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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 $l$?


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closed as too localized by quanta, t.b., Qiaochu Yuan May 14 '11 at 2:57

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@Arturo, what Bill said is for two spaces, $X$ and $Y$. Here $X$ and $Y$ are the same space. The question really is: what if $X$ is not \ell{\infty}^n$? – user4730 Dec 14 '10 at 5:32
Dear Arturo, what did you say a couple of minutes ago? – user4730 Dec 14 '10 at 6:05
@user4730: Something only half right. I said we can assume $||l||=1$, which is true enough (just scale), and something false. – Arturo Magidin Dec 14 '10 at 6:12
@user1736: No. The Hahn-Banach theorem applies only to linear forms (functionals), not to operators. – Florian Mar 14 '11 at 18:51
@Willie: Can't we just close this question? It is both completely uninteresting and unmotivated and pops up every now and again on MO and here e.g. in this version. The OP got an answer by Fedja in the comments on MO in one version of this question, which he didn't understand then deleted the question etc. I'm tired of this. – t.b. May 13 '11 at 22:42

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!.

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I am wondering if $\ell_1^4$ has the metric extension property. – user4730 Dec 15 '10 at 12:33

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...?

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