Trace method is interpolation method Let $X,Y$ be Banach spaces, $1\leq p\leq \infty$ and $\theta\in (0,1)$. Then we want to define the real interpolation space $(X_0,X_1)_{\theta,p}$. This is possible via the trace method: Define the space $V$ of $X+Y$-valued regular distributions $u$ on $(0,\infty)$ such that $t^\theta u(t) \in \mathrm{L}^p_*(0,\infty; X_0)$ and $t^\theta u'(t) \in \mathrm{L}^{p'}_*(0,\infty; X_1)$ where $\mathrm{L}^q_*$ denotes the Lebesgue space with weight $1/t$ and $p'$ is the conjugate exponent to $p$. Then one can show that point evaluation in $0$ is meaningful and consider $V/\{u: u(0)=0\}$. This is a Banach space and via evaluation in $0$ we can define the interpolation space $(X_0,X_1)_{\theta,p}$.
One then needs to show that this construction actually is an interpolation method. Most books do this by showing equivalence to some other real interpolation method, e.g. the K-method.
I want to show this directly. It seems reasonable to try to show that $Tu\in V$ for $T: X_0+X_1\to Y_0+Y_1$. I have difficulties to show first that $Tu$ is a regular distribution and would like to have $(Tu)'=T(u')$ which would show the claim.
Does anybody have ideas on how to resolve those two problems or a reference where this calculation is done?
The only source I have found that directly shows the interpolation property is by Lions and Peetre (http://www.ams.org/mathscinet-getitem?mr=165343, in frensh), however their definitions are different from mine and I was not able to translate them to my setting.
 A: You should use the fact that $u:(0,\infty)\mapsto X_0+X_1$ satisfies the fundamental theorem of calculus, that is, $u(t)=u(t_0)+\int_{t_0}^tu'(r)\,dr$. Then since $T:X_0+X_1\mapsto Y_0+Y_1$ is linear and continuous you get
\begin{align}T(u(t))&=T\left(u(t_0)+\int_{t_0}^tu'(r)\,dr\right)\\&=T(u(t_0))+T\left(\int_{t_0}^tu'(r)\,dr\right)=T(u(t_0))+\int_{t_0}^t T(u'(r))\,dr,\end{align}
where in the last equality you are using the fact that the Bochner integral commutes with linear continuous functions $T$. Since this is true for every $t$ and $t_0$ you get that the function $T(u(t))$ is differentiable for a.e $t$  and $(T(u(t)))'=T(u'(t))$. To prove this you have to use Lebesgue points. The lemma that you want is that if $f:I\mapsto Z$ is Bochner integrable, then for a.e $t$,$$\frac{1}{h} \int_{t}^{t+h}f(r)\,dr\to f(t)$$ as $h\to 0$. This follows by applying the result for real valued functions to $t\mapsto\Vert f(t)-z_n\Vert$ where 
$\{z_n\}$ is dense in $f(I)$.  I am skipping a lot of details but I hope this helps.
