Certain set is dense in $l^p$ if and only if $\{x_n : n \in \mathbb{N}\} \notin l^q$, where $1/p + 1/q = 1$ Assume that $\{x_n : n \in \mathbb{N}\} \subset \mathbb{R}$ is such that $x_n \neq 0$ for some $n$. Let $p \in (1, \infty)$ and$$G := \left\{\{y_n : n \in \mathbb{N}\} \in l^p : \lim_{N \to \infty} \sum_{n=1}^N y_n x_n = 0\right\}.$$Do we have that $G$ is dense in $l^p$ if and only if $\{x_n : n \in \mathbb{N}\} \notin l^q$ where $1/p + 1/q = 1$?
 A: "$\Leftarrow$": Let us assume that $\left(x_{n}\right)_{n\in\mathbb{N}}\notin\ell^{q}$.
We will show that actually $\delta_{n_{0}}\in\overline{G}$ for every
$n_{0}\in\mathbb{N}$. Since $G$ (and hence $\overline{G}$) is obviously
a vector space and since the finitely supported sequences are dense
in $\ell^{p}$ (because of $p<\infty$), this implies density of $G$.
Note that since $\left(x_{n}\right)_{n\in\mathbb{N}}\notin\ell^{q}$,
we also have $\left(x_{n}\right)_{n\geq n_{0}+1}\notin\ell^{q}$.
Now, it is well-known that we can characterize the $\ell^{q}$ norm
by duality (also if it is infinite), i.e. we have
$$
\infty=\left\Vert \left(x_{n}\right)_{n\geq n_{0}+1}\right\Vert _{\ell^{q}}=\sup_{\substack{\left(y_{n}\right)_{n\geq n_{0}+1}\text{ finitely supported},\\
\left\Vert \left(y_{n}\right)_{n\geq n_{0}+1}\right\Vert _{\ell^{p}}=1
}
}\left|\sum_{n=n_{0}+1}^{\infty}x_{n}y_{n}\right|.
$$
Now, let $\varepsilon>0$ be arbitrary and let $\alpha:=x_{n_{0}}$.
By what we just saw, there is a finitely supported sequence $\left(y_{n}\right)_{n\geq n_{0}+1}$
with $\left\Vert \left(y_{n}\right)_{n\geq n_{0}+1}\right\Vert _{\ell^{p}}=1$
and $\left|\gamma\right|>\frac{\left|\alpha\right|}{\varepsilon}$
for $\gamma:=\sum_{n=n_{0}+1}^{\infty}x_{n}y_{n}$. Now, define $z:=\left(z_{n}\right)_{n\in\mathbb{N}}$
by $z_{n}:=-\frac{\alpha}{\gamma}\cdot y_{n}$ for $n\geq n_{0}+1$
and $z_{n}:=0$ otherwise. Note
$$
\sum_{n=1}^{\infty}x_{n}z_{n}=-\frac{\alpha}{\gamma}\cdot\sum_{n=1}^{\infty}x_{n}y_{n}=-\alpha
$$
and hence
$$
\sum_{n=1}^{\infty}\left(\delta_{n_{0}}+z\right)_{n}x_{n}=x_{n_{0}}+\sum_{n=1}^{\infty}z_{n}x_{n}=\alpha-\alpha=0
$$
which implies $\delta_{n_{0}}+z\in G$. But since $\left|\gamma\right|>\frac{\left|\alpha\right|}{\varepsilon}$,
we also have
$$
\left\Vert \delta_{n_{0}}-\left(\delta_{n_{0}}+z\right)\right\Vert _{\ell^{p}}=\left\Vert z\right\Vert _{\ell^{p}}=\left|\frac{\alpha}{\gamma}\right|\cdot\left\Vert \left(y_{n}\right)_{n\geq n_{0}+1}\right\Vert _{\ell^{p}}=\left|\frac{\alpha}{\gamma}\right|<\left|\alpha\right|\cdot\frac{\varepsilon}{\left|\alpha\right|}=\varepsilon.
$$
Since $\varepsilon>0$ was arbitrary, $\delta_{n_{0}}\in\overline{G}$.
"$\Rightarrow$": Let us assume that $G$ is dense. If we had
$\left(x_{n}\right)_{n\in\mathbb{N}}\in\ell^{q}$, then the functional
$$
\Phi:\ell^{p}\to\mathbb{R},\left(y_{n}\right)_{n}\mapsto\sum_{n=1}^{\infty}x_{n}y_{n}
$$
would be well-defined and bounded, so that the kernel ${\rm ker}\,\Phi=G$
would be a closed subset of $\ell^{p}$.
But by assumption, $x_{n}\neq0$ for some $n\in\mathbb{N}$, so that
${\rm ker}\,\Phi\neq\ell^{p}$. All in all, we see that
$$
\overline{G}=\overline{{\rm ker}\,\Phi}={\rm ker}\,\Phi\neq\ell^{p},
$$
in contradiction to density of $G$ in $\ell^{p}$. Hence, $\left(x_{n}\right)_{n\in\mathbb{N}}\notin\ell^{q}$,
as desired.
