Interpolation of $L^p$ spaces (from book by Caffarelli-Cabre) In the book "Fully nonlinear elliptic equations" by L. Caffarelli and X. Cabre, Theorem 4.8 (b) we want to prove the following maximum principle, for any $p>0$,
\begin{align*}
    \sup_{Q_{1/2}}\,u\leq C(p)\left(\|u^+\|_{L^p(Q_{3/4})}+\|f\|_{L^d(Q_1)}\right),
\end{align*}
where $C(p)$ is a constant depending only on $d$, $\lambda$, $\Lambda$ and $p$.
The proof consists of proving this estimate por a particular $p=\varepsilon$, which I have no problem with. Then, the authors claim that the general case $p>0$ follows easily by interpolation, but I don't know how. So, for the purpose of this question, we can assume this inequality holds for a fixed $p=\varepsilon$, and the question is how to generalize this for any $p>0$. Note that $Q_l\subset \mathbb{R}^d$ is a cube of side length $l$. This is my work so far:
If $p>\varepsilon$ we can apply  Holder inequality and get
\begin{align*}
\|u^+\|_{L^\varepsilon(Q_{3/4})}\leq |Q_{3/4}|^\frac{1}{r}\|u^+\|_{L^p(Q_{3/4})}, \quad \frac{1}{r}+\frac{1}{p}=\frac{1}{\varepsilon}. 
\end{align*}
Hence
\begin{align*}
\sup_{Q_{1/2}}u\leq& C\left(|Q_{3/4}|^\frac{1}{r}\|u^+\|_{L^p(Q_{3/4})}+\|f\|_{L^d(Q_1)}\right)\\
\leq &C\left(\|u^+\|_{L^p(Q_{3/4})}+\|f\|_{L^d(Q_1)}\right),
\end{align*}
since $ |Q_{3/4}|^\frac{1}{r}<1$.
The proof for any $p<\varepsilon$ follows by interpolation (I don't know how).
Interpolation of $L^p$ spaces: I only know the generalized Holder inequality: If $0<p_0<p_1\leq \infty$ and $g\in L^{p_0}(X)\cap L^{p_1}(X)$, then for every $\theta\in(0,1)$ and
\begin{align*}
\frac{1}{p_\theta}=\frac{\theta}{p_0}+\frac{1-\theta}{p_1}, 
\end{align*}
it holds
\begin{align*}
\|g\|_{L^{p_\theta}(X)}\leq \|g\|_{L^{p_0}(X)}^{\theta}\|g\|_{L^{p_1}(X)}^{1-\theta}.
\end{align*}
If we choose $p_\theta=\varepsilon$ and $p_1=+\infty$ then
\begin{align*}
p=\theta \varepsilon\in (0,\varepsilon)
\end{align*}
and
\begin{align*}
\|u\|_{L^{\varepsilon}(Q_{3/4})}\leq \|u\|_{L^{p}(Q_{3/4})}^{\theta}\|u\|_{L^{\infty}(Q_{3/4})}^{1-\theta}.
\end{align*}
By Young's inequality, we have for every $\delta>0$
\begin{align*}
\|u\|_{L^{p}(Q_{3/4})}^{\theta}\|u\|_{L^{\infty}(Q_{3/4})}^{1-\theta}\leq &\frac{\left(\delta^{-1} \|u\|_{L^{p}(Q_{3/4})}^{\theta}\right)^r   }{r}+\frac{\left(\delta\|u\|_{L^{\infty}(Q_{3/4})}^{1-\theta}\right)^s}{s}\\
=&\frac{\delta^{-r} \|u\|_{L^{p}(Q_{3/4})} }{r}+\frac{\left(\delta\|u\|_{L^{\infty}(Q_{3/4})}^{1-\theta}\right)^s}{s},
\end{align*}
where
\begin{align*}
\frac{1}{r}+\frac{1}{s}=1 \quad \mbox{ and } \quad r=\frac{1}{\theta}.
\end{align*}
Combining with our estimate for $p=\varepsilon$ we get
\begin{align*}
\sup_{Q_{1/2}}u-C\frac{\left(\delta\|u\|_{L^{\infty}(Q_{3/4})}^{1-\theta}\right)^s}{s}\leq C\left(\frac{\delta^{-r} \|u\|_{L^{p}(Q_{3/4})} }{r}+\|f\|_{L^d(Q_1)}\right).
\end{align*}
Problem: I cant take $\delta$ universally small such that
\begin{align*}
C\frac{\left(\delta\|u\|_{L^{\infty}(Q_{3/4})}^{1-\theta}\right)^s}{s}\leq \frac{1}{2}\sup_{Q_{1/2}}u
\end{align*}
because the domains are different.
 A: The omitted procedure is standard. Before the proof, we first state a useful lemma as following, whose proof can be found in the book (Lemma 4.3) written by Qing Han and Fang-Hua Lin.

Let $f$ be a nonnegative and bounded function defined on $[\tau_0,\tau_1]$ with $\tau_0\geq0.$ Suppose that $$f(t)\leq\theta f(s)+\frac{A}{(s-t)^\alpha}+B$$ for $\tau_0\leq t<s\leq\tau_1,$ where $\theta\in[0,1)$ is a constant. Then, $$f(t)\leq C\left(\frac{A}{(s-t)^\alpha}+B\right)$$ for any $\tau_0\leq t<s\leq\tau_1,$ where $C>0$ is a constant depending only on $\alpha$ and $\theta.$

Now we derive the estimate for $p\in(0,\epsilon).$ It can be shown by the rescaling and covering argument that $$\|u^+\|_{L^\infty(Q_{\theta R})}\leq C(n,\lambda,\Lambda)\left(\frac{1}{(R-\theta R)^{n/\epsilon}}\|u^+\|_{L^\epsilon(Q_R)}+\|f\|_{L^n(Q_1)}\right)$$ for any $\theta\in(0,1)$ and $R\in(0,1].$ Applying the Young inequality, we get $$\|u^+\|_{L^\infty(Q_{\theta R})}\leq\frac{1}{2}\|u^+\|_{L^\infty(Q_R)}+C(n,\lambda,\Lambda,p)\left(\frac{1}{(R-\theta R)^{n/p}}\|u^+\|_{L^p(Q_R)}+\|f\|_{L^n(Q_1)}\right).$$ Set $f(t):=\|u^+\|_{L^\infty(B_t)}$ for any $t\in(0,1].$ Then, $$f(r)\leq\frac12f(R)+\frac{C}{(R-r)^{n/p}}\|u^+\|_{L^p(Q_1)}+C\|f\|_{L^n(Q_1)}$$ for any $0<r<R\leq1.$ Using above lemma and then sending $R\to1-,$ we get $$\|u^+\|_{L^\infty(Q_\theta)}\leq\frac{C}{(1-\theta)^{n/p}}\|u^+\|_{L^p(Q_1)}+C\|f\|_{L^n(Q_1)}$$ for any $\theta\in(0,1).$ Thus, the trick of rescaling and covering yields $$\|u^+\|_{L^\infty(Q_{1/2})}\leq C\left(\|u^+\|_{L^p(Q_{3/4})}+\|f\|_{L^n(Q_1)}\right),$$ where $C>0$ is a constant depending only on $n,~\lambda,~\Lambda,$ and $p.$
