MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A standard Carleman-type estimate is of the form $$ \sum_{|\alpha|<m}{\tau^{2(m-|\alpha|-1)}\int{|D^{\alpha}u|^{2}e^{2\tau\phi}}dx}\leq K\int{|Pu|^{2}e^{2\tau\phi}dx},\quad u\in C_{0}^{\infty} $$ where $\phi$ is some weight function.This formula turn to be very useful in the study of uniqueness of Cauchy problem,and many mathematicians have considered this(such as Calderon,Hormander,Kenig,Sogge,and Tataru...)

For a first look at this inequality,I'm wondering whether the weight fuction makes a essential role,and besides, what's the original idea of it?Are there some very simple but illuminated examples to show the the reasonableness of the Carleman estimates ?

Well,one example in my mind is the first order operator $P=D+ix$,then it's easy to see that $P^*=D-ix$,and $$ P^*P-I=PP^*+I=-\frac{d^2}{dx^2}+x^2 $$ which is the so-called harmonic oscillator,then we have $$ 2\|u\|_{L^2}\leq \|Pu\|_{L^2},\quad u\in C_{0}^{\infty} $$ But in this simple example,there is no need to put a weight function,anyhow, from the proof,I guess the decomposition $P=\frac{P+P^*}{2}+\frac{P-P^*}{2}$ may be one of the general idea.

share|cite|improve this question
How does the operator $D$ look like?, I don't understand how did you get that $P^* P -I = -\frac{d^2}{dx^2}+x^2 $? If $D$ is a derivative operator,like $\frac{d}{dx}$, then we should have: $$(D-ix)(D+ix)=\frac{d^2}{dx^2}+x^2-iI$$ – Alan Jun 22 '13 at 11:25
Dear Alan, here $D=\frac{1}{i}\frac{d}{dx}$ – user23078 Jun 23 '13 at 6:57
Besides the good answer bellow you can get a good overview over some relevant results and the philosophy behind Carleman estimates from these notes of Lerner: – CPJ May 21 '14 at 15:20
up vote 10 down vote accepted

The weight function $\phi$ indeed plays an (I would even say the) essential role in Carleman estimates, and constructing one such that the estimate holds is one of the major challenges in proving and applying Carleman estimates. I will give a very informal overview, and refer to the references below for the technicalities.

The original motivation for introducing Carleman estimates was to prove unique continuation theorems, which can informally be stated as follows: Given a partial differential operator $P$, an oriented hypersurface $\Sigma$, and a function $u$ satisfying $Pu=0$, then if $u=0$ (locally) on one side $\Sigma^+$ of $\Sigma$, it vanishes on the other side $\Sigma^-$ as well. This can be interpreted in the sense that the complete information about $u$ in $\Sigma^-$ can be retrieved from information in $\Sigma^+$, i.e., the complete information flows across $\Sigma$. For linear hyperbolic equations, such information flow is described by the characteristics, and the idea behind Carleman estimates is to extend this concept to general (pseudo-)differential operators via microlocal analysis. This leads to considering so-called bicharacteristic rays, which need to satisfy the geometric property that all such rays passing near $\Sigma$ in $\Sigma^+$ must cross into $\Sigma^-$. Furthermore, solutions to $Pu=0$ should not decay exponentially along such rays towards $\Sigma$ (otherwise the information gets lost before it crosses $\Sigma$). A surface satisfying these conditions is called strongly pseudo-convex (with respect to $P$).

To make these conditions precise, one considers $\Sigma$ as the level set $\{\phi(x)=0\}$ (and $\Sigma^+$ as $\{\phi(x)>0\}$). The pseudo-convexity conditions above then become positivity conditions for the derivative of $\phi$ along the bicharacteristic flow (expressed using Poisson brackets). The connection is now that if $\Sigma=\{\phi(x)=0\}$ is a strongly pseudo-convex surface, then $e^{\lambda\phi}$ is strongly pseudo-convex function (in the sense of these positivity conditions) for sufficiently large $\lambda$. The parameter $\lambda$ is tied to the decay rate along the bicharacteristic rays, and hence the Carleman weight can be seen as accounting for this decay: The weight vanishes as the information flow across $\Sigma$ decays. The weight thus localizes the estimate to that part of $\Sigma^-$ where you still have usable information from $\Sigma^+$.

For example, for elliptic operators of second order, every smooth surface is strongly pseudo-convex (the decay condition is always satisfied); similarly, for hyperbolic operators with constant (real) coefficients, every non-characteristic surface is strongly pseudo-convex. In particular, for the wave operator $Pu=u_{{{tt}}} - c^2\Delta u$, any convex surface is strongly pseudo-convex, as well as the zero level sets for $\phi = x^2 - \beta t^2$ for $\beta < c^2$ (this bound is sharp).

[1] A good overview of unique continuation using Carleman estimates (along the lines I've given) is Daniel Tataru, Unique Continuation Problems for Partial Differential Equations, The IMA Volumes in Mathematics and its Applications Volume 137, 2004, pp 239-255.

[2] Carleman estimates are also useful for inverse problems for partial differential equations; this side is discussed in Victor Isakov, Inverse Problems for Partial Differential Equations, 2nd ed., 2006, Springer. (The example for the wave equation is Theorem 3.4.1.)

[3] David Dos Santos Ferreira has written a habilitation thesis on Carleman estimates.

share|cite|improve this answer

Just a try: I can imagine, that the function $w\equiv 1$ as a weight function would not be sufficient to conclude unique continuation. You should rather think if Carleman holds true for a weight function which is approximately the dirac delta function. Thats why typical carleman estimates have a weight function which have a pole at zero. For example $$ \int w^{-1-2\alpha} (x) f^2 (x) \leq C \int w^{-2-2\alpha} (\Delta f)^2, $$ (I dont know if the exponents are correct, see Bourgain Kenig 2005) where $w (x) = \phi (\lvert x \rvert)$ with $$ \phi (s) = s \exp \left(-\int_0^s \frac{1-e^{-1}}{t} dt \right) . $$ In this case $w^{-1}$ has a pole at zero and I guess this is more useful for application in unique continuation.

share|cite|improve this answer

Your Answer


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.