# A notion of restricted injectivity for Banach spaces

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 $$$$A\subseteq B\subseteq X \tag{*}\label{eq:1}$$$$ and bounded linear $$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.

• 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. Aug 13, 2023 at 13:31
• 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 $X$. Aug 13, 2023 at 13:57
• @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. Aug 13, 2023 at 14:06
• @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. Aug 13, 2023 at 15:33
• 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. Aug 13, 2023 at 16:01

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.