Prove that $X+Y$ is not closed in $\ell^1$

Question

Let $X=\{x\in l^1:x(2k)=0,\forall k=1,2,...\}$ and $Y=\{x\in l^1:\dfrac1ky(2k-1)=y(2k),\forall k=1,2,...\}$. Show that $X+Y$ is not closed in $l^1$

The previous part of this problem ask me to show that $e_k\in X+Y$ where $\{e_k\}$ is the standard Schauder base of $l^1$ and I did it. The hint of the problem quoted is :

Show that $\overline {X+Y}=l^1$ and $z\notin X+Y$ where $z(i)=\begin{cases} 1/n, & \text{if $i=2n$} \\ 0, & \text{otherwise} \end{cases}$

But I do not understand the hint since $z$ not even in $l^1$. Or there are other method to prove $X+Y$ is not closed? Please give me some hints, thank you.

Answer 1

As you correctly noted, the suggested $z$ is not even in $\ell^1$, so it cannot provide a rightful counterexample.

However, the hint is still useful; You just have to come up with a right example.

Notice that, because terms in $X$ are ones whose even terms are $0$, if we say $z=x+y$ where $x \in X$ and $y\in Y$, then it must be necessarily true that even terms of $y$ must coincide with those of $z$. In other words, $z_{2n}=y_{2n}$ for all $n\ge 1$.

Now certainly the sequence $z$ defined by $z_n=\frac{1}{n^2}$ belongs to $\ell^1$. You want to show that $z\notin X+Y$. Suppose otherwise. So $z=x+y$ for some $x,y$.

Then this forces $y_{2n}=\frac{1}{4n^2}$, but $y\in Y$ so $y_{2n-1}=\frac{1}{4n}$. This is clearly a contradiction, so $z\notin X+Y$