$\DeclareMathOperator\Ker{Ker}$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 $(X\cap \Ker(f),\Ker(f))_{s,p}\subset \Ker(f) \cap (X,Y)_{s,p}$ follow 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:

$$(X\cap Z,Z)_{s,p}=Z \cap (X,Y)_{s,p},$$

one 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 (because $Z$ here is not even closed in $Y$).

If what I'm asking is not true in general, 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 $f(X)$.

$\DeclareMathOperator\Ker{Ker}$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:

$$(X\cap Z,Z)_{s,p}=Z \cap (X,Y)_{s,p},$$

one 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 (because $Z$ here is not even closed in $Y$).

If what I'm asking is not true in general, 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 $f(X)$.

$\DeclareMathOperator\Ker{Ker}$Given two Banach spaces $X$ and $Y$ with a continues inclusion $X\subset Y$, and another couple $X’ \subset Y’$ with the same properties. Take $f : Y \longrightarrow Y’$ linear continues, such that $f_{\mid X}$ induce a linear continues 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 $(X\cap \Ker(f),\Ker(f))_{s,p}\subset \Ker(f) \cap (X,Y)_{s,p}$ follow 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 note even close in $Y$).

If what im 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 $f(X)$.

$\DeclareMathOperator\Ker{Ker}$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 $(X\cap \Ker(f),\Ker(f))_{s,p}\subset \Ker(f) \cap (X,Y)_{s,p}$ follow 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:

$$(X\cap Z,Z)_{s,p}=Z \cap (X,Y)_{s,p},$$

one 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 (because $Z$ here is not even closed in $Y$).

If what I'm asking is not true in general, 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 $f(X)$.

Given two Banach spaces $X$ and $Y$ with a continues inclusion $X\subset Y$, and another couple $X’ \subset Y’$ with the same properties. Take $f : Y \longrightarrow Y’$ linear continues, such that $f_{\mid X}$ induce a linear continues 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 $(X\cap Ker(f),Ker(f))_{s,p}\subset Ker(f) \cap (X,Y)_{s,p}$ follow 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 note even close in $Y$).

If what im 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 $f(X)$.

$\DeclareMathOperator\Ker{Ker}$Given two Banach spaces $X$ and $Y$ with a continues inclusion $X\subset Y$, and another couple $X’ \subset Y’$ with the same properties. Take $f : Y \longrightarrow Y’$ linear continues, such that $f_{\mid X}$ induce a linear continues 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 $(X\cap \Ker(f),\Ker(f))_{s,p}\subset \Ker(f) \cap (X,Y)_{s,p}$ follow 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 note even close in $Y$).

If what im 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 $f(X)$.