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## what’s the idea behind Carleman estimate

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.

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What's wrong with my latex ?it refused to work –  Shanlin Huang Oct 16 at 14:08
Thanks,it has been fixed –  Shanlin Huang Oct 16 at 14:31
Try to put a backtick  before and after your dollars. – Salvo Tringali Oct 16 at 14:32
Also, "carets" assume something behind them, so insert something at the position, where I put an X in the last formula: $P=\frac{P+P^}{2}+\frac{P-P^}{2}$` becomes then $P=\frac{P+P^X}{2}+\frac{P-P^X}{2}$ – Gottfried Helms Oct 16 at 14:37

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.