I'm struggling with a part of a proof that the concerning the interpolation of $L_p$ spaces.
Assume that $p_0 \geq 1$, $p_1 \geq 1$ and $0 < \theta < 1$. For $\frac{1}{p} = \frac{1-\theta}{p_0} + \frac{\theta}{p_1}$ and $\frac{1}{r(z)} = \frac{1-z}{p_0} + \frac{z}{p_1}$, define $$ f(z) = \exp(\varepsilon z^2 - \varepsilon \theta^2) |a(x)|^{p/r(z)}a(x)/|a(x)|$$ for $z$ satisfying $0 \leq \text{Re}(z) \leq 1$.
Among other things, the auther claims that $\|f\|_F \leq \exp(\varepsilon)$, where $$\|f\|_F :=~ \max\big( \sup\limits_{t \in \mathbb{R}}\|f(it)\|_{L_{p_0}}, \sup\limits_{t \in \mathbb{R}}\|f(1+it)\|_{L_{p_1}}\big).$$
The thing is I have not been able to write the proof... I'm getting lost on computations.
Some of my computations
$$ \frac{p}{r(z)} ~=~ \frac{\frac{1-z}{p_0}+\frac{z}{p_1}}{\frac{1-\theta}{p_0}+\frac{\theta}{p_1}} ~=~ \frac{p_1-p_1z + p_0z}{p_1-p_1\theta +p_0\theta} ~=~ \frac{p_1+(p_0-p_1)z}{p_1+(p_0-p_1)\theta}$$ Therefore $\frac{p}{r(it)} = \frac{p_1+(p_0-p_1)it}{p_1+(p_0-p_1)\theta}$ and $\frac{p}{r(1+it)} = \frac{p_1+(p_0-p_1)(1+it)}{p_1+(p_0-p_1)\theta}= \frac{p_0 + (p_0-p_1)it}{p_1+ (p_0-p_1)\theta}$.
$$\sup \|f(it)\|_{L_{p_0}} = \sup\Bigg\{\big|\exp\big(-\varepsilon(t^2+\theta^2)\big)\big|\cdot \Big\| |a(x)|^{\frac{p_1 + (p_0-p_1)it}{p_1+(p_0-p_1)\theta} - 1} a(x)\Big\|_{L_{p_0}} \Bigg\}.$$
I'll add some more shortly.