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I apologize in advance if this is well-known.

Let $X$ be a Banach space. Let's call only for this post that $X$ is self-injective if for every closed subspaces \begin{equation} A\subseteq B\subseteq X \tag{*}\label{eq:1} \end{equation} and bounded linear $T:A\to X$, there exists an extension $\widetilde{T}:B\to X$. The difference from the definition of injectivity is the restriction \eqref{eq:1} that $A$, $B$ must be subspaces of $X$. We can define $\lambda$-self-injectivity similarly.

Clearly, injective and separably injective Banach spaces are self-injective in this sense.

Question: Are there other self-injective (especially $1$-self-injective) Banach spaces? Is there a characterization of self-injectivity?


edit/update: Thanks to Jesus Castillo's answer, the common terminology for these spaces is extensible. If I'm allowed, I'd like to be a voluntary advertiser of a group of interesting problems, which I'm eager to learn the answers. The snippet below is from the copy of the book (p.331-332) on books.google.com co-authored by Jesus Castillo.

enter image description here

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    $\begingroup$ Every $ X$ which is a $\mathcal L_\infty$ space satisfying the property, for every closed subspace $A$ and every $ T : A \to X $ a bounded operator $T$ is of the form $ {\lambda}I_{A,X} + K$ with $K :A \to X$ a compact operator, is $\lambda$ - self injective. Argyros Haydon space does not satisfy the property and I do not know any such space. The result follows from a classical result of J. Lindenstrauss that asserts that $ {\mathcal L}_\infty $ spaces are $\lambda $ - injective for compact operators. $\endgroup$
    – S Argyros
    Aug 13, 2023 at 13:31
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    $\begingroup$ As has been pointed out in a deleted answer: you can remove $B$ from the definition, i.e. $X$ is self-injective if and only if every bounded linear map $T:A \to X$ where $A \subseteq X$ is a closed subspace extends with the same norm to a bounded linear map on all of $X$. $\endgroup$
    – Yemon Choi
    Aug 13, 2023 at 13:57
  • $\begingroup$ @SArgyros : Professor Argyros, it is my honor to receive a comment from the co-creator of one of the marvels & artworks of the Banach space theory. The property you've cited, combined with the compact extension property, must give self-injectivity. I'm deadly curious, but I'm not at the level to see why or why not $X_{AH}$ is self-injective. I'd be grateful if you guide. $\endgroup$
    – Onur Oktay
    Aug 13, 2023 at 14:06
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    $\begingroup$ @Onur Oktay:The Argyros Haydon space has closed subspaces $A$ and $S:A \to A$ bounded strictly singular and non compact. Clearly this operator $S$ is not extended to an operator to the whole space since such an extension must be a compact one. $\endgroup$
    – S Argyros
    Aug 13, 2023 at 15:33
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    $\begingroup$ I did not check details, but I think known results and arguments show that a super reflexive space that is self injective must have, for every $\epsilon>0$, type $2-\epsilon$ and cotype $2+\epsilon$. This suggests looking at the $2$-convexification of the Tsirelson space. $\endgroup$ Aug 13, 2023 at 16:01

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Those spaces have been called extensible (and its variations UFO, automorphic) in papers of Moreno-Plichko [On automorphic Banach spaces, Israel J. Math. 169 (2009) 29-45; or Castillo-Ferenczi-Moreno [On Uniformly Finitely Extensible Banach spaces, J.M.A.A. 410 (2014) 670-686] and others. For instance $c_0(I)$ is extensible.

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    $\begingroup$ Probably the best reference is Chapter 7 in "Homological Methods in Banach Spaces", Cambridge Studies in Advanced Math . 203 by Cabello Sánchez and Castillo. $\endgroup$ Aug 13, 2023 at 18:49
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    $\begingroup$ +1 This is amazing, thank you very much for the exact terminology and the references. $\endgroup$
    – Onur Oktay
    Aug 13, 2023 at 18:59
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    $\begingroup$ In particular, Theorem 7.2.10 in Jesus' book with Cabello Sanchez proves something stronger than what I mentioned in my other comment. (Sorry about that, Jesus. I forgot what you et al did.) $\endgroup$ Aug 13, 2023 at 20:00

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