Existence of a complete norm on $\ell^2$ which extends $\|۰\|_1$ (on $\ell^1$) We know that norm-1 is a complete norm on $\ell^1.$ Also $\ell^1$ is a subspace of $\ell^2.$ Now, is there a complete norm
on $\ell^2$ which extends norm-1 (on $\ell^1$)?
 A: Let $V$ be an algebraic complement of  $\ell ^1$ within $\ell ^2$,  so that $\ell ^2 = \ell ^1 \oplus  V$.  It is easy to see that the algebraic
dimension  of $V$ is $2^{\aleph_0}$,  so $V$ is algebraically isomorphic to any separable infinite dimensional Banach space $X$.  Choosing such
an $X$ let
$$
  \varphi :V\to X
  $$
be an algebraic  isomorphism and define a norm on
$\ell ^2$
by
$$
  \|x\| = \|x_1\|_1 +  \|\varphi (x_2)\|,
  $$
where
$$
  x = x_1+x_2
  $$
is the decomposition of any given $x$  in $\ell ^2$ according to
$\ell ^2 = \ell ^1 \oplus  V$.  This norm satisfies the requirements.

EDIT.  I guess my statement above that "the algebraic
dimension  of $V$ is $2^{\aleph_0}$" is not so immediate,  so let me make up for this.
That statement is clearly equivalent to
$$
  \text{dim}(\ell ^2/\ell ^1) = 2^{\aleph_0},
  $$
so let us  prove this instead.
For each $\alpha >0$, consider the vector
$$
  x_\alpha =(1/n^\alpha )_{n\in {\mathbb N}},
  $$
and observe that $x_\alpha \in \ell ^p\Leftrightarrow p\alpha >1$.  In particular
$$
  x_\alpha \in \ell ^2,  \quad \text{and}\quad x_\alpha  \notin  \ell ^1,\quad \forall \alpha \in (1/2,1].
  $$
We next claim that the set
$$
  \{x_\alpha : \alpha \in (1/2,1]\}
  $$
is a linearly independent set in $\ell ^2/\ell ^1$.  To prove  this we must verify  that, whenever
$$
  \lambda _1x_{\alpha _1}+  \lambda _2x_{\alpha _2}+\cdots  + \lambda _nx_{\alpha _n} \in  \ell ^1,
  \tag 1
  $$
one necessarily has $\lambda _1=\lambda _2=\cdots =\lambda _n=0$.  Assuming by contradiction that this is not so, we may discard the zero terms in
the above sum and hence suppose  that the $\lambda _i$ are all nonzero.  We may also assume that  the $\alpha _i$ are organized such that
$$
  \frac 1{\alpha _1}<  \frac 1{\alpha _2}<\cdots <  \frac 1{\alpha _{n-1}}<\frac 1{\alpha _n}.
  $$
Chosing any real number $p$ in between the last to terms above, that is,
$$
  \frac 1{\alpha _{n-1}}<p<\frac 1{\alpha _n},
  $$
we then have that $p\alpha _n<1$, so $x_{\alpha _n} \notin \ell ^p$,  while $p\alpha _k>1$, for all $k=1,2,\ldots ,n-1$,  so $x_{\alpha _k} \in \ell ^p$.
However,   (1) implies  that
$$
  \lambda _nx_{\alpha _n}\in    \ell ^1-  \lambda _1x_{\alpha _1}-  \lambda _2x_{\alpha _2}-\cdots  - \lambda _{n-1}x_{\alpha _{n-1}} \subseteq  \ell ^p,
  $$
a contradiction.
This shows that
$$
  \text{dim}(\ell ^2/\ell ^1) \geq  2^{\aleph_0},
  $$
but since the cardinality of the whole of $\ell ^2$ is
$2^{\aleph_0}$,  it is evident that the opposite inequality holds as well.
