When I learn Carleman estimates by myself, I am confused by the following fact:
Let $I=(-1,1)$ and $\varphi\in C^{1}(\bar{I})$. If there exists $\tau_{0}\in\mathbb{R}$ such that $$\tau^{\mu}\int_{-1}^{1}|u(x)|^{2}e^{2\tau\varphi(x)}dx\leq C\int_{-1}^{1}|u'(x)|^{2}e^{2\tau\varphi(x)}dx$$ for every $u\in C_{0}^{\infty}(I)$ and for every $\tau\geq\tau_{0}$. Let us prove that $\mu$ can not larger than $2$.
Proof: Suppose $\mu>2$. Let $v(x)=u(x)e^{\tau\varphi(x)}$, then $u(x)=v(x)e^{-\tau\varphi(x)}$, thus we have $u'(x)=v'(x)e^{-\tau\varphi(x)}-\tau\varphi'(x)e^{-\tau\varphi(x)}\Rightarrow e^{\tau\varphi(x)}u'(x)=v'(x)-\tau\varphi'(x).$
So, the estimate is$$\tau^{\mu}\int_{-1}^{1}v^{2}dx\leq C\int_{-1}^{1}|v'(x)-\tau\varphi'(x)|^{2}dx\leq C\tau^{2}\int_{-1}^{1}(\tau^{-2}|v'(x)|^{2}+\|\varphi'(x)\|_{L^{\infty}}^{2}|v|^{2})dx.$$ Because $\mu>2$ we thus have $$\int_{-1}^{1}v^{2}dx\leq C\tau^{2-\mu}\int_{-1}^{1}(\tau^{-2}|v'(x)|^{2}+\|\varphi'(x)\|_{L^{\infty}}^{2}|v|^{2})dx.$$Thus we let $\tau\to+\infty$ and get that $v=0$, a contradiction.
Does it right???? In the above calculation, our construction $v$ is depend on $u$ and $\tau$, as we let $\tau\to +\infty$, why $v$ not change as $\tau$ tends to $\infty$? I think that $v$ should be written by $v_{\tau}$.
Any hints are wellcome, thank you!!!