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 $E_\theta=(E_0,E_1)_\theta$?

## 1 Answer

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 $x \in (E_0,E_1)_\theta$.

functorwhich you can apply to the morphism $(P_0,P_1):(X_0,X_1)\to (E_0,E_1)$? $\endgroup$