$G_p=\{\{x_n\}_{n\inℕ} \in l^p(R) :\displaystyle \sum_{k=1}^\infty x_k=0 \}$ is not closed for $p\in (1,\infty)$

Question

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?

Answer 1

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 $G_p\subsetneq \ell^p.$

Answer 2

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.