Suppose \(B>0\) and \(H_B\) is the corresponding Landau operator . For every \(m\in \mathbb{N}\) and every \(f\in \Ran 𝟙_{(-\infty,E]}(H_B)\) the following (magnetic) Bernstein inequality for magnetic derivatives hold
\[ \sum_{\alpha \in \{1,2\}^m} \lVert \widetilde{\partial }{}^\alpha f\rVert^2_{L^2(\mathbb{R}^n)} \le (E+mB)^m \lVert f\rVert^2_{L^2(\mathbb{R}^n)}. \][1, Theorem 7]
Proof
By (magnetic) integration by parts and a density argument we can proof that
\[ \sum_{\alpha \in \{1,2\}^m} \lVert \widetilde{\partial }{}^\alpha f\rVert^2_{L^2(\mathbb{R}^n)} = \langle f, R^m(\id)\rangle, \]where \(R(P)=\widetilde{\partial }_1P \widetilde{\partial }_1 + \widetilde{\partial }_2P \widetilde{\partial }_2\) for some polynomial \(P\).
Indeed, \(R^m(\id)\) is actually a polynomial \(F_m\) in \(H_B\). Pfeifer and Täufer proved in [1] that \(\lvert F_m(t)\rvert \le (t+mB)^m\). Combining this ingredients gives the above mentioned Bernstein inequality.
Remarks
- This result is not useful to proof a spectral inequality, since it is formulated in terms of magnetic derivatives and we do not have a calculus formulated with these ones. Indeed, the authors of [1, Remark 9] provide an example showing that a Bernstein inequality in terms of ordinary derivatives does not exits. By considering a different norm however, they were able to find an useful Bernstein-type inequality .
References Link to heading
- P. Pfeiffer and M. Täufer,
Magnetic Bernstein inequalities and spectral inequality on thick sets for the Landau operator,
2023. doi:10.48550/arXiv.2309.14902