I posted this question on MathOverflow but didn't get any answers, so I'm trying here .
I'm working on the book "Interpolation Spaces" by Bergh and Löfström. I'm interested in solving the exercise 2.8.4 on page 33. which asks:
Let $A$ be an interpolation space of exponent $\theta$ with respect to $\overline{A}$. Prove that there is a minimal interpolation functor $F_\theta$, which is exact and of exponent $\theta$, such that $F_\theta(\overline{A})=A$.
The following hint is given:
Use the fonctional $N_\theta(x)=\sum_j\|T_j\|_{A_0,X_0}^{1-\theta}\|T_j\|_{A_1,X_1}^{\theta}\|a_j\|_A$.
The idea is to rely on the proof of the Aronszajn-Gagliardo theorem (thm 2.5.1 from the Bergh-Löfström), which can be stated as follows
Consider the category $\mathscr{B}$ of all Banach spaces. Let $A$ be an interpolation space with respect to $\overline{A}$. There is a minimal exact interpolation functor $F_0$ such that $F_0(\overline{A})=A$.
If $\overline{X}=(X_0,X_1)$, $X:=F_0(\overline{X})$ is defined by the set of elements $x\in \Sigma(\overline{X})$ which admit a representation $x=\sum_j T_j a_j$ in $\Sigma(\overline{X})$, where $T_j:\overline{A}\rightarrow \overline{X}$, $a_j\in A$. The norm of $x$ in $X$ is defined as the infimum of $N_X(x)$ over all admissible representations of $x$, where $$ N_X(x)=\sum_j\max\left(\|T_j\|_{A_0,X_0},\|T_j\|_{A_1,X_1}\right)\|a_j\|_A. $$ I can prove all the points in the same way by remplacing $N_X$ by $N_\theta$ exept one : the inclusion $F_\theta (\overline{X})\subset \Sigma(\overline{X})$. Indeed, for the classical case, they do it like this : Let $x=\sum_j T_j a_j\in X$, then $$ \begin{split} \|x\|_{\Sigma(\overline{X})}&\leq \sum_j\|T_j a_j\|_{\Sigma(\overline{X})}\\ &\leq \sum_j\inf_{a_j=a_j^{(0)}+a_j^{(1)}}\|T_j\|_{A_0,X_0}\|a_j^{(0)}\|_{A_0}+\|T_j\|_{A_1,X_1}\|a_j^{(1)}\|_{A_1}\\ &\leq \sum_j\max\left(\|T_j\|_{A_0,X_0},\|T_j\|_{A_1,X_1}\right)\|a_j\|_{\Sigma(\overline{A})}\\ &\leq C\sum_j\max\left(\|T_j\|_{A_0,X_0},\|T_j\|_{A_1,X_1}\right)\|a_j\|_{A}, \end{split} $$ since $A\subset \Sigma(\overline{A})$. Therefore, $\|x\|_{\Sigma(\overline{X})}\leq C\|x\|_{X}$. But now for the $\theta$ case, I can't do the same because I can't prove that $$ \inf_{a_j=a_j^{(0)}+a_j^{(1)}}\|T_j\|_{A_0,X_0}\|a_j^{(0)}\|_{A_0}+\|T_j\|_{A_1,X_1}\|a_j^{(1)}\|_{A_1}\leq C\|T_j\|_{A_0,X_0}^{1-\theta}\|T_j\|_{A_1,X_1}^{\theta}\|a\|_{\Sigma(\overline{A})}, $$ and I don't even know if it is true (we need $C$ independant of $j$ and of the representation of $x$)
So my question is: do you have any idea on how to prove the previous inequality or on how to prove that $$ F_\theta(\overline{X})\subset \Sigma(\overline{X}), $$ eventually in another way?
Thanks a lot.