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.

  • 1
    $\begingroup$ I am sure the left hand side must get involved in some way. $p\geq \epsilon$ is correct, but for $p<\epsilon$ mere norm comparison is definitely not right. It seems that there is going to be some inequality between $\sup_{Q_{\frac 12}} u$ and $\|u\|_{p, Q_{3/4}}, \|u\|_{\epsilon,Q_{3/4}}$. This might be an interpolation inequality of some kind. $\endgroup$ Mar 27 at 6:51

1 Answer 1


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 $p.$


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