Show without using density in $l^p$ that $\overline{X+Y}=E$. This question has appeared a few times here on Satck. But, I'm looking to solve it without using density in $l^p$:
Question: Let $E=l^{1}$ (the space of real, absolutely convergent series) and define
$$X=\{x=(x_{n})_{n\geq1}\in E:x_{2n}=0 \ \forall n\geq 1\}$$
and
$$Y=\{y=(y_{n})_{n\geq1}\in E:y_{2n}=\frac {1}{2^{n}}y_{2n-1} \ \forall n\geq 1\}.$$
Show that $X$ and $Y$ are closed linear spaces and $\overline{X+Y}=E$.
My attempt: I believe we can use the following result:
Corollary:  Let $E$ be a normed vector space and $F\subset E$ a linear subspace. Then $\overline{F}=E$
if and only if for all $f \in E^{\ast}$ (dual space) if the restriction  $f|_F =0$  then $f=0$.
But I'm stuck in this problem, if anyone has asked such a question that way and can help, I appreciate it
 A: Hint: $f$ vanishes on $X+Y$ then it vanishes on $X$ and $Y$. It follows that $f(e_k)=0$ for $k$ odd (since $e_k \in X$ in tis case). [$e_n$ is the sequence with $1$ in $n-th$ place and  $0$ elsewhere].
Now consider $(1,\frac 1  2,1,\frac 1  {2^{2}},1,\frac 1  {2^{3}},....1,\frac 1  2,0,0,0...)$ where there are  $2n$ non-zero terms. This sequence belongs to $Y$. Conclude that $\frac 1  {2^{2}}f(e_2)+\frac 1  {2^{4}}f(e_4)+\cdots+\frac 1  {2^{n}}f(e_{2n})=0$. Thsi holds for each  $n$. Use induction to show that $f(e_2)=0, f(e_4)=0$ and so on. It follows that $f \equiv 0$.
A: Assume $f \in (\ell^1)^*$ iz zero on $X+Y$. If $(e_n)_n$ is the canonical basis in $\ell^1$, we wish to show that $f(e_n) = 0$ for all $n \in \Bbb{N}$.
Notice that $e_{2n} \in X \subseteq X+Y$ so $f(e_{2n}) = 0$ for all $n \in \Bbb{N}$. For odd elements we have
$$e_{2n-1} = \underbrace{2^ne_{n-1}+e_n}_{\in Y} + \underbrace{-2^ne_{n-1}}_{\in X} \in X+Y$$
and hence
$$f(e_{2n-1}) = f(2^ne_{n-1}+e_n-2^ne_{n-1}) = 0.$$
Now for arbitrary $x=(x_n)_n \in \ell^1$ by linearity and continuity we gave
$$x = \sum_{n=1}^\infty x_ne_n \implies f(x) = \sum_{n=1}^\infty x_nf(e_n) = 0.$$
Hence $f \equiv 0$.
A: Take $y=(y_1,y_2,\ldots) \in \ell^{\infty}$, and assume that $\langle x,y\rangle = 0$ for all $x\in (X+Y)$. Then, you wish to prove that $y=0$.
If $e_j$ denote the $j^{th}$ "standard basis" vector in $\ell^1$, then $e_{2j-1} \in X$. Thus, $y_{2j-1} = \langle e_{2j-1},y\rangle = 0$ for all $j\in \mathbb{N}$, and $y = (0, y_2, 0, y_4, \ldots)$.
Now consider $x = (x_1, x_2, \ldots) \in \ell^1$ given by
$$
x_{2j} = \frac{1}{2^{2j}}\text{sgn}(y_{2j}), \text{ and } x_{2j-1} = \frac{1}{2^j}\text{sgn}(y_{2j})
$$
where $\text{sgn}(z) = \frac{z}{|z|}$ if $z\neq 0$, and $0$ otherwise. Then, $x\in Y$, and
$$
0 = \langle x,y\rangle = \sum_{j=1}^{\infty} \frac{1}{4^j}|y_{2j}|.
$$
Therefore, $y_{2j} = 0$ for all $j\in \mathbb{N}$, and thus $y = 0$.
