Non-existence of a continuation Let $U \subset \ell^1$ with $$U = \{(x_n)_n \in \ell^1 | x_{2k−1} = 0, \text{ for all } k \in \Bbb{N}\}.$$ I need to show that with one exception (which
one?) every continuous functional on $U$ admits infinitely many continuous linear continuations with the same norm on $\ell^1$.
I know in this situation I should use $(\ell^1)'=\ell^{\infty}$ but not really sure how or where, any help is appreciated!
 A: Let $f \in U'$ be arbitrary nonzero functional. It is not hard to see directly that $$\|f\| = \sup_{k \in \Bbb{N}} |f(e_{2k})|.$$
Now let $a = (\alpha_1,\alpha_3,\alpha_5,\ldots) \in \ell^\infty$ be an arbitrary sequence with $\|a\|_\infty \le \|f\|$. Define $F_a \in (\ell^1)'$ via the $(\ell^1)' = \ell^\infty$ duality as the sequence
$$F_a = (\alpha_1,f(e_2),\alpha_3,f(e_4),\alpha_5,f(e_6),\ldots) \in \ell^\infty,$$
or explicitly as
$$F_a(x) = \sum_{k=1}^\infty (\alpha_{2k-1}x_{2k-1} + f(e_{2k})x_{2k}), \qquad \text{ for all }x = (x_k)_{k=1}^\infty \in \ell^1.$$
Then $F_a$ satisfies $F_a(e_{2k}) = f(e_{2k}), \forall k \in \Bbb{N}$ so $F_a|_U = f$. Furthermore, we have
\begin{align}
\|F_a\| &= \|(\alpha_1,f(e_2),\alpha_3,f(e_4),\alpha_5,f(\alpha_6),\ldots)\|_\infty \\
&= \max\left\{\sup_{k \in \Bbb{N}} |\alpha_{2k-1}|, \sup_{k \in \Bbb{N}} |f(e_{2k})|\right\} \\
&= \max\{\|a\|_\infty,\|f\|\} \\
&= \|f\|.
\end{align}
Therefore, $F_a$ is a Hahn-Banach extension of $f$.
Moreover, if $b = (\beta_1,\beta_3,\beta_5, \ldots) \in \ell^\infty$ is another sequence such that $\|b\|_\infty\le \|f\|$ and $b \ne a$, then there exists $j \in \Bbb{N}$ such that $\beta_{2j-1} \ne \alpha_{2j-1}$ and hence
$$F_b(e_{2j-1}) = \beta_{2j-1} \ne \alpha_{2j-1} = F_a(e_{2j-1}) \implies F_b \ne F_a.$$
Since $\|f\| > 0$, we conclude that $\{F_a : a \in \ell^\infty, \|a\|_\infty\le \|f\|\}$ is an infinite family of Hahn-Banach extensions of $f$ to $\ell^1$.
On the other hand, $0 \in U'$ clearly has a unique extension $0 \in \ell^1$ since this is the only functional with zero norm.
