Non-unique Hahn Banach extension for a functional defined on a closed subspace of $l^1$ I'm trying to solve the following exercise, but I'm stuck on the possible conclusion.

Let $Y=\{ \{x_k\}_k \in l^1: x_{2015} = 0 \} \subset l^1, \quad T \in Y'$. Characterize all the linear and bounded extensions $\Phi \in (l^1)'$ which preserves the dual norm.

I've tried so far to use the duality and write $\Phi: l^1 \rightarrow \mathbb{R}$ as $\Phi(x)= \sum_{k=1}^{\infty} x_k c_k$ where $\{c_k \} \in l^{\infty}$, but I don't know how to get equality of dual norms.
 A: Notice that $\|T\| = \sup_{\substack{n\in\mathbb{N}\\ n\ne 2015}} |Te_n|$. Indeed, if $T$ is bounded then in particular $(Te_n)_n$ is a bounded sequence in the field and for $x = \sum_{\substack{n=1\\ n\ne 2015}}^\infty x_ne_n \in Y$ we have
$$|Tx| = \left|\sum_{\substack{n=1\\ n\ne 2015}}^\infty x_n Te_n\right| \le \sum_{\substack{n=1\\ n\ne 2015}}^\infty |x_n||Te_n|  \le \left(\sup_{\substack{n\in\mathbb{N}\\ n\ne 2015}} |Te_n|\right) \|x\|_1$$
so $\|T\| \le \sup_{\substack{n\in\mathbb{N}\\ n\ne 2015}} |Te_n|$. Conversely, for all $n \in \mathbb{N}, n \ne 2015$ we have
$$\|T\| \ge \frac{|Te_n|}{\|e_n\|_1} = |Te_n|$$
so $\|T\| \ge \sup_{n\in\mathbb{N}, n\ne 2015} |Te_n|$.
Similarly (or using the $(\ell^1)' \leftrightarrow \ell^\infty$ duality) we get that $\|\Phi\| = \sup_{n\in\mathbb{N}} |\Phi e_n|$.
Now, since $\Phi$ extends $T$, for every $n \in \mathbb{N}, n \ne 2015$ it follows that $Te_n = \Phi e_n$.
Now we have
$$\sup_{\substack{n\in\mathbb{N}\\ n\ne 2015}} |\Phi e_n| = \sup_{\substack{n\in\mathbb{N}\\ n\ne 2015}} |Te_n| = \|T\| = \|\Phi\| = \sup_{n\in\mathbb{N}} |\Phi e_n|$$
so $|\Phi e_{2015}| \le \|T\|$.
We also see that $\Phi e_{n} = Te_n, \forall n \ne 2015$ and $|\Phi e_{2015}| \le \|T\|$ is a sufficient condition that $\Phi \in (\ell^1)'$ is a Hahn-Banach extension of $T$.
