that $G_p\not= l^p$, so $G_p$ is not closed.

hbghlyj
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thing
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  • You should explicitly point out that two subspaces $\overline{\ker f}$ showing up in your answer are different. – Ryszard Szwarc Nov 19 '23 at 19:48
  • @RyszardSzwarc, i don't understand your comment. Can you please explain which two subspaces you are talking about? I'm talking about only one kernel of specific functional. – thing Nov 20 '23 at 00:22
  • One kernel but the closures in different spaces. – Ryszard Szwarc Nov 20 '23 at 09:23
  • @RyszardSzwarc, why in different? In $c_{00}$ norm is the induced norm of $l_p$. And closure of $c_{00}$ equal $l_p$ at the same norm. – thing Nov 20 '23 at 09:46
  • You have claimed in your post that $\overline{\ker f}=c_{00}$ and at the same time $\overline{\ker f}=\ell^p.$ In my opinion it is not possible. – Ryszard Szwarc Nov 20 '23 at 19:01
  • @RyszardSzwarc, take the closure in first equality, this gives $\overline{\ker f}=\overline{c}{00}=l_p$, because of densnesness $c{00}$ in $l_p$. All norms here are the same. – thing Nov 21 '23 at 00:29
  • That's correct. But in your answer the line over $c_{00}$ is missing in the first equality. – Ryszard Szwarc Nov 21 '23 at 02:13
  • @RyszardSzwarc, its not missing. First equality is densnesness of kernel of unbounded functional in its domain. I just thought it was obvious to take the second closure. – thing Nov 21 '23 at 08:36