Let us consider $h_n=(2n+1)^{-1/2}\to 0$ as $n\to \infty$ be a small parameter, which we just write as $h$ for convenience, and $u_h : \mathbb{R} \to \mathbb{R}$ be functions satisfying $Pu_h=0$ (I think that $Pu=O(h)$ would be sufficient) where $P = -h^2D_x^2+(x^2-1)$, that is $Op_h(p)u_h=0$ where the principal symbol of $p$ is $p_0(x)=\xi ^2+x^2-1$. We also assume that $\|u_h\|_{L^2}=1$ (again, $\|u_h\|_{L^2}=O(1)$ would suffice I guess).
I want to prove the following:
For $\varepsilon \leq C h^{2/3}$, and $\Omega_{\varepsilon} = \{x : |x-1|<\varepsilon\}$ there holds$\|u_h\|_{L^{2}(\Omega_{\varepsilon})} = O(\varepsilon^{1/4})$.
This statement is in fact proven in [KTZ], Lemma 7.3 (what we want here is a special case), but I think that there should exist a much simpler proof, because in the mentioned article they claimed that intuitively $u_h$ microlocally concentrates on $x^2+\xi^2 =1$, so $\|u_h\|^2_{L^2(\Omega_{\varepsilon}}$ is the portion of the energy coming from it, that is the length of $\{x^2+\xi^2 =1 \} \cap \Omega_{\varepsilon}$ which is $\sim \varepsilon^{1/2}$.
What I know is the following:
- When $A=Op_h(a)$ is a zero order operator of principal symbol $a_0$ then we have $$(Au_h,u_h)_{L^2} \to \langle \mu, a_0 \rangle$$ where $\mu$ is the microlocal defect measure of $u_h$.
- Actually, we know that $\mu$ is the normalized Lebesgue measure on $\{x^2+\xi^2=1\}$.
From that, I want to take something like $a=\mathbf{1}_{|x-1|<\varepsilon}\mathbf{1}_{|\xi|<\varepsilon^{1/2}}$, but the problem is that I need to a smooth substitute for $a$, something like $a(x,\xi)=\chi(x/\varepsilon)\chi(\xi/\varepsilon^{1/2})$, where $\chi$ is a standard cuttof function. But the problem is that this operator is not uniform in $\varepsilon$ so I cannot claim any estimates in terms of $\varepsilon$.
Is there a way of making this intuition work, other than using [KTZ] (which I do not deeply understand)?
Reference: [KT] Koch, Tataru & Zworski : arxiv.org/pdf/math-ph/0603080.pdf
