On uniqueness of extension of linear functional on $\ell^{1}$ Consider the subspace $Y= \{ (x_{n}) : \lim n^{2} x_{n} \text{ exists and is finite} \}$. Show that a linear functional on $Y$ can only be extended in at most one way to a bounded linear functional on $\ell^1$. Furthermore does the $\lim n^{2} x_{n}$ functional have bounded extension?
Own work:
Any functional on $\ell^{1}$ is rep. by element of $\ell^{\infty}$. Hence suppose we have two functionals which are not equal on $\ell^{1}$ then they differ at some index and hence will also differ for some element of $Y$
(This solution assumed that the author means bounded linear functional in the exercise. Maybe this is not solvable for only linear functionals?)
Another idea is to show that $Y$ is dense in $\ell^{1}$ but this I can't either.
The second is harder, I suspect there are no real-valued subadditive function is dominating this functional on $\ell^{1}$ but can't come up with a solid argument.
 A: We regard $Y$ as a linear subspace of $\mathscr{l}^{1}(\mathbb{N})$ and we can characterize it as an $\mathscr{l}^{1}(\mathbb{N})$-sequence $\left\{x_{n}\right\}_{n\geq1}$ belongs to $Y$ if and only if $x_{n}=\Theta(\frac{1}{n^{2}})$, 
i.e vanishes off as $\frac{1}{n^{2}}$ for large $n\geq1$.
The idea is that $Y$ will form a dense subspace of $\mathscr{l}^{1}(\mathbb{N})$, thus given any $\underline{continuous}$ linear functional on $Y$, a continuous extension to all of $\mathscr{l}^{1}(\mathbb{N})$ will be unique. 
To prove density of $Y$, pick an arbitrary element $\left\{x_{n}\right\}_{n\geq1}\in \mathscr{l}^{1}(\mathbb{N})$, so given an $\varepsilon >0$, we can find an $N_{\varepsilon}\in \mathbb{N}$ such that $$\sum_{n\geq N_{\varepsilon}}|x_{n}|< \varepsilon$$ Moreover since $\left\{\frac{1}{n^{2}}\right\}_{n\geq1}\in Y$, we can find $M_{\varepsilon}\in \mathbb{N}$ such that $$\sum_{n\geq M_{\varepsilon}}\frac{1}{n^{2}} < \varepsilon$$ Now setting $K_{\varepsilon}=\max ({M_{\varepsilon},N_{\varepsilon}})$ our candidate is the sequence $\left\{y_{n}\right\}_{n\geq1}\in Y$ defined as 
\begin{equation}
 y_{n} = \begin{cases}
       x_{n} & \text{if} \, , \, 1\leq n <K_{\varepsilon}  \\
       \frac{1}{n^{2}} & \text{if} \, , \, n\geq K_{\varepsilon} \\
       \end{cases}
\end{equation}
which establishes the density of $Y$, since $\left\{x_{n}\right\}_{n\geq 1}\in \mathscr{l}^{1}(\mathbb{N})$ was arbitrary.
For the second part of the question you may use the fact that $\mathscr{l}^{1}(\mathbb{N})^{*}\cong \mathscr{l}^{\infty}(\mathbb{N})$. 
Assume there is a continuous extension, then there should be a $\left\{z_{n}\right\}_{n\geq 1}\in \mathscr{l}^{\infty}(\mathbb{N})$ associated to that linear functional. Then with that in mind you test your continuous linear functional on the standard basis and I am pretty sure that you will reach a contradiction.
