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Feb 24, 2022 at 6:53 comment added M.Oud Thank you a lot @Willie Wong , I think just the condition of the existence of projections is enough for me ( for my original probleme ). But this is very interesting!. And also I found here a paper devoted to this probleme researchgate.net/publication/… . They try to go more general on the conditions that allow the interpolation functor to be compatible with the Kernel.
Feb 23, 2022 at 17:14 history edited Willie Wong CC BY-SA 4.0
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Feb 23, 2022 at 14:30 comment added Willie Wong The proof of that last statement as given in numdam.org/article/AIF_1992__42_4_875_0.pdf is based on relaxing the conditions on $\Pi$: instead of looking for a single projection operator, the author constructs a family of uniformly bounded operators (not necessarily projections!) $\Pi_f: L^1 \to H^1$, indexed by $f\in H^1$, such that $\Pi_f(f) = f$.
Feb 23, 2022 at 14:20 comment added Willie Wong @OUDRANE a quick note, however: the complemented subspace argument is sufficient, but not necessary. An example: the Hardy spaces $H^p(\mathbb{T})$ can be described as those $L^p$ functions (necessarily $\subset L^1$) whose Fourier transforms have no negative modes. $H^1$ is not complemented in $L^1$, but for $p\in (1,\infty)$ we do have $H^p$ is complemented in $L^p$. However the sort of interpolation result you are asking for is true for Hardy spaces. (In this case you can write $H^p$ as the kernel of some $f:L^1\to \ell^\infty$.)
Feb 23, 2022 at 10:32 comment added M.Oud Thank you @Willie Wong. Its very useful,
Feb 23, 2022 at 8:58 comment added Hannes Chapter 1.17 in the Interpolation book of Triebel supports your complemented-subspace/projection suggestion. (And possibly worth a read for OP in any case.)
S Feb 23, 2022 at 5:04 history answered Willie Wong CC BY-SA 4.0
S Feb 23, 2022 at 5:04 history made wiki Post Made Community Wiki by Willie Wong