Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For any real-valued smooth function $u$, we have the Kato inequality

$|D|Du||^2\leq(det(Hess(u)))^2$,

which holds when $|Du|\neq0$.

If moreover $u$ is harmonic (in an open set in $\mathbb{R}^n$), how would the Kato inequality be improved to

$|D|Du||^2\leq\frac{n-1}{n}(det(Hess(u)))^2$ ?

share|improve this question
2  
Am I missing something, or is the latter inequality weaker than the former? Since n > n-1... –  dcs24 Oct 20 '12 at 21:45
    
Oops! Meant $\frac{n-1}{n}$...thanks Any ideas? –  Miranda Oct 20 '12 at 22:50
    
Not sure I believe this. What happens if the hessian is singular? Couldn't left side still be positive? –  Deane Yang Oct 20 '12 at 23:48
2  
Have you checked T. Branson's paper Kato constants in Riemannian geometry? –  Liviu Nicolaescu Oct 21 '12 at 0:51

2 Answers 2

As Matt says, the improved Kato inequality I know about involves the trace of the Hessian squared and not the determinant. The oldest proof I'm aware of for this inequality is in Stein's book, "Singular Integral Operators and Differentiability Properties of Functions", where he uses it to get improved estimates for functions in a Hardy space. But I believe that he does not call it a "Kato inequality".

This inequality was later put to extremely good use by Uhlenbeck, Taubes, Yau, and others and plays a critical role in many major theorems on minimal hypersurfaces, self-dual Yang-Mils fields, and special Riemannian metrics, especially where the PDE involved is "critical" or "scale-invariant". You can, for example, find a rather messy proof of the inequality (Stein's is cleaner) in the paper of Schoen-Simon-Yau, "Curvature estimates for minimal hypersurfaces".

Anyway, here's my take on it. The improved Kato inequality is a pointwise tensor space inequality that holds in many different situations. The desired inequality is, given a section $T$ of a tensor bundle $$ |\nabla |\nabla T|| \le c |\nabla^2T|. $$ With $c = 1$, this follows easily from the identity $$ |\nabla |\nabla T|| = \frac{|\nabla T \cdot \nabla^2T|}{|\nabla T|} $$ and the Cauchy-Schwarz inequality $$ |\langle v, t\rangle| \le c|v||t| $$ with $c = 1$, where $t \in V\otimes W$, $v \in V$, and $V$ and $W$ are vector spaces with inner products, and $\langle\cdot,\cdot\rangle$ is the inner product in $V$.

So here's the key point: If you happen to know that $t$ always lies in a known subspace of $V\otimes W$, then it is sometimes possible that the last inequality holds with a constant less than 1. To identify the optimal constant, it is easiest to differentiate $$ \frac{|\langle v, t\rangle|^2}{|v|^2|t|^2} $$ with respect to $t$ and solve for the critical points of $t$ restricted to the subspace.

In the specific example cited in the question, $W = V$, so $V\otimes W$ is the space of all matrices but $t$ is restricted to the space of traceless matrices. The same argument gives for example a more general improved Kato inequality for divergence free tensor fields. This was used, for example, in the work of Taubes and Uhlenbeck on self-dual Yang-Mills.

share|improve this answer

I think you might mean $tr(Hess(u)^2)$ instead of $det(Hess(u)))^2$ ? The Kato inequality I've seen is $|D|Du||^2 \leq |D^{2} u |^2 = tr(Hess(u)^2)$.

In that case, you might be able to come up with a proof by doing this:

  • Choose coordinates where $D_{1} u = |Du|$ and $D_{j} u = 0 $ for $ j \neq 1 $. Then, you can write out an expression for $|D|Du||^2$ as $\sum_{i=1}^{n}(D_{1i}u)^2$.
  • Note that $tr(Hess(u)^2) = 2\sum_{i < j}|D_{ij} u|^2 + \sum_{i=1}^n |D_{i}^{2}u|^2$.
  • Try to show with some manipulation of these sums that $(n-1)|D^2u|^2 - n|D|Du||^2 \geq 0$, probably using the fact that since $u$ is harmonic, $D_{1}^{2}u = \sum_{i>1} D^{2}_{i}u$ and then, from Jensen's inequality, $ | \sum _{i>1} D^{2} _{i} u |^2 \leq (n-1) \sum _{i>1} |D ^{2} _{i} u|^2 $.
share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.