Can a linear operator from a normed space to itself be extended with operator norm preserved? Let $X$ be a subspace of $\ell_1^4$ (i.e. $\mathbb{R}^4$ equipped with the $\ell^1$ norm). Can one always extend a linear operator $l:X\rightarrow \ell_1^4$
to $L:\ell_1^4\rightarrow \ell_1^4$ such that $L$ has the same operator norm as $l$?
Thanks!
 A: If $X$ is a closed subspace of $\ell_1^4$, then it is true if the latter has the metric extension property.  If you are working over the reals, then Google :"binary intersection property Nachbin". I think Grothendieck proved that if a space has the metric extension property, then it is either finite dimensional or non-separable. I hope this helps!.
A: The answer is (trivially) affirmative if you suppose that $X$ be one of the following subspaces: 
$$\{(x^1, 0, 0, 0) \mid x^1 \in \mathbb{R}\}\quad \{(x^1, x^2, 0, 0) \mid x^i\in \mathbb{R}\},\quad \{(x^1, x^2, x^3, 0) \mid x^i \in \mathbb{R}\}.$$
Then the operator $l \colon X \to \ell^1_4$ can be represented as one of the following matrices
$$\begin{bmatrix} l^1_1 \\ l^2_1 \\ l^3_1 \\ l^4_1 \end{bmatrix},\quad \begin{bmatrix} l^1_1& l^1_2 \\ l^2_1 & l^2_2\\ l^3_1 &l^3_2\\ l^4_1 & l^4_2 \end{bmatrix}, \quad \begin{bmatrix} l^1_1 & l^1_2 & l^1_3 \\ l^2_1 & l^2_2 & l^2_3 \\ l^3_1 & l^3_2 & l^3_3 \\ l^4_1 & l^4_2 & l^4_3\end{bmatrix}$$
meaning that 
$$l(x^1, x^2, x^3, x^4)= \begin{bmatrix}l^i_j\end{bmatrix}\begin{bmatrix} x^1 \\ x^2 \\ x^3 \\ x^4 \end{bmatrix}.$$
Now the operator norm of $l$ is the usual $\ell^1$ norm of the matrix $\begin{bmatrix} l^i_j \end{bmatrix}$, that is 
$$\lVert l \rVert = \max_j \sum_{i=1}^4 \lvert l^i_j \rvert$$
where $j$ ranges over $1..\dim X$. To extend $l$ preserving norm it will be enough to fill up the corresponding matrix with zeroes: 
$$\begin{bmatrix} l^1_1 & 0 & 0 &0 \\ l^2_1 &0 &0 & 0 \\ l^3_1 &0 & 0 & 0\\ l^4_1 &0 &0 &0 \end{bmatrix},\quad \begin{bmatrix} l^1_1& l^1_2 & 0 & 0\\ l^2_1 & l^2_2 & 0 & 0\\ l^3_1 &l^3_2 & 0&0 \\ l^4_1 & l^4_2 &0 & 0\end{bmatrix}, \quad \begin{bmatrix} l^1_1 & l^1_2 & l^1_3 & 0 \\ l^2_1 & l^2_2 & l^2_3 & 0\\ l^3_1 & l^3_2 & l^3_3 & 0\\ l^4_1 & l^4_2 & l^4_3 & 0\end{bmatrix}$$
I don't know if this can be of some help. If the norm were $\ell^2$ then we could isometrically map every subspace $X$ into one of the preceding four and be done with it. With $\ell^1$ can we do the same thing...?
