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

In relation to the question on the Hardy inequality and the answer by Terry Tao, I've always been curious about the following:

Let $U \subset \mathbb{R}^n$ be a bounded domain of class $C^2$, $(e^{-t A})_{t \ge 0}$ be the Dirichlet heat semigroup(s) on $L^p(U)$, $1 \le p \le \infty$. $A$ is the Dirichlet Laplacian (i.e. zero boundary conditions). Compare the following (where $\lesssim$ hides a constant dependent on $p,q,U$):

For $\varphi \in L^p(U)$ and $1\le p \le q < \infty$, we have $$\| e^{-t A} \varphi\|_{q} \lesssim \|\varphi\|_p t^{-\frac{n}{2}(\frac{1}{p}-\frac{1}{q})}, \quad t >0.$$

and now on the weighted spaces $L^p(U,\delta)$ where $\delta(x):=\text{dist}(x,\partial U)$, $1\le p \le \infty$.

For $\varphi \in L^p_\delta(U)$ and $1\le p \le q < \infty$, we have $$\| e^{-t A} \varphi\|_{q,\delta} \lesssim \|\varphi\|_{p,\delta} t^{-\frac{n+1}{2}(\frac{1}{p}-\frac{1}{q})}, \quad t >0.$$

Quittner and Souplet call this the dimension shift phenomenon: the weighted space estimates are similar to those in standard $L^p$-spaces in $n+1$ dimensions.

Question 1: Is there something subtle and interesting happening here?

It seems to be based on the following (sketched) observations: from the estimate$$|e^{-tA} \varphi(x)| \lesssim \|\phi\|_{\infty} \frac{\delta(x)}{\sqrt{t}},\quad x \in U, t > 0,\quad \varphi \in L^\infty(U),$$ and as $e^{-tA}$ is self-adjoint in $(L^2,(\cdot,\cdot))$,
$$\|e^{-tA} \varphi\|_1 = (e^{-tA} \varphi, \chi_{U}) = (\varphi, e^{-tA} \chi_U) \lesssim t^{-1/2}(\varphi,\delta)$$ so $$\|e^{-tA} \varphi\|_{\infty} = \|e^{-(t/2)A}(e^{(t/2)A)} \varphi)\|_\infty \lesssim t^{-n/2} \|e^{-(t/2)A} \varphi\|_1 \lesssim t^{-(n+1)/2}\|\varphi\|_{1,\delta}$$ and the weighted estimate is obtained by Holder's inequality and some additional rigour (see the book by Quittner/Souplet for details).

Question 2: The weight $\delta$ seems very special. What can be said in the case $\delta^\alpha$ where $\alpha > 1$? It seems a different argument is needed.

I would love to hear any insightful or interesting remarks about the above. Thanks.

share|improve this question
1  
I warn you that these inequalities hold only if $p\le q$. Actually, the first one is valid even if $U^is unbounded. *I presume that you assume the Dirichlet boundary condition* $u=0$. –  Denis Serre Dec 5 '10 at 13:16
    
Yes of course! Fixed. Thanks. –  Dale Roberts Dec 5 '10 at 22:01
    
I've decided to roll back my additional comments and accept Terry's answer. Thank you Terry for the "dimensional analysis" insight. –  Dale Roberts Dec 6 '10 at 22:49

1 Answer 1

up vote 5 down vote accepted

It seems dimensional analysis already reveals the exponent behaviour. If we use $m$ (say) to denote the unit of length, then an unweighted $L^p$ norm has units $m^{n/p}$, while a weighted $L^p$ norm has units $m^{(n+1)/p}$. The Laplacian $A$ has units $m^{-2}$, so time should have units $m^2$ in order for the exponent in $e^{tA}$ to be dimensionless. This soon predicts the right exponents for both estimates.

A bit more directly; a ball of radius $r$ has unweighted volume comparable to $r^n$, but has weighted volume comparable to $r^{n+1}$ if it is near the boundary (and the boundary is where all the "action" takes place). This already largely explains the dimension shifting phenomenon (noting also that at time t, a heat flow will have spread things out at the spatial scale of $r \sim t^{1/2}$.)

One can make the above dimensional analysis arguments rigorous by a scaling argument, at least in the model case when U is a half-space. These arguments do not actually prove the above estimates, but they show that the specified powers of $t$ are the only possible choices for such an estimate to be true.

share|improve this answer
    
That's a very nice technique. I generally reason like that when playing around with inequalities (like I'm sure most people do) but I had no idea that it was a mathematical "dimensional analysis". I searched back over your blog posts to see what more you've said about this technique. Is this the most relevant post? terrytao.wordpress.com/2008/12/27/… –  Dale Roberts Dec 5 '10 at 22:20
    
I have not blogged specifically on dimensional analysis (maybe I will do so in the future), but that is probably the closest post I have currently on the subject. Some related topics are also at terrytao.wordpress.com/2007/09/05/… –  Terry Tao Dec 5 '10 at 23:26
    
A blog post on the topic of dimensional analysis would be a fantastic idea :) Especially the concept of "dimensionless" quantities. That's something I've never thought about before when doing heuristics. –  Dale Roberts Dec 6 '10 at 1:56
1  
Terry, I second Dale's sentiments and look forward to seeing what you write about this. I use dimensional analysis all the time in my research in differential and convex geometry. And, as Dale says, it is also worthwhile explaining the difference between dimensioned and dimensionless quantities. I was quite amazed that something my 10th grade chemistry teacher taught me could be so useful in pure mathematics. And, as you point out, it's all about scaling. –  Deane Yang Dec 6 '10 at 2:16
    
Yes that would be an interesting post! I would be especially curious to see examples where one can use dimensional arguments to prove positive results; the typical argument is a negative one where dimensionality is used to get necessary conditions for some inequality to hold. –  Piero D'Ancona Dec 6 '10 at 10:26

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.