Moser iteration in dimension $6$

Question

Let $M$ be a closed Riemannian manifold of dimension $6$. We have a function $f\geq 0$ on $M$ satisfying \begin{align*} \Delta f \leq gf-\frac{3}{4}f^2 \end{align*} Where $g$ is another smooth function on $M$ and our Laplacian is $d^*d$, i.e. locally it looks like $-\sum_i \frac{\partial^2}{\partial x_i^2}$. Using maximum principle we can get a $C^0$ bound of $f$ using $C^0$ bound of $g$, my question is: is it possible to get an $L^\infty$ bound of $f$ if we assume some $L^p$ bound of $g$ using some Moser iteration type argument?

Answer 1

\begin{align*} &\Delta f\leq gf-\frac{3}{4}f^2\\ &\Rightarrow\int f^{p+1}\Delta f\leq \int(gf^{p+2}-\frac{3}{4}f^{p+3})\\ &\Rightarrow\frac{4(p+1)}{(p+2)^2}\int |\nabla(f^{\frac{p+2}{2}})|^2\leq\int |g|f^{p+2}-\frac{3}{4}\int f^{p+3}\\ &\Rightarrow\frac{4(p+1)}{(p+2)^2}\int |\nabla(f^{\frac{p+2}{2}})|^2\leq \int \frac{(c|g|)^{p+3}}{p+3}+\int \frac{f^{p+3}}{c^{\frac{p+3}{p+2}}\frac{p+3}{p+2}}-\frac{3}{4}\int f^{p+3}\hspace{1 ex}[c \text{ any positive constant}] \end{align*} Notice that for any given $p\geq -1,\exists \hspace{0.4 ex}c,$ such that \begin{align*} &c^{\frac{p+3}{p+2}}\geq \frac{4}{3},\text{ and for that $c$ we get,}\\ &c^{\frac{p+3}{p+2}}>\frac{4}{3}\times\frac{p+2}{p+3}\Rightarrow\bigg(\frac{1}{c^{\frac{p+3}{p+2}}\frac{p+3}{p+2}}-\frac{3}{4}\bigg)< 0 \end{align*} Hence, we can write, \begin{align*} \int f^{p+3}\leq C(p)\int |g|^{p+3} \end{align*} Where the constant $C(p)$ depends on $p$ but not on $f$ or $g$. Hence, we get an $L^{p+3}$ bound of $f$ from an $L^{p+3}$ bound of $g$ for any $p\geq -1$.

We can actually do better! We start with the inequality: \begin{align*} &\frac{4(p+1)}{(p+2)^2}\int |\nabla(f^{\frac{p+2}{2}})|^2\leq\int |g|f^{p+2}-\frac{3}{4}\int f^{p+3}\\ &\Rightarrow\frac{4(p+1)}{(p+2)^2}\big(\int |\nabla(f^{\frac{p+2}{2}})|^2+\int (f^{\frac{p+2}{2}})^2\big)\leq \frac{4(p+1)}{(p+2)^2}\int |\nabla(f^{\frac{p+2}{2}})|^2+4\int f^{p+2}\leq\int (|g|+4)f^{p+2}-\frac{3}{4}\int f^{p+3} \end{align*} We do the same trick on the right hand side as before but we do it with $(|g|+4)$ and $f$ instead of $|g|$ and $f$ (the trick is nothing but Young's inequality) and we end up with \begin{align*} \int |\nabla(f^{\frac{p+2}{2}})|^2+\int \big(f^{\frac{p+2}{2}}\big)^2\leq \tilde{C}(p)\int (|g|+4)^{p+3} \end{align*} Where the constant $\tilde{C}(p)$ depends on $p$ but not on $f$ or $g$. Now we use Sobolev embedding $L^2_1\hookrightarrow L^3$ in dimension $6$ and this gives us: \begin{align*} &\big(\int f^{\frac{3(p+2)}{2}}\big)^{\frac{1}{3}}\leq C_S\big(\int |\nabla(f^{\frac{p+2}{2}})|^2+\int (f^{\frac{p+2}{2}})^2\big)^{\frac{1}{2}} \end{align*} So, we can control the $L^{\frac{3(p+2)}{2}}$ norm of $f$ using the $L^{p+3}$ norm of $g.$ Using elliptic regularity, we get \begin{align*} ||f||_{L^q_2}&\leq C\big(||\Delta f||_{L^q}+||f||_{L^q}\big)\\ &\leq C\big(||gf||_{L^q}+||f^2||_{L^q}\big)\\ &\leq C\big(||\frac{g^m}{m}||_{L^q}+||\frac{f^n}{n}||_{L^q}+||f^2||_{L^q}\big) \hspace{1 ex}[\frac{1}{m}+\frac{1}{n}=1,m,n>1] \end{align*} For $q>3,L^q_2\hookrightarrow C^0$ (using Sobolev embedding). If we take $m=n=2,$ then an $L^s$ bound of $g$ for $s>6$ gives us an $L^{\frac{s}{2}}$ bound for $f^2$ and $g^2$ and hence an $L^q_2$ bound of $f$ for $q>3$ and hence a $C^0$ bound of $f.$