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.

  • $\begingroup$ I've decided to roll back my additional comments and accept Terry's answer. Thank you Terry for the "dimensional analysis" insight. $\endgroup$ Dec 6, 2010 at 22:49

1 Answer 1


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.

  • $\begingroup$ 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/… $\endgroup$ Dec 5, 2010 at 22:20
  • $\begingroup$ 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/… $\endgroup$
    – Terry Tao
    Dec 5, 2010 at 23:26
  • $\begingroup$ 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. $\endgroup$ Dec 6, 2010 at 1:56
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    $\begingroup$ 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. $\endgroup$
    – Deane Yang
    Dec 6, 2010 at 2:16
  • $\begingroup$ 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. $\endgroup$ Dec 6, 2010 at 10:26

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