Let $(X_0,X_1)$ be an interpolation couple of Banach spaces. Using complex interpolation we can form Banach spaces $X_\theta:=(X_0,X_1)_\theta$ where $0<\theta<1.$ Let $E_\theta\subseteq X_\theta$ be a closed subspace. Let there be a contractive projection $P:X_\theta\to X_\theta$ for all $0\leq\theta\leq 1$ such that image of $P$ is $E_\theta.$ Is it true that $E_\theta=(E_0,E_1)_\theta$?
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asked Feb 21, 2023 at 16:31
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1$\begingroup$ Maybe the assumptions on $P$ should also hold for $\theta=0$ and $\theta=1$ (otherwise, $E_0$ and $E_1$ might be quite arbitrary and yield easy counter-examples). $\endgroup$– cs89Feb 21, 2023 at 22:51
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$\begingroup$ Yes. That is true. $\endgroup$– A beginner mathmaticianFeb 22, 2023 at 10:26
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2$\begingroup$ Did you check in Triebel, 'Interpolation theory and function spaces', Theorem 1.17.1? $\endgroup$– HannesFeb 22, 2023 at 10:55
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2$\begingroup$ Isn't complex interpolation a functor which you can apply to the morphism $(P_0,P_1):(X_0,X_1)\to (E_0,E_1)$? $\endgroup$– Jochen WengenrothFeb 22, 2023 at 14:41
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1 Answer
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I believe this is indeed true, if, as mentioned in my comment, one assumes that $P$ is a contractive projection from $X_0 \to X_0$ and $X_1 \to X_1$ and defines $E_\theta := P X_\theta$ for all $\theta \in [0,1]$.
I use the notations of Wikipedia's page on Interpolation spaces for complex interpolation.
Let $x \in (E_0,E_1)_\theta$. It is clear that $x \in X_\theta$. Moreover, by definition, there exists a function $f \in \mathcal{F}(E_0,E_1)$ such that $x = f(\theta)$. In particular, since $f : \mathbb{C} \to E_0 + E_1$, $Pf = f$, so $Px = x$ and $x \in E_\theta$.
Conversely, let $x \in E_\theta$. Thus, there exists $z \in X_\theta$ such that $x = Pz$. By definition, there exists $f \in \mathcal{F}(X_0,X_1)$ such that $z = f(\theta)$. Since $P$ is contractive, one checks that $Pf \in \mathcal{F}(E_0,E_1)$. Since $x = (Pf)(\theta)$, this proves that $x \in (E_0,E_1)_\theta$.
