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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 $l$?

Thanks!

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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$?2010-12-14
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    Dear Arturo, what did you say a couple of minutes ago?2010-12-14
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    @user4730: Something only half right. I said we can assume $||l||=1$, which is true enough (just scale), and something false.2010-12-14
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    Dear Arturo, that is generally a good strategy. I like it.2010-12-14
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    To be simple, concrete, and computable, the space is set to be just $\ell_1^4$.2010-12-14
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    If $l$ is bounded, then I think you can use Hahn-Banach if I'm not mistaken.2011-01-13
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    @user1736: No. The Hahn-Banach theorem applies only to linear forms (functionals), not to operators.2011-03-14
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    I cleaned up the comment thread a bit to remove some orphaned and no-longer useful ones.2011-05-13
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    @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](http://mathoverflow.net/questions/50600/an-existence-question-on-linear-map). 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.2011-05-13
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    @user4730: If you deface the question again, I will lock it.2011-10-31

2 Answers 2

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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.2010-12-15
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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...?