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Let $M$ be a Riemannian manifold. Assume $u \in C^\infty(M)$ such that $u>0$ and

$\Delta u + \lambda u = 0,$

where $\lambda \geq 0$. There is a poinwise estimate for $|\nabla u|$ in Peter Li's book (Harmonic functions and applications to complete manifolds): Suppose

$Ric \geq -K.$

Then

$ | \nabla \log u |^2 \leq \frac{(n-1)K}{2} - \lambda + \sqrt{\frac{(n-1)^2K^2}{4} - (n-1) \lambda K}.$

My question is the following: if we assume

$\Delta u + \lambda u \geq 0$

do we still have a similar gradient estimate?

Note that by de Giorgi-Nash-Moser theory we still have (non-sharp) a Harnack inequality.

Thanks!

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2  
I suspect not. It appears that the inequality was proved using the maximum principle. Proving a bound for $u$ using the maximum principle still works for an elliptic inequality. However, to prove a bound on the gradient, you have to differentiate the original elliptic equation. But if $u$ satisfies only an inequality, then the inequality does not imply one for the gradient. –  Deane Yang May 19 '13 at 17:17
    
Deane Yang, thank you for your answer. I have one more question. The local version of this gradient estimate imply a Harnack inequality: for $x,y \in B(R)$ $u(x) \leq C_1 u(y) e^{C_2 R \sqrt{K+\lambda}}.$ Is it possible to obtain a explicit Harnack inequality like this for subsolutions: $\Delta u + \lambda u \geq 0?$ –  user34154 May 19 '13 at 17:39
1  
Unfortunately, I don't know the answer to that but I also suspect not. The inequality is two-sided, since you can switch $x$ and $y$, and I'm not sure that you can prove a two-sided inequality from a one-sided elliptic inequality. In fact, I suspect if you tried, you could come up with a counterexample (set $K = 0$). –  Deane Yang May 19 '13 at 21:02

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