$\ell^1$ extension for sequences zero on odds Write $G \subset \ell^1$ for the set of sequences $(x_n)$ with $x_1=x_3=...=0.$ I am trying to show that any linear functional $L : G \to \mathbb{R}$ has infinitely many norm-preserving extensions $\tilde{L}:\ell^1 \to \mathbb{R}$. 
So there is an obvious Hamel basis $\{y_1,y_2,...\}$ for $\ell^1/G$. I can write any $x \in \ell^1$ as $x=(\sum a_ix_i$)+($\sum b_jy_j$). Being an extension I must have $\tilde{L}(x)= \sum a_i L(x_i) + \sum b_j \tilde{L}(\sum y_j)$. I could get infinitely many extensions by defining, for each $j$, $$\tilde{L}(y_k) = \begin{cases} 1, & k=j \\ 0, & k \neq j\end{cases}$$ 
but this does not preserve the norm. Perhaps somebody could recommend an alternative line of attack?
 A: One issue you're facing is that you've defined $\tilde{L}$ without any regard to the properties of $L$, so there's no reason to believe that $\tilde{L}$ is related to $L$ at all. Another is that a Hamel (i.e. algebraic) basis is (as far as I can tell) entirely useless for defining anything that hopes to have nice norm properties. In fact, any Hamel basis on an infinite-dimensional Banach space is guaranteed to be uncountable, so it will be impossible to define arbitrary sums over the Hamel basis.
I suspect, however, that you meant to refer to a Schauder basis, and we have a nice one in $\ell^1$. Let $e_j\in\ell^1$ be defined the sequence $(e_j)_k = \delta_{jk}$, where $\delta_{jk}$ denotes the Kronecker delta; i.e. $e_j$ has $1$ in its $j$-th entry, and $0$-s everywhere else. Then the collection $e_1,e_2,\ldots$ forms a Schauder basis of $\ell^1$, and moreover every element has unit norm.
Here is how I would come up with infinitely many extensions: for odd $j$,
$$
\tilde{L}_k(e_j) = \|L\|\delta_{jk},
$$
where $\delta_{jk}$ is the Kronecker delta. That is $\tilde{L}_k(e_k) = \|L\|$, and $\tilde{L}_k(e_j) = 0$ for $j\neq k$. This induces an extension of $L$ to $\ell^1$.  Unlike your definition of $\tilde{L}_k(e_j) = \delta_{jk}$, this gives me a chance at hoping my extension preserves norm if $\|L\|<1$.
To see that $\tilde{L}_k$ is norm-preserving, suppose $v\in G$ and $a\in\mathbb{C}$ such that $v+ae_k$ has norm $1$. Note by how we defined $e_k$ that
$$
\|v + ae_k\| = \|v\| + |a|\|e_k\| = \|v\|+|a|.
$$
Then
$$
|\tilde{L}_k(v+ae_k)| = |Lv + a\tilde{L}_ke_k| \leq |Lv| + |a||\tilde{L}_ke_k| \leq \|L\|(\|v\|+|a|) = \|L\|\|v+ae_k\|.
$$
Therefore $\|\tilde{L}_k\|\leq \|L\|$. The reverse inequality $\|L\| \leq \|\tilde{L}_k\|$ is trivial. Therefore $\|L\| = \|\tilde{L}_k\|$.
A: For $n\in\mathbb N$, set $$L_n(x_1,x_2,\dots)=\|L\|\sum_{k=n}^{\infty}x_{2k+1}+L(0,x_2,0,x_4,\dots).$$ Then $L_n$ is a linear extension of $L$ for all $n$, therefore $\|L_n\|\geq \|L\|$. Also, the $L_n$ are all different, since for $m>n$, $$L_n(e_{2n+1})=\|L\|,\quad L_m(e_{2n+1})=0.$$ Moreover, for $x\in\ell^1$,
$$\begin{align}
\|L_nx\|&\leq\|L\|\sum_{k=n}^{\infty}|x_{2k+1}|+|L(0,x_2,0,x_4,\dots)|\\
&\leq\|L\|\sum_{k=n}^{\infty}|x_{2k+1}|+\|L\|\sum_{k=1}^{\infty}|x_{2k}|\\
&\leq \|L\|\sum_{k=1}^{\infty}|x_k|,
\end{align}$$ therefore $\|L_n\|=\|L\|$ for all $n\in\mathbb N$.
