14

Consider the following problem:

Suppose $(M, g_{ij})$ is a compact Riemannian manifold. Assume $u$ is a smooth, nonnegative function which satisfies the differential inequality $$\Delta u \geq -cu$$ where $c$ is a constant.

Show that $$||u||_{L^{\infty}(M)} \leq C_{M}||u||_{L^{2}}(M)$$ for any constant $C_{M}$ depending only on the manifold $M$ and the constant $c$.

Could someone get me started on this, or give me some suggestions on how to tackle it?

A full solution would be great, but not necessary.

Context

Eventually, I need to take this inequality and extend it so that it's true for any $L^{p}$ norm on the RHS.

I'm not sure how to start; I think this might be a Moser iteration problem, but the only ones I've seen like this have been ones where we have a differential equality, or we have been working only on balls of various radii and not on whole manifolds.

Thank you in advance.

Community
  • 1

1 Answers1

1

We will consider the case where $c>0$.

(1) Formula : Assume that $f$ is a function on star-shaped domain $U$ in $\mathbb{R}^n$ wrt a point $O$. Then $$ \int_{x\in U} f(x) d{\rm vol}(x) =\frac{1}{n}\int_{y\in \partial U } \int_{0}^{|y-O|} f\bigg(O+ t\frac{y-O}{|y-O|}\bigg)\ dt dA(y) $$

(2) $\Delta u +cu \geq 0$ for some $c>0$ Since $M$ is a closed manifold so by divergence theorem $$ \int_M |\nabla u|^2 d{\rm vol} \leq c\int_M u^2 d{\rm vol} $$

(3) Fix $y\in \partial B(p,\epsilon )$. Then by FTC and Schwartz inequality, $$ u(p)-u(y) \leq \int_0^\epsilon |\nabla u|(c(t)) dt\leq \frac{1}{2} \int_0^\epsilon |\nabla u|^2(t)dt + \frac{1}{2}\epsilon$$ where $c$ is a unit speed geodesic from $p$ to $y$

Define $S:=\exp^{-1}_p\ \partial B(p,\epsilon )$ so that we have $$ u(p) V -\int_S u d{\rm vol}_g \leq \frac{1}{2} \int_S \int_0^\epsilon |\nabla u|^2(t) dtd{\rm vol}_g + \frac{1}{2}\epsilon V$$ where $V$ is $(n-1)$-dimensional volume of $S$ wrt $d{\rm vol}_g$ and $d{\rm vol}_g$ is Lebesgue measure of $T_pM$.

Hence since $M$ is locally Euclidean, so $$ u(p) - u_{{\rm ave}} (p) \leq C_M\int_{B(p,\epsilon )} |\nabla u|^2 d{\rm vol} + \frac{1}{2}\epsilon\ \ast $$ where $u_{ave} (p):= \frac{1}{V}\int_S u d{\rm vol}_g$

(4) Note that we suffice to consider the function $u$ with $\| u\|_2=1$

We will prove by a contradiction. Assume that there is sequence $u_i$ s.t. $$\Delta u_i+ cu_i\geq 0,\ \| u_i\|_2=1,\ \|u_i\|_\infty =u_i(p_i)>i $$

Note that $\int_{B(p_i,\epsilon_i) } u_i^2 d{\rm vol}\leq 1$ implies $ \epsilon_i\rightarrow 0$ Hence there is $i>C_M$ s.t. $(u_i)_{{\rm ave}} (p_i)< \frac{1}{2} u_i (p_i)$.

So from $\ast$ and (2) $ u_i (p_i) \leq C_M $. So it is a contradiction

HK Lee
  • 20,532