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When reading the paper Control for Schrödinger operators on tori by N.Burq and M.Zworski (Math. Res. Lett., 19(2):309–324, 2012), an inequality confused me:

Define the flat torus $\mathbb{T}^2=\mathbb{R}^2/A\mathbb{Z}\times B\mathbb{Z}$. Suppose $u_0\in L^2(\mathbb{T}^2)$. $\Delta$ is the Laplace operator and $V\in\mathcal{C}^\infty(\mathbb{T}^2)$ is a smooth potential.

We define $$ v_{\varepsilon,0}=\frac{1}{\varepsilon} \left(e^{-i\varepsilon(-\Delta+V)}-I \right) u_0. $$ We can write $u_0$ in terms of orthonormal eigenvectors of Schrödinger operator $-\Delta+V: u_0=\sum_{\lambda\in\sigma(-\Delta+V)}u_{0,\lambda}e_\lambda$, then we have $$ \left \| v_{0,\alpha}-v_{0,\beta} \right \| _{H^{-4}}^2 \leq \sum_{\lambda\in\sigma(-\Delta+V)} \left | \frac{e^{-i\alpha \lambda}-1}{\alpha}-\frac{e^{-i\beta \lambda}-1}{\beta} \right | ^2 (1+\lambda)^{-2}\left | u_{0,\lambda} \right |^2. $$ What confused me is the term $(1+\lambda)^{-2}$.

I know the definition of the norm $\left \| u_0 \right \| _{H^{-4}}=\left \| (1-\Delta)^{-2}u_0 \right \|_{L^2}$, but what is the relationship between the eigenvalues of $-\Delta$ and $-\Delta+V$?

Why we can "substitute" the eigenvalues of $-\Delta+V$ into the position of $-\Delta$?

Thanks in advance!

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  • $\begingroup$ I don't think this can be right. If we take $u_0$ as an eigenfunction of $-\Delta$ and $V$ as a large constant and $\alpha,\beta$ such that $e^{i\alpha V}=e^{i\beta V}=1$, then the RHS almost computes the norm exactly. More specifically, we are off by a factor of $(1+\lambda)^2/(1+\lambda+V)^2$, and this we can make small. $\endgroup$
    – Christian Remling
    Commented Nov 28 at 23:52

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I have asked Professor Burq why the inequality holds.

He told me that we can add the potential and have an equivalent norm.

So it should be $\lesssim$ rigorously.

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  • $\begingroup$ Nice, but could you elaborate a little more? After all the question was upvoted so every nice answer is interesting for at least few other members. $\endgroup$
    – Daniele Tampieri
    Commented 6 hours ago
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    $\begingroup$ @DanieleTampieri OK. Thanks for the reminder. I'll add some details later. $\endgroup$
    – Tzy
    Commented 5 hours ago

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