Given two Banach spaces $X$ and $Y$ with a continuous inclusion $X\subset Y$, and another couple $X’ \subset Y’$ with the same properties. Take $f : Y \longrightarrow Y’$ linear continuous, such that $f_{\mid X}$ induces a linear continuous map from $X$ to $X'$. My question is for $0<s<1$ and $p>1$, is the following true?
$$(X\cap Ker(f),Ker(f))_{s,p}=Ker(f) \cap (X,Y)_{s,p}$$
Here I'm considering the K-method for the interpolation.
The inclusion $(X\cap Ker(f),Ker(f))_{s,p}\subset Ker(f) \cap (X,Y)_{s,p}$ follows directly from the definition. My problem is the other inclusion?
It's clear that if we have $Z\subset Y$ then in general the following is not true:
$(X\cap Z,Z)_{s,p}=Z \cap (X,Y)_{s,p}$,
someone can take $X=H^{2}(U)$, $Z=H^{1}(U)$, $Y=L^{2}(U)$, $s=\frac{1}{2}$, and $p=2$. But this does not contradict our case (cause $Z$ here is not even closed in $Y$).
If what I'm asking is not true, is it true under the following assumptions:
-$f(X)$ is closed in $X'$, and $f(Y)$ is closed in $Y'$, and
-$f$ is open onto $f(Y)$, and $f_{\mid X}$ is open onto $f(X)$.