# Complex interpolation of subspaces

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

• 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).
– cs89
Feb 21 at 22:51
• Yes. That is true. Feb 22 at 10:26
• Did you check in Triebel, 'Interpolation theory and function spaces', Theorem 1.17.1? Feb 22 at 10:55
• Isn't complex interpolation a functor which you can apply to the morphism $(P_0,P_1):(X_0,X_1)\to (E_0,E_1)$? Feb 22 at 14:41

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 $$\theta \in [0,1]$$.
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 $$x \in (E_0,E_1)_\theta$$.