Proof of 5. Let us introduce a partition of unity \[\begin{align} \label{eq:partition32of32unity32for32h32theta32lambda}\Set{\chi_\nu}_{\nu=0}^{N_\Lambda}\, \end{align}\tag{13}\] where \(N_\Lambda<\infty\) is to be specified below. We pick the partition as follows. For any \(\nu\in\Set{0,\cdots,N_\Lambda}\), \(\chi_\nu\in C^\infty(\mathbb{R}^2\to[0,1])\) and \(1=\sum_{\nu=0}^{N_\Lambda}\chi_\nu\). We divide the members of the partition into three categories:

1. \(\nu=0\): \(\chi_0\) is centered at the origin and has support of diameter of order \(\delta\to0\) as \(\Lambda\to\infty\), for some \(\delta\) to be chosen below. To be concrete, we choose \(\chi_0\) to equal \(1\) on \(B_{C \delta}(0)\) for some sufficiently large order \(1\) constant \(C\).

2. \(\nu=N_\Lambda\): \(\chi_{N_\Lambda}\) is supported in the complement of a sufficiently large disc of radius \(R\) of order \(1\) (in \(\Lambda\)) chosen so that the disc entirely covers \(\operatorname{supp}(\theta)\).

3. \(\nu=1,\cdots,N_\Lambda-1\): \(\chi_\nu\) is supported in an annulus with radii between \(2^{\nu-1} \delta\) to \(2^{\nu+1} \delta\). Hence we choose \[\begin{align} N_\Lambda := \lceil\log_2(R/\delta)\rceil\;. \end{align}\]

The construction of this partition is presented below in 6.

In addition to \(\Set{\chi_\nu}_\nu\), we introduce a slightly more generous collection \(\Set{\psi_\nu}_\nu\) such that \(\psi_\nu=1\) on \(\operatorname{supp}(\chi_\nu)\) and such that for any \(x\in\mathbb{R}^2\), no more than, say, four values of \(\nu\) have \(x\in\operatorname{supp}(\psi_\nu)\). We shall also make use of open subsets \(U_\nu\) of \(\mathbb{R}^2\) below, each defined so it contains the support of \(\psi_\nu\). We shall take \[\begin{align} U_\nu \subseteq \Set{ 2^{\nu-3}\delta<\left\lVert x\right\rVert< 2^{\nu+3}\delta}\,. \end{align}\]

We then write a putative expression for the analytic continuation of \(\left(h_\lambda^\theta-z_\lambda\mathbb{1}\right)^{-1}\) as \[\begin{align} \label{eq:putative32expression32for32the32inverse32of32h95lambdatheta} \widetilde{A^\theta_\lambda} := \psi_0 \left(h_{\lambda}^{\text{MHO}}-\left(z_\lambda+\lambda^2\right)\mathbb{1}\right)^{-1}\chi_0 + \sum_{\nu=1}^{N_\Lambda-1}T_{\lambda,\nu}+\psi_{N_\Lambda} \left(P^2-z_\lambda\mathbb{1}\right)^{-1}\chi_{N_\Lambda}\,.\nonumber\\ \end{align}\tag{14}\] The operators \(T_{\lambda,\nu} \in \mathcal{B}(L^2(\mathbb{R}^2))\) should roughly be considered as being given by \[\begin{align} T_{\lambda,\nu} "=" \psi_\nu \left(h^\theta_\lambda-z_\lambda\mathbb{1}\right)^{-1}\chi_\nu\, , \end{align}\] where \(z_\lambda=z_\lambda(\xi)\) is given 10 . The issue is that we do not know a-priori that \(h^\theta_\lambda-z_\lambda\mathbb{1}\) is invertible for \(\lambda\in\Omega_{\Lambda,\varepsilon}\) (indeed, that’s precisely what we’re trying to prove) so we cannot write down such an expression, even if it’s restricted between \(\chi_\nu\) and \(\psi_\nu\). To deal with this problem, we use the formalism of pseudo-differential operators (see 5 below). Using this formalism, \(T_{\lambda,\nu}\in\mathcal{B}(L^2(\mathbb{R}^2))\) is constructed as follows. We work with the symbol \(s_\lambda^\theta\) associated with \(h_\lambda^\theta-z_\lambda\mathbb{1}\), but consider it as a symbol where the position variable ranges over \(U_\nu\) rather than the whole of \(\mathbb{R}^2\). If we can show ellipticity of the operator associated with this symbol, general pseudo-differential arguments (26 below, applied to a rescaling of \(s_\lambda^\theta\)) guarantee the existence of an approximate analytic inverse \(T_{\lambda,\nu}\) with appropriate bounds, in particular, we will derive \[\begin{align} \label{eq:estimate32on32approximate32resolvent32in32mesoscopic32annuli} \left\lVert T_{\lambda,\nu}\right\rVert_{\mathcal{B}(L^2(U_\nu))} \lesssim \frac{1}{\Lambda^2 2^{2\nu} \delta^2} \end{align}\tag{15}\] and \[\begin{align} \label{eq:estimate32on32error32term32in32approximate32resolvent32in32mesoscopic32annuli} \left\lVert\left(h_\lambda^\theta-z_\lambda\mathbb{1}\right)T_{\lambda,\nu}-\chi_\nu\right\rVert_{\mathcal{B}(L^2(U_\nu))} \lesssim \frac{1}{\Lambda 2^{2\nu} \delta^2}\,. \end{align}\tag{16}\] This prompts us to choose \(\delta\) so that \[\begin{align} \label{eq:constraint32on32delta} \lim_{\Lambda\to\infty}\Lambda\delta^2 = \infty\,. \end{align}\tag{17}\] Indeed, we take \[\begin{align} \label{eq:the32choice32of32delta}\delta = \Lambda^{\eta-1/2} \end{align}\tag{18}\] for some small \(\eta>0\). Let \[r_{\lambda,\nu}(z_\lambda) := \begin{cases} \left(h_{\lambda}^{\text{MHO}}-\left(z_\lambda+\lambda^2\right)\mathbb{1}\right)^{-1} & \nu=0 \\ T_{\lambda,\nu} & \nu \in \Set{1,\cdots,N_\Lambda-1}\\ \left(P^2-z_\lambda\mathbb{1}\right)^{-1} & \nu=N_\Lambda \end{cases}\] We calculate \(\left(h_\lambda^\theta-z_\lambda\mathbb{1}\right)\widetilde{A^\theta_\lambda}\), where \(\widetilde{A^\theta_\lambda}\) is given by 14 : \[\begin{align} \left(h_\lambda^\theta-z_\lambda\mathbb{1}\right)\widetilde{A^\theta_\lambda} &= \sum_{\nu=0,N_\Lambda} \left(h_\lambda^\theta-z_\lambda\mathbb{1}\right) \psi_\nu r_{\lambda,\nu}(z_\lambda) \chi_\nu+\sum_{\nu=1,\cdots,N_\Lambda-1} \left(h_\lambda^\theta-z_\lambda\mathbb{1}\right) T_{\lambda,\nu}\\ &=: \mathbb{1}+K_\lambda^\theta\;. \end{align}\] The error term is explicitly given by: \[\begin{align} \label{eq:error32term32for32theta-resolvent} K_\lambda^\theta &:= \left[h_\lambda^\theta ,\psi_0\right] r_{\lambda}^{\mathrm{MHO}}(z_\lambda) \chi_0 + \psi_0 \left(h_\lambda^\theta+\lambda^2\mathbb{1}-h_\lambda^{\mathrm{MHO}}\right)r_{\lambda}^{\mathrm{MHO}}(z_\lambda) \chi_0+\\ &+\sum_{\nu=1}^{N_\Lambda-1} \left(\left(h_\lambda^\theta-z_\lambda\mathbb{1}\right) T_{\lambda,\nu}-\chi_\nu\right)+\left[h_\lambda^\theta ,\psi_{N_\Lambda}\right] \left(P^2-z_\lambda\mathbb{1}\right)^{-1} \chi_{N_\Lambda}\nonumber \end{align}\tag{19}\]

We will show that \(\Omega_{\Lambda,\varepsilon}\ni\lambda\mapsto K_\lambda^\theta\) is analytic and moreover, with the choice 18 , \[\begin{align} \left\lVert K_\lambda^\theta\right\rVert_{\mathcal{B}(L^2(\mathbb{R}^2))} \lesssim N_\lambda\Lambda^{-\eta} \sim \log\left(\Lambda\right)\Lambda^{-\eta} \end{align}\] so that the Neumann series may be inverted to yield the sought after analytic inverse \[\begin{align} \boxed{A_\lambda^\theta := \widetilde{A^\theta_\lambda} \left(\mathbb{1}+K_\lambda^\theta\right)^{-1}}\,. \end{align}\]

We note that it suffices to bound the norm of each summand within \(K_\lambda\) separately since there are only \(N_\Lambda\sim\log(\Lambda)\) terms, each of which decays at least like \(\Lambda^{-\eta}\).

2.1.1.1 The \(\nu=0\) term: the MHO.

We deal with the two terms in 19 separately. For the commutator, \[\begin{align} \left[h_\lambda^\theta ,\psi_0\right] r_{\lambda}^{\mathrm{MHO}}(z_\lambda) \chi_0 = \left[H_\lambda^{\mathrm{Landau},\theta} ,\psi_0\right] r_{\lambda}^{\mathrm{MHO}}(z_\lambda) \chi_0\,. \end{align}\] Since by 12 , we have \(H_\lambda^{\mathrm{Landau},\theta} \equiv \left(P-\frac{1}{2}b\lambda \theta(X)X^{\perp}\right)^{2}\) we learn that \[\begin{align} \left[H_\lambda^{\mathrm{Landau},\theta} ,\psi_0\right] = \left[P^2,\psi_0\right] - b \lambda \theta(X) X \wedge \left[P, \psi_0\right] = -\left(\Delta \psi_0\right) -2 \operatorname{i}\left(\nabla \psi_0\right) \cdot P + \operatorname{i}b \lambda \theta(X) X \wedge \left(\nabla \psi_0\right)\,. \end{align}\]

Now, thanks to 30, all terms are indeed analytic in \(\lambda\in\Omega_{\Lambda,\varepsilon}\). Moreover, since \(\psi_0\) is equal to \(1\) on a somewhat enlarged neighborhood containing \(\operatorname{supp}(\chi_0)\), we have \[\begin{align} \mathrm{dist}(\operatorname{supp}(\chi_0),\operatorname{supp}(\partial_i\psi_0)) \gtrsim c\delta\,. \end{align}\]

But now we invoke the estimate proven in 30 \[\begin{align} \left\lVert\chi_S(X) P_j^\alpha r_{\lambda}^{\mathrm{MHO}}(z_\lambda)\chi_T(X)\right\rVert_{\mathcal{B}(L^2(\mathbb{R}^2))} \lesssim \exp\left(-c \Lambda\mathrm{dist}(S,T)^2\right)\qquad(\lambda\in\Omega_{\Lambda,\varepsilon}\text{ and } S,T\subseteq\mathbb{R}^2\nonumber\\j=1,2,\alpha=0,1)\nonumber\\ \end{align}\] which holds as soon as \(\mathrm{dist}(S,T)>\Lambda^{-1/2}\).

Thanks to 18 we may conclude \(\left\lVert\left[h_\lambda^\theta ,\psi_0\right] r_{\lambda}^{\mathrm{MHO}}(z_\lambda) \chi_0\right\rVert\lesssim\exp\left(-c \Lambda^{\eta}\right)=o(1)\).

For the second term, \[\begin{align} \psi_0 \left(h_\lambda^\theta+\lambda^2\mathbb{1}-h_\lambda^{\mathrm{MHO}}\right)r_{\lambda}^{\mathrm{MHO}}(z_\lambda) \chi_0 \end{align}\] we have, using the fact that \(\theta=1\) on \(\operatorname{supp}(\psi_0)\), \[\begin{align} \psi_0\left(h_\lambda^\theta+\lambda^2\mathbb{1}-h_\lambda^{\mathrm{MHO}}\right) &= \psi_0\left(\lambda^2\left(v(X)+1\right) - \frac{1}{2}\lambda^2\langle X, \left(\mathbb{H} v\right)(0)X \rangle\right)\,. \end{align}\] We recall \(\delta\equiv\delta_\Lambda\) was the support radius of \(\psi_0\). Below in 30 we prove the estimate \[\begin{align} \left\lVert r_{\lambda}^{\mathrm{MHO}}(z_\lambda) \right\rVert_{\mathcal{B}(L^2(\mathbb{R}^2))} \lesssim \Lambda^{-1} \end{align}\] so that all together we estimate \[\begin{align} \left\lVert\psi_0 \left(h_\lambda^\theta-h_\lambda^{\mathrm{MHO}}\right)r_{\lambda}^{\mathrm{MHO}}(z_\lambda) \chi_0\right\rVert \lesssim \Lambda \delta^3\,. \end{align}\] The bound depends also on the third derivatives of \(v\).

We see that if we pick \(\delta\) smaller than \(\Lambda^{-1/3}\) we can make this term arbitrarily small; this is compatible with 18 .

2.1.1.2 The term \(\nu=N_\Lambda\): the free Laplacian.

Here the support of \(\chi_\nu\) has no intersection with \(\operatorname{supp}(\theta)\). Since \(\operatorname{supp}(\theta)\supseteq\operatorname{supp}(v)\), this means that in that region, our operator agrees with the free Laplacian \(P^2\), which is why we approximate its resolvent by \(\left(P^2-z_\lambda\mathbb{1}\right)^{-1}\).

We shall use freely the trivial estimate \[\begin{align} \label{eq:upper32bound32on32free32resolvent} \left\lVert\left(P^2-z_\lambda\mathbb{1}\right)^{-1}\right\rVert &= \frac{1}{\inf_{E\in[0,\infty)}|E-z_\lambda|}\lesssim \frac{1}{\sin(\varepsilon)\Lambda^2}\,. \end{align}\tag{20}\] Moreover, we also have the Combes-Thomas estimate: For any \(S,T\subseteq\mathbb{R}^2\), \(j=1,2\) and \(\alpha=0,1\), \[\begin{align} \left\lVert\chi_S P_j^\alpha \left(P^2-z_\lambda\mathbb{1}\right)^{-1} \chi_T\right\rVert \lesssim \exp\left(-\sqrt{\sin(\varepsilon)\Lambda^2}\mathrm{dist}(S,T)\right)\,. \end{align}\]

As a result, we see the commutator term \[\begin{align} \left[h_\lambda^\theta ,\psi_{N_\Lambda}\right] \left(P^2-z_\lambda\mathbb{1}\right)^{-1} \chi_{N_\Lambda} \end{align}\] has exponentially decaying norm. The distance between \(\operatorname{supp}(\left[h_\lambda^\theta ,\psi_{N_\Lambda}\right])\) and \(\operatorname{supp}(\chi_{N_\Lambda})\) is of order \(1\).

2.1.1.3 The terms \(\nu=1,\cdots,N_\Lambda-1\).

We postpone the construction of \(T_{\lambda,\nu}\) which obey the estimates 15 16 to 4 below.

The proof of 5 is now complete. ◻

Proof. If \(\alpha=0\) the claim is covered by the analysis above (collecting all the various estimates on the terms in \(\widetilde{A_\lambda^\theta}\) yields \(\left\lVert A_\lambda^\theta\right\rVert\lesssim \Lambda^{-1}\)).

Next, thanks to \[\begin{align} \left\lVert P_j A_\lambda^\theta\right\rVert&=\left\lVert P_j\widetilde{A_\lambda^\theta}\left(\mathbb{1}+K_\lambda^\theta\right)^{-1}\right\rVert\leq \left\lVert P_j \widetilde{A_\lambda^\theta}\right\rVert\left\lVert\left(\mathbb{1}+K_\lambda^\theta\right)^{-1}\right\rVert \end{align}\] it suffices to bound \(P_j\) acting on each individual term in 14 . To that end, we proceed in much the same fashion as we have above.

For \(\nu=0\) or \(\nu=N_\Lambda\), we first take care of the commutator term \([P_j,\psi_\nu] r_{\lambda,\nu}(z_\lambda)\chi_\nu\), which boils down to the boundedness of the resolvent of either the MHO or the free Laplacian. Next, \(P_j r_{\lambda,\nu}(z_\lambda)\) is also bounded: for the MHO this is proven below in 30 and for the free Laplacian it is standard.

Finally for \(\nu=1,\cdots,N_{\Lambda-1}\), 26 below yields also the estimates \[\begin{align} \left\lVert P_j T_{\lambda,\nu}\right\rVert_{\mathcal{B}(L^2(U_\nu))} \lesssim \frac{1}{\Lambda^2 2^{2\nu} \delta^3} \end{align}\] and hence the result. ◻

Proof of 7. Let \(\chi\in C^\infty(\mathbb{R}^2\to[0,1])\) be some cut-off function to be determined below. We assume \(\chi=1\) on \(B_a(0)\), so \(\chi Q=Q\), and moreover, that \(\theta=1\) on \(\operatorname{supp}(\chi)\).

We begin by calculating \[\begin{align} \left(h_\lambda-z_\lambda\mathbb{1}\right)\chi A_\lambda^\theta Q &= \left[\left(h_\lambda-z_\lambda\mathbb{1}\right),\chi\right]A_\lambda^\theta Q+\chi \left(h_\lambda-h_\lambda^\theta\right)A_\lambda^\theta Q+Q \\ &= \left[h_\lambda,\chi\right]A_\lambda^\theta Q+Q \,. \end{align}\] Now we would like to argue that the first term is small so that we can find a right inverse on \(QL^2(\mathbb{R}^2)\). However, the problem is the commutator \(\left[h_\lambda,\chi\right]\) "leaks" outside of \(Q\), i.e., \[\begin{align} \left[h_\lambda,\chi\right]A_\lambda^\theta Q = Q\left[h_\lambda,\chi\right]A_\lambda^\theta Q+Q^\perp \left[h_\lambda,\chi\right]A_\lambda^\theta Q \end{align}\] and \(Q^\perp \left[h_\lambda,\chi\right]A_\lambda^\theta Q\neq 0\).

To deal with this issue, we apply the Landau resolvent on the left and rewrite it as \[\begin{align} \left[h_\lambda,\chi\right]A_\lambda^\theta Q &= \left(h_\lambda-z_\lambda\mathbb{1}\right)R^{\mathrm{Landau}}_\lambda(z_\lambda) \left[h_\lambda,\chi\right]A_\lambda^\theta Q-\lambda^2v(X)R^{\mathrm{Landau}}_\lambda(z_\lambda) \left[h_\lambda,\chi\right]A_\lambda^\theta Q \,. \end{align}\] This makes sense, even though \(\lambda\) may not be real, thanks to 28 down below, which yields an extension of \(R^{\mathrm{Landau}}_\lambda(z_\lambda)\) when acting on functions with (fixed) compact support; \(\left[h_\lambda,\chi\right]\) is an operator of the form \(f(X)+g(X)P\) where both \(f,g\) have compact support within \(\operatorname{supp}(\partial_j \chi)\).

Re-arranging and using the fact that \(Q v(X) = v(X)\) we find \[\begin{align} \left(h_\lambda-z_\lambda\mathbb{1}\right) \widetilde{A_\lambda} &= Q + K_\lambda \end{align}\] with \[\begin{align} \widetilde{A_\lambda} := \left(\chi-R^{\mathrm{Landau}}_\lambda(z_\lambda) \left[h_\lambda,\chi\right]\right) A_\lambda^\theta Q \end{align}\] and \[\begin{align} K_\lambda := -\lambda^2 Q v(X)R^{\mathrm{Landau}}_\lambda(z_\lambda)\left[h_\lambda,\chi\right]A_\lambda^\theta Q\,. \end{align}\] We remark that \(K_\lambda\) is now an operator \(QL^2\to QL^2\) whereas \(A_\lambda:QL^2\to L^2\).

Hence if we can show that \(\left\lVert K_\lambda\right\rVert<1\) we would be finished via \[\begin{align} \boxed{A_\lambda := \widetilde{A_\lambda} \left(\mathbb{1}+K_\lambda\right)^{-1}}\,. \end{align}\] To do so we will use two basic facts: ?? and ?? . The latter, which is applicable thanks to our small \(b\) assumption, means we have exponential decay in \(\Lambda\) since \(\mathrm{dist}(\operatorname{supp}(v),\partial_j\chi)\gtrsim O(1)\), but the former means there can’t be any term in \(P_j A_\lambda^\theta\) which explodes faster (in fact the norm of this term decays too). Hence \(\left\lVert K_\lambda\right\rVert\) is indeed arbitrarily small for large \(\lambda\).

Finally, we tend to establishing ?? . Let \(S\subseteq\mathbb{R}^2\) be such that \(\mathrm{dist}(S,\operatorname{supp}(\chi))>0\). Then \[\begin{align} \label{eq:estimate32on32continuation32of32resolvent} \left\lVert\chi_S P_j^\alpha A_\lambda \right\rVert &= \left\lVert\chi_S P_j^\alpha \left(\chi-R^{\mathrm{Landau}}_\lambda(z_\lambda) \left[h_\lambda,\chi\right]\right) A_\lambda^\theta Q \left(\mathbb{1}+K_\lambda\right)^{-1} \right\rVert \\ &\leq \left\lVert\chi_S P_j^\alpha R^{\mathrm{Landau}}_\lambda(z_\lambda) \left[h_\lambda,\chi\right] \right\rVert\left\lVert A_\lambda^\theta \right\rVert\frac{1}{1-\left\lVert K_\lambda\right\rVert} \end{align}\tag{21}\] and we then invoke ?? again. We now choose \(\chi\) to have support within \(B_{2a}(0)\) to get the statement.

2.1.2.1 Intermediate conclusion.

So far we managed to define an operator \(A_\lambda\in\mathcal{B}(QL^2(\mathbb{R}^2))\) such that

1. \(\left(h_\lambda-z_\lambda\mathbb{1}\right)A_\lambda = Q\) for all \(\lambda\in\Omega_{\Lambda,2\Lambda,\varepsilon}\).

2. \(\Omega_{\Lambda,2\Lambda,\varepsilon}\ni\lambda\mapsto A_\lambda\in\mathcal{B}(QL^2(\mathbb{R}^2))\) is analytic.

3. We have the bound \(\left\lVert A_\lambda\right\rVert_{\mathcal{B}(QL^2(\mathbb{R}^2))}\lesssim \Lambda^{\sharp}\exp\left(Cb\Lambda\right)\) for some \(C\in\mathbb{R}\), for all \(\lambda\in\Omega_{\Lambda,2\Lambda,\varepsilon}\).

4. We have off-diagonal exponential decay in the sense that \[\begin{align} \left\lVert\chi_S A_\lambda \right\rVert_{\mathcal{B}(QL^2(\mathbb{R}^2)\to L^2(\mathbb{R}^2))} \lesssim \exp\left(-c \left|\lambda\right| \mathrm{dist}(S,B_{2a}(0))\right)\qquad(S\subseteq\mathbb{R}^2) \end{align}\] for some \(c>0\), for all \(\lambda\in\Omega_{\Lambda,2\Lambda,\varepsilon}\).

2.1.2.2 Patching different regions in \(\Omega_{\Lambda,\varepsilon}\).

Fix a number \(r\in(1,2)\) and set \[\Lambda_k:=r^k\Lambda,\qquad \mathcal{D}_k:=\Omega_{\Lambda_k,2\Lambda_k,\varepsilon} =\bigl\{z\in\mathbb{C}:\Lambda_k<|z|<2\Lambda_k,\;\;|\arg z|<\tfrac{\pi}{2}-\varepsilon\bigr\}.\] Then \(\bigcup_{k\ge0}\mathcal{D}_k=\Omega_{\Lambda,\varepsilon}\) and consecutive domains have a nontrivial overlap: \[\mathcal{D}_k\cap\mathcal{D}_{k+1} =\bigl\{z:\Lambda_{k+1}<|z|<2\Lambda_k,\;\;|\arg z|<\tfrac{\pi}{2}-\varepsilon\bigr\}\neq\varnothing\] because \(r<2\).

For each \(k\) the construction above produces an analytic map \[\mathcal{D}_k\ni\lambda\mapsto A^{(k)}_\lambda\in \mathcal{B}(QL^2(\mathbb{R}^2)\to L^2(\mathbb{R}^2))\] and satisfies \[\bigl(h_\lambda-z_\lambda\mathbb{1}\bigr)A^{(k)}_\lambda=Q \qquad(\lambda\in\mathcal{D}_k),\] and obeys \(\|A^{(k)}_\lambda\|\lesssim \Lambda_k^{\sharp}\exp(Cb\Lambda_k)\) as well as \(\left\lVert\chi_S A_\lambda \right\rVert_{\mathcal{B}(QL^2(\mathbb{R}^2)\to L^2(\mathbb{R}^2))} \lesssim \exp\left(-c \Lambda_k \mathrm{dist}(S,B_{2a}(0))\right)\) (all bounds uniform in \(\xi\in\partial B_1(0)\)).

Fix \(k\) and consider \(\lambda\in\mathcal{D}_k\cap\mathcal{D}_{k+1}\) with \(\lambda>0\) real. Then by construction \(z_\lambda(\xi)\in\rho(h_\lambda)\) (the resolvent set) for all \(\xi\in\partial B_1(0)\); hence \(\bigl(h_\lambda-z_\lambda\mathbb{1}\bigr)^{-1}\) exists as a bounded operator on \(L^2(\mathbb{R}^2)\). Since both \(A^{(k)}_\lambda\) and \(A^{(k+1)}_\lambda\) are right inverses on \(QL^2\), \[\bigl(h_\lambda-z_\lambda\mathbb{1}\bigr)A^{(k)}_\lambda =\bigl(h_\lambda-z_\lambda\mathbb{1}\bigr)A^{(k+1)}_\lambda =Q,\] and the true inverse on \(L^2\) is unique, we must have \[A^{(k)}_\lambda=\bigl(h_\lambda-z_\lambda\mathbb{1}\bigr)^{-1}Q =A^{(k+1)}_\lambda \qquad(\lambda\in(\Lambda_{k+1},2\Lambda_k)\subset\mathbb{R}).\] Thus the two analytic \(\mathcal{B}(QL^2)\)-valued maps \(A^{(k)}\) and \(A^{(k+1)}\) agree on the real interval \((\Lambda_{k+1},2\Lambda_k)\), which has accumulation points in the overlap \(\mathcal{D}_k\cap\mathcal{D}_{k+1}\). By the identity theorem for Banach–valued holomorphic functions, they coincide on all of \(\mathcal{D}_k\cap\mathcal{D}_{k+1}\).

Define \(A_\lambda\) on \(\Omega_{\Lambda,\varepsilon}\) by choosing any \(k\) with \(\lambda\in\mathcal{D}_k\) and setting \(A_\lambda:=A^{(k)}_\lambda\). The overlap consistency shows this is well defined. The resulting map \[\Omega_{\Lambda,\varepsilon}\ni\lambda\longmapsto A_\lambda\in\mathcal{B}(QL^2(\mathbb{R}^2))\] is analytic, and for \(\lambda\in\mathcal{D}_k\) the bounds \[\|A_\lambda\|=\|A^{(k)}_\lambda\|\;\lesssim\Lambda_k^\sharp\;\exp\!\bigl(Cb\,\Lambda_k\bigr) \;\le\left|\lambda\right|^\sharp\;\exp\!\bigl(Cb\,|\lambda|\bigr)\] as well as \[\begin{align} \left\lVert\chi_S A_\lambda \right\rVert = \left\lVert\chi_S A^k_\lambda \right\rVert\lesssim \exp\left(-c \Lambda_k \mathrm{dist}(S,B_{2a}(0))\right)\leq \exp\left(-\tilde{c} \left|\lambda\right| \mathrm{dist}(S,B_{2a}(0))\right) \end{align}\] hold because \(|\lambda|\in(\Lambda_k,2\Lambda_k)\).

If one repeats the construction on \(\mathcal{D}_k\) with different auxiliary data \((\theta,\chi,\delta,\{\chi_\nu\},\{\psi_\nu\})\), the two resulting analytic right inverses agree for all real \(\lambda\in(\Lambda_k,2\Lambda_k)\) (both equal \((h_\lambda-z_\lambda\mathbb{1})^{-1}Q\)), hence agree on \(\mathcal{D}_k\) by the identity theorem. Therefore the global \(A_\lambda\) is canonical and independent of these choices.

The proof of 7 is now complete. ◻

Proof. We shall use the magnetic translation operators given in 5 , here factorized as \[\begin{align} \widehat{R}^z =\exp\left(-\operatorname{i}z\cdot P\right)\exp\left(-\operatorname{i}\frac{1}{2}b\lambda z\cdot X^\perp\right)\qquad(z\in\mathbb{R}^2)\,. \end{align}\] These operators are unitary if \(\lambda\in\mathbb{R}\) and otherwise generally not even bounded; in factorized form the two exponentials commute. \(\widehat{R}^z\) commutes with the magnetic kinetic energy \(H^{\rm Landau}_{b,\lambda}\) and obeys \[\begin{align} \label{eq:magnetic32translations32shift32potential} \widehat{R}^{\pm d} v(X) \widehat{R}^{\mp d} = v(X\mp d) \end{align}\tag{28}\] so we have the relation \[\begin{align} \label{eq:magnetic32translations32shift32Hamiltonian} \widehat{R}^{\pm d} \left(H^{\rm Landau}_{b,\lambda} + \lambda^2v(X)\right) \widehat{R}^{\mp d} = H^{\rm Landau}_{b,\lambda} + \lambda^2v(X\mp d)\,. \end{align}\tag{29}\]

Consequently, via the analytic extension of \(\left(h_\lambda-z_\lambda\mathbb{1}\right)^{-1}Q\), \(A_\lambda\), constructed above, we now have right inverse of the translates of the one-well Hamiltonians too: \[\begin{align} \left(H^{\rm Landau}_{b,\lambda} + \lambda^2v(X\mp d)-z_\lambda\mathbb{1}\right) \widehat{R}^{\pm d} A_\lambda \widehat{R}^{\mp d} = Q_{\pm d} \,. \end{align}\] It is clear that the map \[\begin{align} \Omega_{\Lambda,\varepsilon} \ni \lambda \mapsto A_\lambda^{\pm d}:=\widehat{R}^{\pm d} A_\lambda \widehat{R}^{\mp d} \in \mathcal{B}(Q_{\pm d}L^2(\mathbb{R}^2)\to L^2(\mathbb{R}^2)) \end{align}\] is also analytic, but if \(\lambda\) is not real, due to the magnetic translations not being unitary, we have a deteriorated bound. First note that in the magnetic translations, for complex \(\lambda\), \(\exp\left(-\operatorname{i}z\cdot P\right)\) is still unitary and commutes with the other term, so it becomes irrelevant in the operator norm. We thus only keep the factor \(\exp\left(-\operatorname{i}\frac{1}{2}b\lambda z \cdot X^\perp\right)\): \[\begin{align} \left\lVert A_\lambda^{\pm d}\right\rVert &= \left\lVert\exp\left(\mp\operatorname{i}\frac{1}{2}b\lambda d\cdot X^\perp\right)A_\lambda \exp\left(\pm\operatorname{i}\frac{1}{2}b\lambda d\cdot X^\perp\right)\right\rVert_{\mathcal{B}(Q_{\pm d}L^2(\mathbb{R}^2)\to L^2(\mathbb{R}^2))} \\ &\leq \left\lVert\exp\left(\mp\operatorname{i}\frac{1}{2}b\lambda d\cdot X^\perp\right) \left(\chi_{B_{3a}(0)}(X)+\chi_{B_{3a}(0)}(X)^\perp\right) A_\lambda \right\rVert \exp\left(\frac{b \left|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}\right|}{2} \left\lVert d\right\rVert a\right) \\ &\leq \exp\left(3\frac{b \left|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}\right|}{2} \left\lVert d\right\rVert a\right)\left\lVert A_\lambda \right\rVert \exp\left(\frac{b \left|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}\right|}{2} \left\lVert d\right\rVert a\right) + \\ &+ \left\lVert\exp\left(\mp\operatorname{i}\frac{1}{2}b\lambda d\cdot X^\perp\right) \chi_{B_{3a}(0)}(X)^\perp A_\lambda \right\rVert \exp\left(\frac{b \left|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}\right|}{2} \left\lVert d\right\rVert a\right)\,. \end{align}\] Now using ?? controls the first term. To deal with the second term we rewrite \[\begin{align} B_{3a}(0)^c = \bigsqcup_{n=0}^\infty \Set{x\in\mathbb{R}^2 | 3a + n \leq \left\lVert x\right\rVert < 3a + n + 1} \end{align}\] so that, invoking ?? , we get \[\begin{align} \left\lVert\exp\left(\mp\operatorname{i}\frac{1}{2}b\lambda d\cdot X^\perp\right) \chi_{B_{3a}(0)}(X)^\perp A_\lambda \right\rVert &\leq \sum_{n=0}^\infty \exp\left(\frac{b \left|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}\right|}{2} \left\lVert d\right\rVert\left(3a+n+1\right)-c \left|\lambda\right| \left(a+n\right)\right) \end{align}\] so that invoking again the assumption that \(b\) is sufficiently small, we find \[\begin{align} \left\lVert A_\lambda^{\pm d}\right\rVert \lesssim \left|\lambda\right|^{\sharp } \exp\left(C b \left|\lambda\right|\right) \end{align}\] indeed.

Using \(A_\lambda^{\pm d}\), we construct an approximate \(R_\lambda(z_\lambda)Q_{-d,d}\) as \[\begin{align} \widetilde{A_{\lambda}^{-d,d}} := A_\lambda^{d} Q_d + A_\lambda^{-d} Q_{-d}\,. \end{align}\]

Then \[\begin{align} \left(H_\lambda -z_\lambda\mathbb{1}\right)\widetilde{A_{\lambda}^{-d,d}} = Q_{-d,d}+\lambda^2 v(X+d) A_\lambda^{d} Q_d + \lambda^2 v(X-d) A_\lambda^{-d} Q_{-d}\,. \end{align}\] The correction terms \[\begin{align} \lambda^2 v(X+d) A_\lambda^{d} Q_d + \lambda^2 v(X-d) A_\lambda^{-d} Q_{-d} \end{align}\] can be made arbitrarily small thanks to ?? as \(v(X+d)\) is supported within \(Q_{-d}\). Hence the desired operator \[\begin{align} A_\lambda^{-d,d} := \widetilde{A_{\lambda}^{-d,d}}\left(Q_{-d,d}+\lambda^2 v(X+d) A_\lambda^{d} Q_d + \lambda^2 v(X-d) A_\lambda^{-d} Q_{-d}\right)^{-1} \end{align}\] exists and is analytic.

The bound on its derivatives is obtained from the \(\alpha>0\) case of ?? . ◻

Proof. For simplicity we ignore \(\chi\) in the definition of \(\Phi_{\pm d}\), since it is obvious how to correct for it. Using 46 below we see that for real \(\lambda\), \[\begin{align} \left\lVert\Pi_\lambda^\perp \widehat{R}^{\pm d}\varphi_\lambda^{\rm MHO}\right\rVert \lesssim \lambda^{-1/2}\,. \end{align}\] Hence \[\begin{align} \left\lVert\Pi_\lambda^\perp \Phi_{\pm d}\right\rVert \lesssim \lambda^{-1/2} + \left\lVert\Pi_\lambda^\perp Q_{\pm d}^\perp\widehat{R}^{\pm d}\varphi_\lambda^{\rm MHO}\right\rVert\lesssim \lambda^{-1/2}\,. \end{align}\]

We thus learn that \[\begin{align} \left|1-\left\lVert\psi_{\pm d}\right\rVert^2\right| \lesssim \lambda^{-1/2}\,. \end{align}\]

Moreover, \[\begin{align} \left|\langle \psi_{-d}, \psi_{d} \rangle\right| &= \left|\langle \Phi_{-d}, \Pi_\lambda \Phi_{d} \rangle\right| \\ &= \left|\langle Q_{-d} \widehat{R}^{- d}\varphi_\lambda^{\rm MHO}, \Pi_\lambda Q_{d} \widehat{R}^{ d}\varphi_\lambda^{\rm MHO} \rangle\right| \\ &= \left|\langle Q_{-d} \widehat{R}^{- d}\varphi_\lambda^{\rm MHO}, \Pi_\lambda^\perp Q_{d} \widehat{R}^{ d}\varphi_\lambda^{\rm MHO} \rangle\right| \\ &\lesssim \lambda^{-1/2}\,. \end{align}\] ◻

Proof. It is convenient to introduce, for \(x>0\), the shorthand: \[\begin{align} \mathfrak{m}\left(x\right) := \min\Set{x,\frac{1}{x}}= x\; \chi_{(0,1)}(x) + \frac{1}{x}\; \chi_{[1,\infty)}(x)\;. \end{align}\] Let us write \(a =: \alpha+\operatorname{i}\beta\) with \(\alpha>0\), since \(a\in\mathbb{H}\). Then \[\begin{align} \label{eq:the32equation32before32EE1}1-\left|B_a(t)\right|^2 = \frac{4\alpha t}{(t+\alpha)^2+\beta^2}\leq\frac{4\alpha t}{(t+\alpha )^2 }= 1-\left|B_\alpha(t)\right|^2 \end{align}\tag{35}\] from which ?? follows. To prove ?? , we first observe a scaling property of \(B_x(t)\):
\[\begin{align} \textrm{For x>0},\quad \left|B_x(t)\right| =\left|\frac{t-x}{t+x}\right| = \left|\frac{1-x/t}{1+x/t}\right|= \left|B_{\frac{x}{t}}(1)\right|.\label{eq:scaling32property32of32B} \end{align}\tag{36}\] Therefore, to prove the bound ?? it suffices to bound \(-\log\left(B_x\right)\) for \(x=\operatorname{\mathbb{R}\mathbb{e}}\left\{a\right\}/t\).

First let \(x\in(0,\frac{1}{2})\). Then, \(\left|B_x(1)\right| = \frac{1-x}{1+x}\). Since over this range of \(x\), \(\log(1+x)\le x\) and \(-\log(1-x)=\int_0^x(1-s)^{-1}\operatorname{d}{s}\le x(1-x)^{-1}=-1+ (1-x)^{-1}\), we have \[\begin{align} -\log\left(\left|B_x(1)\right|\right)=\log\left(1+x\right)-\log\left(1-x\right)\leq x -1+\frac{1}{1-x}=\frac{x(2-x)}{1-x}\leq 2x(2-x)\leq 4x\,. \end{align}\]

The case \(x>2\) follows at once from the case \(x<1/2\) since \(B_x(1) = B_{1/x}(1)\). ◻

Proof. First we note that \[\begin{align} \label{eq:mfun32decomp} \mathfrak{m}\left(t/\alpha\right) \equiv \min\Big\{\frac{t}{\alpha},\frac{\alpha}{t}\Big\} = \frac{1}{\alpha}\;t\; \chi_{(0,\alpha]}(t) + \alpha\;\frac{1}{t}\; \chi_{(\alpha,\infty)}(t)\;. \end{align}\tag{38}\] For a fixed \(\delta>0\), consider the three cases: (i) \(\alpha<\delta\), (ii) \(\alpha> 2\delta\) and (iii) \(\delta \le \alpha\le 2\delta\).

For the first case, when \(\alpha<\delta\le t\le 2\delta\), \(\mathfrak{m}\left(t/\alpha\right)=\alpha/t\) and we get \[\begin{align} \frac{1}{\delta}\int_{t=\delta}^{2\delta}\mathfrak{m}\left(t/\alpha\right)\operatorname{d}{t} &= \frac{\alpha}{\delta} \left[\log(2\delta)-\log(\delta)\right] = \log(2)\frac{\alpha}{\delta} {=} \log 2 \;\mathfrak{m}\left(\delta/\alpha\right). \end{align}\]

For the second case, when \(\delta\le t \le 2\delta<\alpha\), \(\mathfrak{m}\left(t/\alpha\right)=t/\alpha\) and we get \[\begin{align} \frac{1}{\delta}\int_{t=\delta}^{2\delta}\mathfrak{m}\left(t/\alpha\right)\operatorname{d}{t} &= \frac{1}{\delta\alpha} \frac{1}{2}\left((2\delta)^2-\delta^2\right) = \frac{3}{2}\frac{\delta}{\alpha} =\frac{3}{2}\;\mathfrak{m}\left(\delta/\alpha\right)\,. \end{align}\]

Finally, let \(\delta\le \alpha\le 2\delta\). Then, we get: \[\begin{align} \frac{1}{\delta}\int_{\delta}^{2\delta}\mathfrak{m}\left(t/\alpha\right)\operatorname{d}{t} &= \frac{1}{\delta}\int_{\delta}^{\alpha} \frac{t}{\alpha}\operatorname{d}{t} + \frac{1}{\delta}\int_{\alpha}^{2\delta} \frac{\alpha}{t} \operatorname{d}{t} \\ &= \frac{1}{2\delta\alpha}(\alpha^2-\delta^2)+\frac{\alpha}{\delta}\log\left(\frac{2\delta}{\alpha}\right)\\ &\leq 2-\frac{1}{2}\left(\frac{\alpha}{\delta}+\frac{\delta}{\alpha}\right)\\ &\leq \frac{3}{2}\frac{\delta}{\alpha}=\frac{3}{2}\mathfrak{m}\left(\delta/\alpha\right)\,. \end{align}\] The bounds over regions (i), (ii) and (iii) imply ?? . ◻

Proof. Using the elementary bound \[\begin{align} \label{eq:id32for32Poisson32kernel} \frac{t}{t^2+\psi^2}\leq \frac{1}{t} \frac{1}{1+\psi^2},\qquad\textrm{for 0<t<1,} \end{align}\tag{45}\] we obtain ?? as follows: For \(0<t<1\), \[\begin{align} u(t) &\equiv \int_{\psi\in\mathbb{R}}\frac{t}{t^2+\psi^2}\operatorname{d}{\mu(\psi)} \le \frac{1}{t}\int_{\psi\in\mathbb{R}}\frac{1}{1+\psi^2}\operatorname{d}{\mu(\psi)} \leq \frac{\beta}{t}\,. \end{align}\]

We next turn the proof of ?? . Let \(\eta>0\) be given. Then we decompose \[\begin{align} \mathbb{R}= \Set{0} \sqcup \left([-\eta,\eta]\setminus\Set{0}\right) \sqcup \left(\mathbb{R}\setminus[-\eta,\eta]\right)\,. \end{align}\] and, correspondingly, in the Poisson integral ?? defining \(u\). We obtain, for \(0<t<1\): \[\begin{align} u(t) \leq \frac{1}{t}\mu(\Set{0})+\frac{1}{t}\int_{0<|\psi|\le\eta}\frac{1}{1+\psi^2}\operatorname{d}{\mu(\psi)}+\frac{t(1+\eta^2)}{t^2+\eta^2}\beta\;. \end{align}\] For the second term we used 45 and for the third term we used \[\begin{align} \frac{t(1+\psi^2)}{t^2+\psi^2}\leq \frac{t(1+\eta^2)}{t^2+\eta^2}\Longleftrightarrow \left(1-t^2\right)\left(\eta^2-\psi^2\right)\leq 0,\quad \textrm{for \left|\psi\right|\geq\eta and 0<t<1}\,. \end{align}\] Next, given any \(\varepsilon>0\), we can pick \(\eta_\varepsilon>0\) sufficiently small so that \[\begin{align} \int_{0<|\psi|\le\eta_\varepsilon}\frac{1}{1+\psi^2}\operatorname{d}{\mu(\psi)}\leq \varepsilon/2\,. \end{align}\] Fixing \(\eta_\varepsilon\), we now choose \(t>0\) sufficiently small, so that \(\frac{t(1+\eta_\varepsilon^2)}{t^2+\eta_\varepsilon^2}\beta\leq\varepsilon/2\). Hence, for all \(t>0\) sufficiently small, \(0\leq u(t) - t^{-1}\mu(\Set{0})\;\leq\; \varepsilon t^{-1}\). The proof of 19 is now complete. ◻

Proof of 12. Recall that

1. By 13, \(F\) factors into a Blaschke product, \(B\), and a non-vanishing analytic function, \(\operatorname{e}^{-G}\), both analytic on \(\mathbb{H}\): \[\begin{align} F(z) = \operatorname{e}^{-G(z)}\prod_{\nu\ge 1}B_{a_\nu}(z) = \operatorname{e}^{-G(z)}B(z), \end{align}\] where \(\Set{a_\nu}_{\nu\in\mathbb{N}}\subseteq\mathbb{H}\) are the zeros of \(F\) in \(\mathbb{H}\) (with multiplicity listed).

2. The assumed lower bound ?? , \(|F(1)|\ge \operatorname{e}^{-\beta}\), implies the upper bounds 33 : \[\begin{align} -\log\left(\left|B_a(1)\right|\right) &= \sum_{\nu=1}^{\infty}-\log\left(\left|B_{a_\nu}(1)\right|\right)<\beta,\; \textrm{and} \tag{46} \\ 0< \operatorname{\mathbb{R}\mathbb{e}}\left\{G(1)\right\} &= G_R(1)<\beta \tag{47} \end{align}\]

We first show that 46 implies the summability ?? . Write \(a_\nu=:\alpha_\nu+\operatorname{i}\beta_\nu\). Using ?? and the bound \(-\log y^2\ge 1-y^2\) and 35 for \(t=1\), i.e. that \(1-|B_a(1)|^2= 4\alpha/[(1+\alpha)^2+\beta^2]\), \[\begin{align} \nonumber \beta &> \frac{1}{2}\sum_{\nu=1}^{\infty}-\log\left(\left|B_{a_\nu}(1)^2\right|\right) \ge \frac{1}{2}\sum_{\nu=1}^{\infty}\left(1-|B_{a_\nu}(1)|^2\right)\\ \label{eq:beta-sum} &\ge \frac{1}{2}\sum_{|a_\nu|\le 1}\left(1-|B_{a_\nu}(1)|^2\right)=\sum_{|a_\nu|\le 1} \frac{2\operatorname{\mathbb{R}\mathbb{e}}\left\{a_\nu\right\}}{(1+\operatorname{\mathbb{R}\mathbb{e}}\left\{a_\nu\right\})^2+\operatorname{\mathbb{I}\mathbb{m}}\left\{a_\nu\right\}^2}\ge \frac{1}{2}\;\sum_{|a_\nu|\le 1} \operatorname{\mathbb{R}\mathbb{e}}\left\{a_\nu\right\}\;. \nonumber\\ \end{align}\tag{48}\] Here, we used that \[\begin{align} \label{eq:z-ratio} \frac{2\operatorname{\mathbb{R}\mathbb{e}}\left\{z\right\}}{(1+\operatorname{\mathbb{R}\mathbb{e}}\left\{z\right\})^2+\operatorname{\mathbb{I}\mathbb{m}}\left\{z\right\}^2}\ge \frac{1}{2}\;\operatorname{\mathbb{R}\mathbb{e}}\left\{z\right\}, \quad\textrm{if |z|^2\le 1 and \operatorname{\mathbb{R}\mathbb{e}}\left\{z\right\}>0.} \end{align}\tag{49}\] This proves the first summability bound in ?? .

Next, using the first line of 48 and the scaling property \(|B_a(1)|=|B_{1/a}(1)|\), we have \[\begin{align} \label{eq:1-over-a-sum}\beta> \frac{1}{2}\sum_{\nu=1}^{\infty}-\log\left(\left|B_{a_\nu}(1)^2\right|\right) & \ge \frac{1}{2}\sum_{|a_\nu|\ge 1}\left(1-|B_{a_\nu}(1)|^2\right)\; =\;\frac{1}{2}\sum_{|1/a_\nu|\le 1}\left(1-|B_{1/a_\nu}(1)|^2\right)\;. \end{align}\tag{50}\] We may now apply the bound 48 , with \(a_\nu\) replaced by \(1/a_\nu\), to 50 and we obtain \[\frac{1}{2}\;\sum_{|a_\nu|\geq1} \operatorname{\mathbb{R}\mathbb{e}}\left\{\frac{1}{a_\nu}\right\} \le \beta.\] This proves the second summability bound in ?? .

Thus, we may apply the bounds: ?? , ?? , and ?? , ?? . Recall \(-\log|F|=G_R-\log|B|\). Using ?? and ?? we have, for \(\delta\in(0,1/4)\), \[\begin{align} \frac{1}{\delta}\int_{t=\delta}^{2\delta}-\log\left(\left|F(t)\right|\right)\operatorname{d}{t} &= \frac{1}{\delta}\int_{t=\delta}^{2\delta}G_R(t)\operatorname{d}{t} + \frac{1}{\delta}\int_{t=\delta}^{2\delta}-\log\left(\left|B(t)\right|\right)\operatorname{d}{t} \\ &\leq \frac{1}{\delta}\int_{t=\delta}^{2\delta}\left(\beta t+\frac{\beta}{t}\right)\operatorname{d}{t} + 18 \frac{\beta}{\delta} = (18+\log(2)+\frac{3}{2}\delta^2)\frac{\beta}{\delta}, \end{align}\] where the term \(\beta t\) comes from \(Ax\) in 43 . This is the first conclusion of 12. The second conclusion of 12 is obtained from ?? and ?? as follows: , \[\begin{align} \frac{1}{\delta}\int_{t=\delta}^{2\delta}\left|-\log\left(\left|F(t)\right|\right)-\frac{1}{t}\mu(\Set{0})\right|\operatorname{d}{t} &= \frac{1}{\delta}\int_{t=\delta}^{2\delta}\left|-\log\left(\left|B(t)\right|\right)+G_R(t)-\frac{1}{t}\mu(\Set{0})\right|\operatorname{d}{t} \\ &\leq \frac{1}{\delta}\int_{t=\delta}^{2\delta}-\log\left(\left|B(t)\right|\right)\operatorname{d}{t} + \frac{1}{\delta}\int_{t=\delta}^{2\delta}\left(G_R(t)-\frac{1}{t}\mu(\Set{0})\right)\operatorname{d}{t} \\ &= o\left(\frac{1}{\delta}\right),\quad \textrm{as \delta\to 0^+}. \end{align}\] This completes the proof of 12. ◻

Proof. Given \(F:\Gamma_\alpha\to\mathbb{D}\) analytic, introduce its composition with \(\mathbb{H}\ni z\mapsto z^{-\alpha}\in\Gamma_\alpha\): \[\begin{align} \widetilde{F} (z) := F(z^{-\alpha})\qquad(z\in\mathbb{H})\, , \end{align}\] yielding an analytic function \(\widetilde{F}:\mathbb{H}\to\mathbb{D}\) satisfying \(\left|\widetilde{F}(1)\right|= \left|F(1)\right|\geq \operatorname{e}^{-\beta}\). We may apply 12 to \(\widetilde{F}\) to obtain an upper bound for the average of \(-\log\left(\left|\widetilde{F}\right|\right)\) over small \(t-\) intervals in \(\mathbb{H}\): \[\begin{align} 20\frac{\beta}{\delta}\ge \frac{1}{\delta}\int_{t=\delta}^{2\delta}-\log\left(\left|\widetilde{F}(t)\right|\right)\operatorname{d}{t},\quad \textrm{for 0<\delta<1/4.} \end{align}\] We then convert the average of \(\widetilde{F}\) to one over appropriate large real intervals in \(\Gamma_\alpha\) via the change of variables \(s:=t^{-\alpha}\): \[\begin{align} \frac{1}{\delta}\int_{\delta}^{2\delta}-\log\left(\left|\widetilde{F}(t)\right|\right)\operatorname{d}{t} &= \frac{1}{\delta}\int_{\delta}^{2\delta}-\log\left(\left|F(t^{-\alpha})\right|\right)\operatorname{d}{t}\\ &= \frac{1}{\alpha\delta}\int_{(2\delta)^{-\alpha}}^{\delta^{-\alpha}}-\log\left(\left|F(s)\right|\right)s^{-1-1/\alpha}\operatorname{d}{s} \\ &\geq \frac{2^{\alpha+1}}{\alpha}\delta^\alpha\int_{(2\delta)^{-\alpha}}^{\delta^{-\alpha}}-\log\left(\left|F(s)\right|\right)\operatorname{d}{s}\,. \end{align}\] Hence, \[\begin{align} \frac{2}{\alpha} (2\delta)^\alpha\int_{(2\delta)^{-\alpha}}^{\delta^{-\alpha}}-\log\left(\left|F(s)\right|\right)\operatorname{d}{s} \leq 20\frac{\beta}{\delta}\, ,\quad \textrm{for 0<\delta<1/4}. \end{align}\]

Set \(R:=(2\delta)^{-\alpha}\) (\(R^{-1}=(2\delta)^\alpha\)) and correspondingly we take \(R>2^\alpha\). Then, \[\begin{align} \frac{1}{(2^\alpha-1)R}\int_{R}^{2^\alpha R}-\log\left(\left|F(t)\right|\right)\operatorname{d}{t} \leq 20 \frac{\alpha\;\beta}{2^{\alpha}-1}R^{1/\alpha}\, ,\quad R>2^\alpha. \end{align}\]

Since \(\alpha\leq 1\), we may cover \([R,2 R]\) by \(m:=\lceil \frac{1}{\alpha} \rceil\) segments, i.e., \([R,2R] = \bigcup_{j=0}^{m-1}[2^{\alpha j}R,2^{\alpha \left(j+1\right)}R]\) and so we immediately get the bound, valid for all \(R>2^\alpha\), \[\begin{align} \frac{1}{R}\int_{s=R}^{2 R}-\log\left(\left|F(t)\right|\right)\operatorname{d}{t} \leq 80\alpha\beta 2^{1/\alpha}R^{1/\alpha}\,. \end{align}\] This completes the proof of ?? .

Finally, let \(\widetilde{\mu}\) denote the positive measure associated with \(z\mapsto\widetilde{G}_R(z)\equiv \operatorname{\mathbb{R}\mathbb{e}}\left\{G(z^{-\alpha})\right\}\), which is non-negative and harmonic on \(\mathbb{H}\). Then, via the above change of variables, \(s=t^{-\alpha}\), we get \[\begin{align} \frac{1}{\delta}\int_{\delta}^{2\delta}\left|-\log\left(\left|\widetilde{F}(t)\right|\right)-\frac{1}{t}\widetilde{\mu}\left(\Set{0}\right)\right|\operatorname{d}{t} &= \frac{1}{\alpha\delta}\int_{(2\delta)^{-\alpha}}^{\delta^{-\alpha}}\left|-\log\left(\left|F(s)\right|\right)-s^{1/\alpha}\widetilde{\mu}\left(\Set{0}\right)\right|s^{-1-1/\alpha}\operatorname{d}{s} \\ &\geq \frac{2}{\alpha} (2\delta)^\alpha\int_{(2\delta)^{-\alpha}}^{\delta^{-\alpha}}\left|-\log\left(\left|F(s)\right|\right)-s^{1/\alpha}\widetilde{\mu}\left(\Set{0}\right)\right|\operatorname{d}{s} \end{align}\] and the proof of ?? is completed using the above relation between \(\delta\) and \(R\).

The proof of 20 is now complete. ◻

Proof. We apply 20 on the function \(\widetilde{F}(z) := F(z) / U(z)\) which clearly obeys \(\left|\widetilde{F}(z)\right|\leq 1\) for all \(z\in\Gamma_\alpha\). The lower bound becomes \[\begin{align} \left|\widetilde{F}(1)\right|\geq \operatorname{e}^{-\left(\beta+\log\left(\left|U(1)\right|\right)\right)}\,. \end{align}\] Then \[\begin{align} 80 \alpha 2^{1/\alpha}\left(\beta+\log\left(\left|U(1)\right|\right)\right) R^{1/\alpha} &\geq \frac{1}{R}\int_{t=R}^{2 R}-\log\left(\left|\widetilde{F}(t)\right|\right)\operatorname{d}{t}\\ &= \frac{1}{R}\int_{t=R}^{2 R}-\log\left(\left|F(t)\right|\right)\operatorname{d}{t}-\frac{1}{R}\int_{t=R}^{2 R}-\log\left(\left|U(t)\right|\right)\operatorname{d}{t} \,. \end{align}\] ◻

Proof. Let \(\chi\in C^\infty_c(U\to[0,1])\) with \(\chi=1\) on \(V\).

We then apply [19] on the symbol \(\mathbb{R}^n\times\mathbb{R}^n\ni (x,p)\mapsto a(x,p)\chi(x)\). ◻

Proof. Fix \(\alpha\in\mathbb{N}^n\) with \(|\alpha|\le m\). By iterating the commutation rule \[P_j \operatorname{Op}\left(a\right) = \operatorname{Op}\left(p_j a(x,p)\right) - \operatorname{i}\,\operatorname{Op}\left((\partial_{x_j}a)(x,p)\right) }\] we obtain an expansion \[\label{eq:Palpha-expansion} P^\alpha \operatorname{Op}\left(a\right) \;=\; \operatorname{Op}\left(p^\alpha a(x,p)\right) \;+\; \sum_{0\neq\gamma\le \alpha} c_{\alpha,\gamma}\,\operatorname{Op}\left(p^{\alpha-\gamma}\,\partial_x^\gamma a(x,p)\right),\tag{58}\] for suitable combinatorial constants \(c_{\alpha,\gamma}\in\mathbb{C}\).

Set \[b_0(x,p):=p^\alpha a(x,p),\qquad b_\gamma(x,p):=p^{\alpha-\gamma}\,\partial_x^\gamma a(x,p)\quad(0\neq\gamma\le\alpha).\] Since \(a\in S^{-m}_M(U)\), we have \[b_0\in S^{|\alpha|-m}_M(U),\qquad b_\gamma\in S^{|\alpha|-|\gamma|-m}_M(U)\subset S^{-1}_M(U)\quad(0\neq\gamma\le\alpha),\] and because \(|\alpha|\le m\) these orders are \(\le 0\). In particular, \[b_0,\;b_\gamma\in S^0_M(U)\qquad(\gamma\le\alpha).\]

Define the zero-order symbol \[\tilde{a}_\alpha(x,p)\;:=\;M^{m-|\alpha|}\,p^\alpha a(x,p)\in S^0_M(U).\] Then \(b_0 = M^{|\alpha|-m}\tilde{a}_\alpha\), and the symbol estimates in 22 (Leibniz rule and the bounds for multiplication by \(p\) and differentiation in \(x\)) imply that, for all multi-indices \(|\tilde{\alpha}|,|\beta|\le 2N\), \[C^{\,0}_{M,\tilde{\alpha}\beta}(b_\gamma) \;\le\; C_{\alpha,m,N}\,M^{|\alpha|-m} \sum_{|\tilde{\alpha}'|,|\beta'|\le 2N} C^{\,0}_{M,\tilde{\alpha}'\beta'}(\tilde{a}_\alpha), \qquad 0\le\gamma\le\alpha,\] for some constant \(C_{\alpha,m,N}\) depending only on \(\alpha,m,N,n\) (but not on \(a\)).

Apply 23 to each term in 58 . Using that \(V\subseteq U\) and the triangle inequality, we get \[\begin{align} \left\lVert P^\alpha \operatorname{Op}\left(a\right)\right\rVert_{\mathcal{B}(L^2(\mathbb{R}^n)\to L^2(V))} &\le \sum_{\gamma\le\alpha} |c_{\alpha,\gamma}|\, \left\lVert\operatorname{Op}\left(b_\gamma\right)\right\rVert_{\mathcal{B}(L^2(\mathbb{R}^n)\to L^2(V))}\\[1mm] &\le C_{n,N}(U,V)\! \sum_{\gamma\le\alpha} |c_{\alpha,\gamma}| \sum_{|\tilde{\alpha}|,|\beta|\le 2N} C^{\,0}_{M,\tilde{\alpha}\beta}(b_\gamma)\\[1mm] &\le C_{n,N}(U,V)\,M^{|\alpha|-m} \sum_{|\tilde{\alpha}|,|\beta|\le 2N} C^{\,0}_{M,\tilde{\alpha}\beta}(\tilde{a}_\alpha), \end{align}\] after enlarging \(C_{n,N}(U,V)\) by a factor depending only on \(\alpha,m,N,n\) (which we absorb into the notation).

Recalling that \(\tilde{a}_\alpha(x,p)=M^{m-|\alpha|}p^\alpha a(x,p)\) gives the stated estimate. ◻

Proof. Let \(l\) be the symbol associated to \(\mathcal{L}\equiv \operatorname{Op}\left(l\right)\). Define the symbol \[\begin{align} a(x,p) := \frac{\chi(x)}{l(x,p)}\qquad(x\in U,p\in\mathbb{R}^n) \end{align}\] and the associated operator \(A:=\operatorname{Op}\left(a\right)\). First, we note \(a\in S^{-2}_M(\mathbb{R}^n)\) thanks to the ellipticity of \(\mathcal{L}\). Moreover, 56 implies that for any \(f\in L^2(\mathbb{R}^n)\), \(Af\) will be supported within \(V\) since \(x\mapsto a(x,p)\) is supported there, by its construction.

Next, let \(\tilde{\chi}\in C_0^\infty(U)\) satisfy \(\tilde{\chi} =1\) on \(V\). Clearly \(\tilde{\chi}\in S^0_M(U)\), and so, thanks to ?? , \(\tilde{\chi}l\) is in \(S^0_M(U)\). Applying the decomposition 57 on the product of symbols \(\tilde{\chi}l\) and \(a\) we find there must exist some symbol \(\varepsilon\in S^{-1}_M(U)\) which obeys \[\begin{align} \tilde{\chi}(X) \mathcal{L}A = \operatorname{Op}\left(\tilde{\chi} l a\right)+\operatorname{Op}\left(\varepsilon\right) \end{align}\] with \(\varepsilon\) the associated symbol, which must be in \(S^{-1}_M\) as the first term is in the top degree. By construction, \(\tilde{\chi} l a = \tilde{\chi}\chi=\chi\), so we get \[\begin{align} \label{eq:definition32of32remainder32term} \tilde{\chi}(X) \mathcal{L}A = \chi(X) +\operatorname{Op}\left(\varepsilon\right)\,. \end{align}\tag{59}\] Let \(\mathcal{E}:= \operatorname{Op}\left(\varepsilon\right)\).

Since \(\mathcal{L}\) acts by differentiation and multiplication with smooth coefficient functions, its application does not change the support of a function it acts on. This implies that \(\operatorname{supp}(\mathcal{L}Af)\subseteq V\) for any \(f\in L^2(\mathbb{R}^n)\) and \(\tilde{\chi} \mathcal{L}A f = \mathcal{L}A f\). Then 59 implies that \(\mathcal{E}f\) is also supported within \(V\).

Applying 24 on \(A\) and \(\mathcal{E}\) implies their co-domains are \(H^2\) and \(H^1\) respectively, with the stated estimates.

Next, the definition of \(A,\mathcal{E}\) via the symbols, implies that if \(l\) depends analytically on a parameter \(\lambda\), then so do the symbols \(a,\varepsilon\) and hence so do the operators \(A,\mathcal{E}\). ◻

Proof of Lemma 27. Choose a nonincreasing \(\vartheta\in C^\infty([0,\infty)\to[0,1])\) such that \[\vartheta(s)= \begin{cases} 1,& 0\le s\le 1,\\ 0,& s\ge 2, \end{cases} \qquad\text{and}\qquad \vartheta \text{ decreases on } (1,2).\] We also pick smooth companions \(\sigma_0,\sigma,\sigma_\infty\in C^\infty([0,\infty)\to[0,1])\) with \[\sigma_0\equiv 1 \text{ on }[0,2],\;\;\operatorname{supp}\sigma_0\subset[0,3];\qquad \sigma\equiv 1 \text{ on }[1/2,2],\;\;\operatorname{supp}\sigma\subset[1/3,3];\] and \[\begin{align} \sigma_\infty\equiv 0 \text{ on }[0,1/2],\;\;\sigma_\infty\equiv 1 \text{ on }[1,\infty). \end{align}\]

With \(r:=\left\lVert x\right\rVert\), set \[\chi_0(x):=\vartheta\!\left(\frac{r}{\delta}\right),\qquad \chi_\nu(x):=\vartheta\!\left(\frac{r}{2^\nu\delta}\right)-\vartheta\!\left(\frac{r}{2^{\nu-1}\delta}\right)\quad(1\le \nu\le N-1),\] \[\chi_N(x):=1-\vartheta\!\left(\frac{r}{2^N\delta}\right).\] Then for any \(x\) (writing \(t:=r/\delta\)) we have the telescoping identity \[\sum_{\nu=0}^{N}\chi_\nu =\vartheta(t) + \sum_{\nu=1}^{N-1}\big(\vartheta(2^{-\nu}t)-\vartheta(2^{-(\nu-1)}t)\big) +\big(1-\vartheta(2^{-N}t)\big)=1,\] so \(\{\chi_\nu\}\) is a partition of unity with \(0\le \chi_\nu\le 1\).

• For \(\nu=0\), \(\chi_0\equiv 1\) on \(B_\delta(0)\) and \(\operatorname{supp}\chi_0\subset B_{2\delta}(0)\). If one wishes the radii to be \(C\delta\) and \(2C\delta\) (with \(C\gg1\) fixed), simply replace \(\chi_0(x)\) by \(\vartheta(r/(C\delta))\).

• For \(1\le \nu\le N-1\), since \(\chi_\nu(x)=\vartheta(s)-\vartheta(2s)\) with \(s:=r/(2^\nu\delta)\), we have \(\operatorname{supp}\chi_\nu\subset\{\,s\in(1/2,2)\,\}\), i.e. \[\operatorname{supp}\chi_\nu\subset\big\{\,2^{\nu-1}\delta<\left\lVert x\right\rVert<2^{\nu+1}\delta\,\big\}.\]

• For \(\nu=N\), \(\chi_N\equiv 0\) on \(B_{2^N\delta}(0)\) and hence \(\operatorname{supp}\chi_N\subset\{\,\left\lVert x\right\rVert>2^N\delta\,\}\). Since \(2^N\delta\ge R\), this is contained in the complement of the disc of radius \(R\).

By the chain rule and scaling, for any multi-index \(\alpha\) there exists \(C_\alpha<\infty\) such that \[\|\partial^\alpha \vartheta(r/a)\|_{L^\infty(\mathbb{R}^2)}\le C_\alpha\,a^{-|\alpha|}.\] Therefore, for \(1\le \nu\le N-1\), \[\|\partial^\alpha \chi_\nu\|_\infty \le \|\partial^\alpha \vartheta(r/\delta_\nu)\|_\infty + \|\partial^\alpha \vartheta(r/\delta_{\nu-1})\|_\infty \le C_\alpha\,\delta_\nu^{-|\alpha|},\] and the same bound holds for \(\nu=0,N\).

Define \[\psi_0(x):=\sigma_0\!\left(\frac{r}{\delta}\right),\qquad \psi_\nu(x):=\sigma\!\left(\frac{r}{2^\nu\delta}\right)\;\;(1\le \nu\le N-1),\qquad \psi_N(x):=\sigma_\infty\!\left(\frac{r}{2^N\delta}\right).\] By construction, \(\psi_\nu\equiv 1\) on \(\operatorname{supp}\chi_\nu\) for every \(\nu\).

To bound the overlap, fix \(x\) with \(t:=r/\delta>0\). For \(1\le\nu\le N-1\), \[x\in\operatorname{supp}\psi_\nu \iff 2^{-\nu}t\in[1/3,3] \iff \nu\in\big[\log_2 t-\log_2 3,\;\log_2 t+\log_2 3\big],\] an interval of length \(2\log_2 3<3.2\), hence containing at most \(4\) integers. Moreover, \(\psi_0\) can only be nonzero when \(r\lesssim \delta\) and \(\psi_N\) only when \(r\gtrsim 2^N\delta\); in those regimes the other indices contribute none or at most a couple more, so at any \(x\) there are no more than four indices \(\nu\) with \(x\in\operatorname{supp}\psi_\nu\).

For \(1\le \nu\le N-1\) set \[U_\nu:=\big\{\,2^{\nu-3}\delta<\left\lVert x\right\rVert<2^{\nu+3}\delta\,\big\},\] which are open and satisfy \(\operatorname{supp}\psi_\nu\subset U_\nu\) by the choice of supports of \(\sigma\). For the edge indices, take \[U_0:=\{\,\left\lVert x\right\rVert<2^3\delta\,\},\qquad U_N:=\{\,\left\lVert x\right\rVert>2^{N-3}\delta\,\}.\] These choices match the inclusion required in the statement. ◻

Proof. Set \(A := 1/2-z_\lambda/ (2\lambda b)\in\mathbb{C}\) with \(\operatorname{\mathbb{R}\mathbb{e}}\left\{A\right\}>0\) and \(w := \frac{1}{2}\lambda b\left\lVert x-y\right\rVert^2\) with \(\operatorname{\mathbb{R}\mathbb{e}}\left\{w\right\}>0\) (assuming \(x\neq y\) for the moment). Then \[\begin{align} \left|\Gamma(A) U(A,1,w)\right| &= \left|\int_{t=0}^{\infty}\exp\left(-t w\right)t^{A-1}(1+t)^{-A}\operatorname{d}{t}\right| \\ &\leq \int_{t=0}^{\infty}\exp\left(-t \operatorname{\mathbb{R}\mathbb{e}}\left\{w\right\}\right)t^{\operatorname{\mathbb{R}\mathbb{e}}\left\{A\right\}-1}(1+t)^{-\operatorname{\mathbb{R}\mathbb{e}}\left\{A\right\}}\operatorname{d}{t} \\ &\leq \int_{t=0}^{1}t^{\operatorname{\mathbb{R}\mathbb{e}}\left\{A\right\}-1}\operatorname{d}{t} + \int_{t=1}^{\infty}\exp\left(-t \operatorname{\mathbb{R}\mathbb{e}}\left\{w\right\}\right)t^{-1}\operatorname{d}{t} \\ &\leq \frac{1}{\operatorname{\mathbb{R}\mathbb{e}}\left\{A\right\}} + \int_{t=1}^{1/\operatorname{\mathbb{R}\mathbb{e}}\left\{w\right\}}t^{-1}\operatorname{d}{t} + \int_{t=1/\operatorname{\mathbb{R}\mathbb{e}}\left\{w\right\}}^{\infty}\exp\left(-t \operatorname{\mathbb{R}\mathbb{e}}\left\{w\right\}\right)t^{-1}\operatorname{d}{t} \\ &\leq \frac{1}{\operatorname{\mathbb{R}\mathbb{e}}\left\{A\right\}} + \left|\log\left(\operatorname{\mathbb{R}\mathbb{e}}\left\{w\right\}\right)\right| + 1\\ &\leq \left(1+\frac{1}{\operatorname{\mathbb{R}\mathbb{e}}\left\{A\right\}}\right)\left(1+\left|\log\left(\operatorname{\mathbb{R}\mathbb{e}}\left\{w\right\}\right)\right|\right)\,. \end{align}\]

Moreover, we also have \[\begin{align} \left|\exp\left(-\operatorname{i}\frac{\lambda b}{2}x\wedge y \right)\right| = \exp\left(\frac{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\} b}{2}x\wedge y\right)\,. \end{align}\] Now if \(y\in K\) then, defining \[D:=\;D_K\;=\;\sup_{y\in K}\left\lVert y\right\rVert,\] and using that \(\left|x\wedge y\right| = \left|\left(x-y\right)\wedge y\right| \leq D \left\lVert x-y\right\rVert\) we find that the whole kernel is upper bounded by \[\begin{align} \left|\left(H^{\mathrm{Landau}}_{\lambda b}-z_\lambda\mathbb{1}\right)^{-1}(x,y)\right| &\leq \frac{1+\frac{1}{\operatorname{\mathbb{R}\mathbb{e}}\left\{A\right\}}}{4\pi}\left(1+\left|\log\left(\frac{1}{2}\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\} b\left\lVert x-y\right\rVert^2\right)\right|\right) \times \\ &\qquad \times \exp\left(-\frac{1}{4} b \operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\left\lVert x-y\right\rVert^2 + \frac{1}{2} b \left|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}\right|D\left\lVert x-y\right\rVert\right) \,. \end{align}\]

At this point it is convenient to "forget" the fact that \(y\) only ranges in \(K\) (as an upper bound) to simplify the invocation of Schur’s test \[\begin{align} \left\lVert G\right\rVert^2 \leq \left(\sup_x \int_{y} \left|G(x,y)\right|\operatorname{d}{y}\right)\left(\sup_y \int_{x}\left|G(x,y)\right|\operatorname{d}{x}\right) \end{align}\] to \[\begin{align} \left\lVert R^{\rm Landau}_{\lambda b}(z_\lambda)Q\right\rVert_{\mathcal{B}(QL^2(\mathbb{R}^2)\to L^2(\mathbb{R}^2))} &\leq \frac{1+\frac{1}{\operatorname{\mathbb{R}\mathbb{e}}\left\{A\right\}}}{4\pi}\int_{x\in\mathbb{R}^2}\operatorname{d}{x}\left(1+\left|\log\left(\frac{1}{2}\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\} b\left\lVert x\right\rVert^2\right)\right|\right) \times \\ &\qquad \times \exp\left(-\frac{1}{4} b \operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\left\lVert x\right\rVert^2 + \frac{1}{2} b \left|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}\right|D\left\lVert x\right\rVert\right)\,. \end{align}\] Now using \[\begin{align} \int_{r=0}^\infty r\left(1+\left|\log\left(\alpha r^2\right)\right|\right)\exp\left(-\frac{1}{2}\alpha r^2 + \beta r \right)\operatorname{d}{r} &\leq \int_{r=0}^\infty r\left(1+2\sqrt{\alpha} r + \frac{2}{\sqrt{\alpha} r}\right)\exp\left(-\frac{1}{2}\alpha r^2 + \beta r \right) \operatorname{d}{r} \\ &= \frac{\beta +1}{\alpha }+\frac{\sqrt{2\pi} \left(\alpha +\beta ^2+\beta +1\right) \operatorname{e}^{\frac{\beta ^2}{2 \alpha } } }{\alpha ^{3/2}} \end{align}\] we find the estimate ?? .

To get the rate of off-diagonal decay, we return to the integral expression for \(\left|\Gamma(A) U(A,1,w)\right|\). Now \(\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\) is a large parameter: \(A = \frac{1}{2}-\frac{-\nu\lambda^2+\mu\lambda}{2\lambda b} = \frac{\nu\lambda+b-\mu}{2b}\) so \(2b\operatorname{\mathbb{R}\mathbb{e}}\left\{A\right\} = \nu\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}+b-\operatorname{\mathbb{R}\mathbb{e}}\left\{\mu\right\}\) and then, with \(\gamma:=\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\), \(r := \left\lVert x-y\right\rVert\), \(w := \frac{1}{2}\lambda b\left\lVert x-y\right\rVert^2\), we get, \[\begin{align} \left|\Gamma(A) U(A,1,w)\right| &\leq \int_{t=0}^{\infty}\frac{\left(1+\frac{1}{t}\right)^{-\frac{1}{2}+\frac{\operatorname{\mathbb{R}\mathbb{e}}\left\{\mu\right\}}{2b}}}{t}\exp\left(-\gamma\left(\frac{1}{2} b t r^2+\frac{\nu}{2b}\log\left(1+\frac{1}{t}\right)\right)\right)\operatorname{d}{t}\\ &\leq \int_{t=0}^{\infty}\frac{\left(1+\frac{1}{t}\right)^{-\frac{1}{2}+\frac{\operatorname{\mathbb{R}\mathbb{e}}\left\{\mu\right\}}{2b}}}{t}\exp\left(-\left(\frac{1}{2} b t r^2+\frac{\nu}{2b}\log\left(1+\frac{1}{t}\right)\right)\right)\operatorname{d}{t}\times \\ &\times \exp\left(-\left(\gamma-1\right) \left(\frac{1}{4} r \sqrt{b^2 r^2+4\nu}+\frac{\nu\log \left(\frac{\frac{1}{2} b r \left(\sqrt{b^2 r^2+4\nu}+b r\right)}{2\nu}+1\right)}{2 b}-\frac{b r^2}{4}\right)\right)\,. \end{align}\] Here we have calculated the minimal value of the function (of \(t\)) in the exponential and pulled it out of the integral, akin to a Laplace asymptotic calculation. Now since the logarithm term within the exponential is positive and \(\sqrt{b^2 r^2 + 4\nu}\geq2\sqrt{\nu}\) we see that \[\begin{align} \left|\left(H^{\mathrm{Landau}}_{\lambda b}-z_\lambda\mathbb{1}\right)^{-1}(x,y)\right| &\leq \left(1+\frac{2b}{\nu+b-\operatorname{\mathbb{R}\mathbb{e}}\left\{\mu\right\}}\right)\left(1+\left|\log\left(\frac{1}{2} b \left\lVert x-y\right\rVert^2\right)\right|\right) \times \\ &\times \exp\left(-\frac{\sqrt{\nu}}{2}\left(\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}-1\right) \left\lVert x-y\right\rVert +\frac{1}{2} b \left|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}\right| D\left\lVert x-y\right\rVert \right) \end{align}\] but now \(\left\lVert x-y\right\rVert\geq \mathrm{dist}(S,K)\) so Schur’s test again yields the claimed bound.

For the case \(\alpha\neq0\), we use \[\begin{align} \left(P_j^\alpha R^{\rm Landau}_{\lambda b}(z_\lambda)\right)(x,y) &= -\operatorname{i}\partial_{x_j} R^{\rm Landau}_{\lambda b}(z_\lambda)(x,y) \\ &= -\operatorname{i}\partial_{x_j} K_{\lambda b,z_\lambda}\left(\left\lVert x-y\right\rVert\right)\exp\left(-\operatorname{i}\frac{\lambda b}{2}x\wedge y \right) \\ &= -\operatorname{i}\left(\frac{x_j-y_j}{\left\lVert x_j-y_j\right\rVert}K'_{\lambda b, z_\lambda}(\left\lVert x-y\right\rVert)-\operatorname{i}\frac{\lambda b}{2}\varepsilon_{ji}y_jK_{\lambda b,z_\lambda}\left(\left\lVert x-y\right\rVert\right)\right)\exp\left(-\operatorname{i}\frac{\lambda b}{2}x\wedge y \right)\,. \end{align}\] The second term is dealt with in the same manner as before, so we only study \(K'\): \[\begin{align} \label{eq:derivative32of32Landau32positive32kernel} K'_{B, z}(r) &= -\frac{B r}{4\pi}\Gamma\left(\frac{1}{2}-\frac{z}{2B}\right)\operatorname{e}^{-\frac{1}{4} B r^2}\left(\left(\frac{1}{2}-\frac{z}{2B}\right)U\left(\frac{3}{2}-\frac{z}{2B},2,\frac{1}{2} B r^2\right) + \frac{1}{2} U\left(\frac{1}{2}-\frac{z}{2B},1,\frac{B}{2}r^2\right) \right)\,. \end{align}\tag{65}\] The second term is identical to what was studied above, hence we only need to study the term involving \(U\left(\frac{3}{2}-\frac{z}{2B},2,\frac{1}{2} B r^2\right)\): \[\begin{align} \Gamma\left(A\right)U\left(1+A,2,w\right) = \frac{1}{A}\int_{t=0}^\infty\operatorname{e}^{-w t}t^{A}\left(1+t\right)^{-A}\operatorname{d}{t} \end{align}\] and we see that this leads to the same asymptotics of the integral as above, for the large parameter \(\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\). ◻

Proof. We shall invoke the Weyl criterion for the spectrum, i.e., we shall show that for any \(\delta>0\) there exists some \(\psi_\delta\in L^2(\mathbb{R}^2)\) such that \[\begin{align} \left\lVert\left(H^{\rm Landau}_{b \lambda}+\lambda^2 v(X)-z_\lambda\mathbb{1}\right)\psi_\delta\right\rVert\leq\delta\left\lVert\psi_\delta\right\rVert\,. \end{align}\]

The plan will be to choose define a smooth bump function, \(\widetilde{\psi_\delta}\), supported at a large distance from the support of \(v(x)\), in such a way that the magnitude of the phase appearing in the Landau resolvent kernel, \(\exp\left(\frac{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\} b}{2}x\wedge y\right)\) can be made arbitrarily large on the support of \(\widetilde{\psi_\delta}\). Then, \(\psi_\delta := R_{\lambda b}^{\rm Landau}(z_\lambda)\widetilde{\psi_\delta}\), which by 28 is well-defined for complex \(\lambda\) since \(\widetilde{\psi_\delta}\) is supported on a compact set, will be large.

On the other hand, \[\begin{align} \left(H^{\rm Landau}_{b \lambda}+\lambda^2 v(X)-z_\lambda\mathbb{1}\right)\psi_\delta = \widetilde{\psi_\delta} + \lambda^2 v(X) R_{\lambda b}^{\rm Landau}(z_\lambda)\widetilde{\psi_\delta} \end{align}\] will be bounded, thanks to the off-diagonal decay bound on \(R_{\lambda b}^{\rm Landau}(z_\lambda)\), ?? .

Let \(\xi\in\mathbb{R}^2\) be chosen below with \(\left\lVert\xi\right\rVert\gg 1\). Let \(\widetilde{\psi_\delta}\geq0\) be a smooth bump function supported within \(K:=B_{\beta \left|\lambda\right|^{-1}\left\lVert\xi\right\rVert^{-1}}(\xi)\) for some \(\beta>0\) and obeying \(\int_{\mathbb{R}^2}\widetilde{\psi_\delta} = 1\) so that \(\left\lVert\widetilde{\psi_\delta}\right\rVert_2\lesssim \beta^{-1}\left|\lambda\right|\left\lVert\xi\right\rVert\). We find, by the off-diagonal bound ?? , for sufficiently large \(\left|\lambda\right|\). : \[\begin{align} \left\lVert\left(H^{\rm Landau}_{b \lambda}+\lambda^2 v(X)-z\mathbb{1}\right)\psi_\delta\right\rVert &\leq C \beta^{-1}\left|\lambda\right|\left\lVert\xi\right\rVert\;. \end{align}\]

To obtain a lower bound on \(\left\lVert\psi_\delta\right\rVert\), we shall require \(U\subseteq\mathbb{R}^2\) to be a subset on which the Landau phase factor is large, i.e., for some \(c>0\) \[\begin{align} \label{eq:Landau32phase32is32large}\left|\exp\left(-\operatorname{i}\frac{\lambda b}{2}x\wedge \xi\right)\right| \geq \exp\left(\frac{1}{2}c \left|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}\right|\left\lVert\xi\right\rVert\right). \end{align}\tag{66}\] Then \[\begin{align} \left\lVert\psi_\delta\right\rVert^2 & = \int_{x\in \mathbb{R}^2}\left|\psi_\delta(x)\right|^2\operatorname{d}{x} \\ &\geq \int_{x\in U}\left|\left(R_{\lambda b}^{\rm Landau}(z_\lambda)\widetilde{\psi_\delta}\right)(x)\right|^2\operatorname{d}{x} \\ &\geq \exp\left(c \left|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}\right|\left\lVert\xi\right\rVert\right) \int_{x\in U}\left|\exp\left(\operatorname{i}\frac{\lambda b}{2}x\wedge\xi\right)\left(R_{\lambda b}^{\rm Landau}(z_\lambda)\widetilde{\psi_\delta}\right)(x)\right|^2\operatorname{d}{x}\,. \end{align}\] We claim we can choose the set \(U\) such that both 66 holds and the latter integral over \(U\) satisfies a lower bound which is independent of \(\xi\). We make the following choice: \[\begin{align} U := \Set{x \in \mathbb{R}^2 | C < \left\lVert x-\xi\right\rVert < C' } \cap \Set{x \in \mathbb{R}^2 | \operatorname{sgn}(\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\})\;( x\wedge \xi) \geq c \left\lVert\xi\right\rVert} \end{align}\] The set \(U\) has an area, which is strictly positive and uniformly bounded away from zero, independently of \(|\lambda|\) and \(\xi\in\mathbb{R}^2\).

We then rewrite, via 63 and abusing \(K(x-y)\equiv K(\left\lVert x-y\right\rVert)\), \[\begin{align} \exp\left(\operatorname{i}\frac{\lambda b}{2}x\wedge\xi\right)\left(R_{\lambda b}^{\rm Landau}(z_\lambda)\widetilde{\psi_\delta}\right)(x) &= \int_{y\in K} K_{B,z}(x-y)\operatorname{e}^{-\operatorname{i}\frac{B}{2}x\wedge \left(y-\xi\right) }\widetilde{\psi_\delta}(y)\operatorname{d}{y} \\ &= \int_{y\in K} K_{B,z}(x-\xi)\widetilde{\psi_\delta}(y)\operatorname{d}{y} +\\ &+\int_{y\in K} \left(K_{B,z}(x-y)\operatorname{e}^{-\operatorname{i}\frac{B}{2}x\wedge \left(y-\xi\right) }-K_{B,z}(x-\xi)\right)\widetilde{\psi_\delta}(y)\operatorname{d}{y} \\ &= K_{B,z}(x-\xi)+\\&+\int_{y\in K} \left(K_{B,z}(x-y)\operatorname{e}^{-\operatorname{i}\frac{B}{2}x\wedge \left(y-\xi\right) }-K_{B,z}(x-\xi)\right)\widetilde{\psi_\delta}(y)\operatorname{d}{y}\,. \end{align}\]

Now we may establish, via 65 and the lines below it, that \[\begin{align} \left|K_{B,z}(\left\lVert x\right\rVert)-K_{B,z}(\left\lVert y\right\rVert)\right| \leq C \left|B\right|\left\lVert x\right\rVert\left\lVert x-y\right\rVert\left|K_{B,z}(\left\lVert x\right\rVert)\right| \end{align}\] for all \(\left\lVert x-y\right\rVert\leq c \left|B\right|^{-1}\left\lVert x\right\rVert^{-1}\). Using these estimates one may prove \[\begin{align} \left|\left(K_{B,z}(\left\lVert x-y\right\rVert)\operatorname{e}^{-\operatorname{i}\frac{B}{2}x\wedge \left(y-\xi\right) }-K_{B,z}(\left\lVert x-\xi\right\rVert)\right)\right| &\leq C \beta \left|K_{B,z}(\left\lVert x-\xi\right\rVert)\right| \end{align}\] and hence we have the desired bound \[\begin{align} \left\lVert\psi_\delta\right\rVert^2 \gtrsim \exp\left(c \left|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}\right|\left\lVert\xi\right\rVert\right) C_\lambda \end{align}\] for some constant \(C_\lambda\) independent of \(\xi\). By choosing \(\xi\) sufficiently large we get our result. ◻

Proof. \[\begin{align} &\bigg|\frac{1}{\lambda}\psi_0(x,\lambda)\psi_0^*(y,\lambda^*)\bigg| = \sqrt{\frac{\eta\zeta}{\pi^2}} \Bigg|\exp\left( -\frac{\lambda}{2}(\eta x_1^2+\zeta x_2^2-i\xi x_1x_2)-\frac{\lambda}{2}(\eta y_1^2+\zeta y_2^2+i\xi y_1y_2)\right)\Bigg|\notag\\ &= \sqrt{\frac{\eta\zeta}{\pi^2}} \exp\left( -\frac{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}}{2}\Big(\eta (x_1^2 + y_1^2)+\zeta (x_2^2 + y_2^2)\Big) + \frac{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda \right\}\xi}{2}\Big(x_1x_2-y_1y_2\Big)\right)\notag\\ &\leq C\exp\Big(-c\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\big(|x|^2+|y|^2\big) - c'|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}| |B|\big(|x_1x_2|+|y_1y_2|\big)\Big). \end{align}\] Now by Cauchy-Schwarz, \[|x_1x_2|+|y_1y_2|\leq 2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)} \leq |x|^2+|y|^2.\] Thus, for \(|B|<c_B\) small enough, we have \[|\psi_0(x,\lambda)\psi_0^*(y,\lambda)| \leq \exp\left(-c(\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}-|B||\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}|)(|x|^2+|y|^2) \right) \leq C\exp(-c|\lambda|(|x|^2+|y|^2)),\] provided \(|B|<c_B\) is small enough. ◻

Proof. We first move to \(w\) coordinates defined by \[\begin{align} w_1 &= x_1 - y_1, \qquad w_2 = x_2 - y_2,\notag\\ w_3 &= x_1 + y_1, \qquad w_4 = x_2 + y_2. \end{align}\] The function \(\phi(x,y,s)\) can be rewritten as \[\begin{align} \label{eq:phi95w95coordinates} \phi(w,s) &= \frac{\lambda f_+f_- a_{12}(s)}{2K(s)}w_1^2 + \frac{\lambda f_+f_- a_{21}(s)}{2K(s)}w_2^2 \notag\\ &+ \frac{\lambda f_+f_- b_{12}(s)}{2K(s)}w_3^2 + \frac{\lambda f_+f_- b_{21}(s)}{2K(s)}w_4^2 \notag\\ &+ \frac{i\lambda B g_-(s)}{2K(s)}w_1w_4 +\frac{i\lambda B g_+(s)}{2K(s)}w_2w_3, \end{align}\tag{79}\] where \(K(s),f_\pm\) are defined as in 72 and \[\begin{align} a_{ij}(s) &= k_i\left[\cosh\left(\frac{f_+s}{2}\right) + \cosh\left(\frac{f_-s}{2}\right)\right]\notag\\ &\qquad\qquad \times\left[f_-(k_1+k_2)\sinh\left(\frac{f_+s}{2}\right)+ f_+(k_j-k_i)\sinh\left(\frac{f_-s}{2}\right)\right],\notag\\ b_{ij}(s) &= k_i\left[\cosh\left(\frac{f_+s}{2}\right) - \cosh\left(\frac{f_-s}{2}\right)\right]\notag\\ &\qquad\qquad \times\left[f_-(k_1+k_2)\sinh\left(\frac{f_+s}{2}\right)- f_+(k_j-k_i)\sinh\left(\frac{f_-s}{2}\right)\right],\notag\\ g_\pm(s) &= (k_1^2-k_2^2)\left[f_+^2(\cosh(f_-s)-1)-f_-^2(\cosh(f_+s)-1)\right]\notag\\ &\qquad\qquad \pm 8f_+f_-k_1k_2\sinh\left(\frac{f_+s}{2}\right)\sinh\left(\frac{f_-s}{2}\right). \end{align}\] Let us factor out the leading order behavior in small \(s\): \[\begin{align} \label{eq:small95s95factors} a_{ij}(s) &= \frac{f_+f_-k_is}{2}\left[\cosh\left(\frac{f_+s}{2}\right)+ \cosh\left(\frac{f_-s}{2}\right)\right]\notag\\ &\qquad\qquad \times\left[(k_1+k_2)\frac{\sinh(f_+s/2)}{f_+s/2}+(k_j-k_i)\frac{\sinh(f_-s/2)}{f_-s/2}\right]=s\cdot \hat{a}_{ij}(s),\notag\\ b_{ij}(s) &= \frac{f_+f_-k_is^3}{2}\bigg[\frac{\cosh(f_+s/2)-\cosh(f_-s/2)}{s^2}\bigg]\notag\\ &\qquad\qquad \times\bigg[(k_1+k_2)\frac{\sinh(f_+s/2)}{f_+s/2}+(k_j-k_i)\frac{\sinh(f_-s/2)}{f_-s/2}\bigg]=s^3\cdot \hat{b}_{ij}(s),\notag\\ g_\pm(s) &= f_+^2f_-^2(k_1^2-k_2^2) s^2 \bigg[\frac{\cosh(f_-s)-1}{f_-^2s^2} - \frac{\cosh(f_+s)-1}{f_+^2s^2}\notag \\ &\qquad\qquad \pm 2k_1k_2\frac{\sinh(f_+s/2)}{f_+s/2}\frac{\sinh(f_-s/2)}{f_-s/2}\bigg]=s^2\cdot \hat{g}_\pm(s),\notag\\ K(s) &= f_+^2f_-^2s^2\left[(k_1+k_2)^2\left(\frac{\cosh(f_+s)-1}{f_+^2s^2}\right)-(k_1-k_2)^2\left(\frac{\cosh(f_-s)-1}{f_-^2s^2}\right)\right]= s^2\cdot \hat{K}(s), \end{align}\tag{80}\] and the real part of \(\phi\) is \[\begin{align} \operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}(w,s) &= \frac{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda \right\}f_+f_-}{2}\left[\frac{a_{12}(s)}{K(s)}w_1^2 + \frac{a_{21}(s)}{K(s)}w_2^2 + \frac{b_{12}(s)}{K(s)}w_3^2 + \frac{b_{21}(s)}{K(s)}w_4^2\right]\notag\\ &+\frac{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda \right\}B}{2}\left[\frac{g_-(s)}{K(s)}w_1w_4 + \frac{g_+(s)}{K(s)}w_2w_3\right]. \end{align}\]

First note that \(f_\pm\) are positive constants and \(f_+ > f_-\) by definition. Also, \(\hat{a}_{ij}(s), \hat{b}_{ij}(s)\) and \(\hat{K}(s)\) are positive and finite for small \(s>0\). Indeed, \(\cosh(f_+s/2)\pm\cosh(f_-s/2)\geq0\), and \[\begin{align} &\frac{\sinh(f_\pm s/2)}{f_\pm s/2},\; \frac{\cosh(f_\pm s/2)-1}{f_\pm^2 s^2},\;\text{and } \frac{\cosh(f_+s/2)-\cosh(f_-s/2)}{s^2}\;\text{are finite and positive},\notag\\ &\frac{\sinh(f_+ s/2)}{f_+ s/2}\geq \frac{\sinh(f_- s/2)}{f_- s/2},\qquad \frac{\cosh(f_+ s)-1}{f_+^2 s^2}\geq \frac{\cosh(f_- s)-1}{f_-^2 s^2},\;\text{and }\notag\\ &k_1+k_2 > k_i-k_j\;(i,j=1,2), \end{align}\] for all \(s\geq0\), where we define \(\sinh(x)/x=1\) and \(\frac{\cosh(x)-1}{x^2}=1/2\) for \(x=0\).

Therefore, \(\hat{a}_{ij}(s), \hat{b}_{ij}(s)\) and \(\hat{K}(s)\) are uniformly continuous in \((s,B)\) and positive on the compact set \(\{s\in[0,c_1]\}\times \{|B|\leq c_B\}\) and thus attain a positive minimum, so that \(a_{ij}(s)/K(s)\geq c/s\) and \(b_{ij}(s)/K(s)\geq cs\). Similarly, \(\hat{g}_\pm(s)\) is finite for small \(s>0\) and uniformly continuous in \((s,B)\) on \(\{s\in[0,c_1]\}\times \{|B|\leq c_B\}\), so that \(|g_\pm(s)/K(s)| \leq \hat{C}\).

Therefore, we have \[\begin{align} &\operatorname{\mathbb{R}\mathbb{e}}\left\{\phi \right\}\geq \operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\left[\frac{c}{s}(w_1^2+w_2^2)+cs(w_3^2+w_4^2)\right]-\frac{|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}||B|}{2}\Big[C'|w_1w_4| + C'|w_2w_3|\Big]\notag\\ &\geq \tilde{c}\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\left[\frac{w_1^2+w_2^2}{s}+s(w_3^2+w_4^2)\right]-\hat{C}|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}||B|\Big[|w_1w_4| + |w_2w_3|\Big]. \end{align}\] Now by Cauchy-Schwarz, \[|w_1w_4| + |w_2w_3| \leq \sqrt{(w_1^2+w_2^2)(w_3^2+w_4^2)}\leq 2\sqrt{(w_1^2+w_2^2)(w_3^2+w_4^2)},\] and for any two positive numbers \(N,M\), we have \(2\sqrt{NM}\leq N+M\). Thus, for \(N=\frac{w_1^2+w_2^2}{s}\) and \(M=s(w_3^2+w_4^2)\), we have \[\label{eq:cauchy95schwarz95application} |w_1w_4| + |w_2w_3| \leq \frac{w_1^2+w_2^2}{s} + s(w_3^2+w_4^2).\tag{81}\] Putting everything together, we have \[\begin{align} \operatorname{\mathbb{R}\mathbb{e}}\left\{\phi \right\}&\geq \left(\tilde{c}\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda \right\}- \hat{C}|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}||B|\right)\left[\frac{w_1^2+w_2^2}{s} + s(w_3^2+w_4^2)\right]\notag\\ &\geq c''|\lambda|\left[\frac{w_1^2+w_2^2}{s} + s (w_3^2+w_4^2)\right] = c''|\lambda|\left[\frac{|x-y|^2}{s} + s|x+y|^2\right]\notag\\ &= c''|\lambda|\left[2s(|x|^2+|y|^2)+\left(\frac{1}{s}-s\right)|x-y|^2\right] \geq c|\lambda|\left(s(|x|^2+|y|^2)+\frac{|x-y|^2}{s} \right), \end{align}\] since \(\frac{1}{s}-s\geq\frac{1}{2s}\) for \(0<s<1/2\), so \(c_1\) is chosen small enough. This holds provided \(|B|<c_B\), for a small enough controlled constant \(c_B\).

Moving to the prefactor, we saw above that \(\hat{K}(s)\) attains a positive minimum on \(\{s\in[0,c_1]\}\times \{|B|\leq c_B\}\), so that \(|K(s)|\geq c''s^2\). Remembering \(P(s)=\frac{f_+f_-}{2\pi}\sqrt{\frac{2k_1k_2}{K(s)}}\), we get \(|P(s)|\leq C'/s\). The proof of the lem is thus complete. ◻

Proof. We first set \(B=0\). In this case, we have \[\begin{align} \phi_{B=0}(w,s) &= \frac{\lambda k_1}{4}\coth\left(\frac{k_1s}{2}\right)w_1^2 + \frac{\lambda k_2}{4}\coth\left(\frac{k_2s}{2}\right)w_2^2 \notag\\ &+ \frac{\lambda k_1}{4}\tanh\left(\frac{k_1s}{2}\right)w_3^2 + \frac{\lambda k_2}{4}\tanh\left(\frac{k_2s}{2}\right)w_4^2. \end{align}\] For \(c_1<s<C_1\), \(\coth\) and \(\tanh\) are continuous and bounded from below by a positive constant, so that \[\text{Re}(\phi_{B=0}(x,y,s))\geq c'|\lambda|\left(w_1^2+w_2^2+w_3^2+w_4^2 \right) = 2c'|\lambda| \left(|x|^2+|y|^2\right).\] Now we move to \(|B|\) small. We remind the reader that \(f_\pm=\sqrt{(k_1\pm k_2)^2+4B^2}\to|k_1\pm k_2|\) as \(B\to0\), and \(a_{ij},b_{ij},g_\pm,K\) depend on \(B\) only through \(f_\pm\). Specifically, \[\begin{align} K(s)|_{B=0} &= (k_1^2-k_2^2)^2\big[\cosh\big((k_1+k_2)s\big)-\cosh\big((k_1-k_2)s\big)\big] \notag\\ &= 2(k_1^2-k_2^2)^2 \sinh(k_1s)\sinh(k_2s) > 0, \end{align}\] for \(c_1\leq s\leq C_1\). By uniform continuity in \((s,B)\) on the compact set \([c_1,C_1]\times\{|B|\leq c_B\}\), there exists \(\delta>0\) such that whenever \(|B|<\delta\), we have, for some \(c\), \[\label{eq:uniform95bound95on95K} K(s) \geq c > 0,\tag{87}\] uniformly for \(c_1\leq s\leq C_1\). We now reduce \(c_B\) so that \(c_B <\delta\). Hence 87 holds for \((s,B)\in[c_1,C_1]\times\{|B|\leq c_B\}\), so no singularities arise in the denominators. Therefore, we have uniform continuity in \((s,B)\) on the compact set \([c_1,C_1]\times\{|B|\leq C_B\}\) for the following quantities: \[\frac{f_+f_-a_{ij}(s)}{2K(s)},\;\frac{f_+f_-b_{ij}(s)}{2K(s)},\;\frac{Bg_\pm(s)}{2K(s)},\] and in particular, they are all uniformly bounded.

Thus, there exists \(\tilde{c}>0\) such that for \(|B|<\tilde{c}\), we have \[|\phi(w,s) - \phi_{B=0}(w,s)| \leq \frac{c'}{2}|\lambda| (w_1^2+w_2^2+w_3^2+w_4^2),\] so that \[\begin{align} \text{Re}(\phi(w,s)) &\geq \text{Re}(\phi_{B=0}(w,s)) - |\phi(w,s) - \phi_{B=0}(w,s)| \notag\\ &\geq \frac{c'}{2}|\lambda| (w_1^2+w_2^2+w_3^2+w_4^2) = c'|\lambda|(|x|^2+|y|^2). \end{align}\] We again reduce \(c_B\) so that \(c_B < \tilde{c}\).

We now move to the prefactor \(P(s)\). Since \(K(s)\) is bounded away from zero for \(c_1\leq s\leq C_1\) and \(|B|<c_B\), we have \(|P(s)|\leq C\) for some constant \(C>0\). This completes the proof of the lem. ◻

Proof. For the first estimate, we write the explicit expression for the limit of \(\phi(s,x,y)\) as \(s\to\infty\): \[\lim_{s\to\infty}\phi(s,x,y) = \frac{\lambda f_+k_1}{2(k_1+k_2)}(x_1^2+y_1^2) + \frac{\lambda f_+k_2}{2(k_1+k_2)}(x_2^2+y_2^2) - \frac{i\lambda B (k_1-k_2)}{k_1+k_2}(x_1x_2-y_1y_2).\] Then we have \[\text{ERROR}_1=\lambda[M_1(x_1^2+y_1^2) + M_2x_1y_1 + M_3(x_2^2+y_2^2) + M_4x_2y_2+M_5(x_1x_2-y_1y_2) + M_6(x_1y_2-x_2y_1)]\] for \(M_i=M_i(B,k_1,k_2)\).

We want to show that \[\label{eq:matrix95size95region955} |M_i|\leq C \exp(-c s),\tag{90}\] for some constants \(C,c>0\) and large enough \(s\). Indeed, if so, then \[|\text{ERROR}_1| \leq |\lambda|C\exp(-c s)(|x|^2+|y|^2 + |x_1y_1| + |x_2y_2| + |x_1x_2| + |y_1y_2| + |x_1y_2| + |x_2y_1|).\] By Cauchy-Schwarz, we have \[\begin{align} |x_1y_1| + |x_2y_2| &\leq 2\sqrt{|x|^2+|y|^2} \leq |x|^2+|y|^2,\notag\\ |x_1y_2| + |x_2y_1| &\leq 2\sqrt{|x|^2+|y|^2} \leq |x|^2+|y|^2, \notag\\ |x_1x_2| + |y_1y_2| &\leq 2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)} \leq |x|^2+|y|^2, \end{align}\] and we get the desired result \[|\text{ERROR}_1| \leq C'|\lambda|\exp(-c s)(|x|^2+|y|^2).\]

By definition we have \(f_+>f_->0\). Then \[\begin{align} \sinh(f_+s) &= \frac{1}{2}e^{f_+s}\left(1-e^{-2f_+s}\right),\notag\\ |\sinh(f_-s)| &\leq e^{f_-s} = \frac{1}{2}e^{f_+s}\left(2e^{(f_+-f_-)s}\right). \end{align}\] Therefore, for large enough \(s\), we have \[\alpha_{ij}(s) = k_if_-(k_1+k_2)\frac{1}{2}e^{f_+s}\left(1+\mathcal{O}(e^{-(f_+-f_-)s})\right).\] Similarly, for large enough \(s\), we have \[\begin{align} \sinh\left(\frac{f_-+f_+}{2}s\right) &= \frac{1}{2}e^{f_+s}\left[e^{-(f_+-f_-)s/2}-e^{-(3f_++f_-)s/2}\right] = e^{f_+s}\mathcal{O}\left(e^{-(f_+-f_-)s/2}\right),\notag\\ \sinh\left(\frac{f_--f_+}{2}s\right) &= \frac{1}{2}e^{f_+s}\left[e^{-(3f_+-f_-)s/2}-e^{-(f_++f_-)s/2}\right] = e^{f_+s}\mathcal{O}\left(e^{-(f_++f_-)s/2}\right). \end{align}\] Then, \[\beta_{ij}(s) = e^{f_+s}\mathcal{O}(e^{-(f_+-f_-)s/2}).\] Also, \[\begin{align} \cosh(f_+s)-1 &= \frac{1}{2}e^{f_+s}\left(1-e^{-f_+s}\right)^2,\notag\\ \cosh(f_-s)-1 &= e^{f_+s}\mathcal{O}(e^{-(f_+-f_-)s}). \end{align}\] Then, \[\begin{align} \gamma_1(s) &= (k_1^2-k_2^2)\left[f_+^2e^{f_+s}\mathcal{O}\left(e^{-(f_+-f_-)s}\right) - f_-^2\frac{1}{2}e^{f_+s}\left(1-e^{-f_+s}\right)^2\right] ,\notag\\ &= e^{f_+s}\left[-\frac{f_-^2(k_1^2-k_2^2)}{2}+\mathcal{O}\left(e^{-(f_+-f_-)s}\right)\right],\notag\\ K(s) &= f_-^2(k_1+k_2)^2\frac{1}{2}e^{f_+s}\left(1+\mathcal{O}(e^{-(f_+-f_-)s})\right) -f_+^2(k_1-k_2)^2e^{f_+s}\mathcal{O}(e^{(f_+-f_-)s})\notag\\ &= e^{f_+s}\left[\frac{f_-^2(k_1+k_2)^2}{2} + \mathcal{O}(e^{-(f_+-f_-)s})\right]. \end{align}\] And finally, \[\begin{align} \sinh\left(\frac{f_+s}{2}\right)\sinh\left(\frac{f_-s}{2}\right) &= \frac{1}{4}e^{f_+s}\left[e^{-(f_+-f_-)s/2}-e^{-(f_++f_-)s/2}-e^{-(3f_+-f_-)s/2}+e^{-(3f_++f_-)s/2}\right]\notag\\ &= e^{f_+s}\mathcal{O}\left(e^{-(f_+-f_-)s/2}\right), \end{align}\] so that we have \[\gamma_2(s) = e^{f_+s}\mathcal{O}\left(e^{-(f_+-f_-)s/2}\right).\]

For the \(x_1^2+y_1^2\) term, we have \[\begin{align} |M_1| &= \bigg|\frac{f_+f_- \alpha_{12}(s)}{2K(s)} - \frac{f_+k_1}{2(k_1+k_2)}\bigg|\\ &= \bigg|\frac{f_+f_-}{2} \frac{k_1f_-(k_1+k_2)}{f_-^2(k_1+k_2)^2}\frac{1+\mathcal{O}(e^{-(f_+-f_-)s})}{1 + \mathcal{O}(e^{-(f_+-f_-)s})} - \frac{f_+k_1}{2(k_1+k_2)}\bigg|\\ &= \bigg|\frac{f_+k_1}{2(k_1+k_2)}\left(1 + \mathcal{O}(e^{-(f_+-f_-)s})\right) - \frac{f_+k_1}{2(k_1+k_2)}\bigg| \notag\\ &= |\mathcal{O}(e^{-(f_+-f_-)s})|\leq C\exp(-c s). \end{align}\] The same result applies for \(M_3\) (the \(x_1y_1\) term), since they are defined in the same way as \(M_1\), but with \(k_1\) and \(k_2\) interchanged.

For the \(x_2^2+y_2^2\) term, we have \[\begin{align} |M_2| &= \bigg|\frac{f_+f_- \beta_{12}(s)}{K(s)}\bigg| = \Bigg|\frac{f_+f_- e^{f_+s}\mathcal{O}(e^{-(f_+-f_-)s})}{e^{f_+s}\left[\frac{f_-^2(k_1+k_2)^2}{2} + \mathcal{O}(e^{-(f_+-f_-)s})\right]}\Bigg| \notag\\ &= \bigg|\mathcal{O}(e^{-(f_+-f_-)s}) \bigg| \leq C\exp(-c s). \end{align}\] The same result applies for \(M_4\) (the \(x_2y_2\) term), since they are defined in the same way as \(M_2\), but with \(k_1\) and \(k_2\) interchanged.

Next is the cross term \(x_1x_2-y_1y_2\). For this case we have \[\begin{align} |M_5| &= \bigg|\frac{iB \gamma_1(s)}{K(s)} + \frac{iB (k_1-k_2)}{k_1+k_2}\bigg| = |B|\Bigg| \frac{e^{f_+s}\left[-\frac{f_-^2(k_1^2-k_2^2)}{2}+\mathcal{O}\left(e^{-(f_+-f_-)s}\right)\right]}{e^{f_+s}\left[\frac{f_-^2(k_1+k_2)^2}{2} + \mathcal{O}(e^{-(f_+-f_-)s})\right]} + \frac{k_1-k_2}{k_1+k_2}\Bigg|\notag\\ &= |B| \Bigg| \frac{-f_-^2(k_1^2-k_2^2)}{f_-^2(k_1+k_2)^2} \frac{1+\mathcal{O}(e^{-(f_+-f_-)s})}{1+\mathcal{O}(e^{-(f_+-f_-)s})} + \frac{k_1-k_2}{k_1+k_2} \Bigg|\notag\\ &= |B| \Bigg|-\frac{k_1-k_2}{k_1+k_2}\left(1 + \mathcal{O}(e^{-(f_+-f_-)s})\right) + \frac{k_1-k_2}{k_1+k_2}\Bigg|\notag\\ &= |B||\mathcal{O}(e^{-(f_+-f_-)s})| \leq C\exp(-c s), \end{align}\] provided \(|B|\) is bounded.

Finally, we have the last cross term \(x_1y_2-x_2y_1\). For this case we have \[\begin{align} |M_6| &= \bigg|-\frac{iB \gamma_2(s)}{K(s)}\bigg| = |B| \Bigg| \frac{e^{f_+s}\mathcal{O}\left(e^{-(f_+-f_-)s/2}\right)}{e^{f_+s}\left[\frac{f_-^2(k_1+k_2)^2}{2} + \mathcal{O}(e^{-(f_+-f_-)s})\right]} \Bigg|\notag\\ &= |B| \big|\mathcal{O}\left(e^{-(f_+-f_-)s/2}\right) \big| \leq C\exp(-c s), \end{align}\] again provided \(|B|\) is bounded. This completes the proof of 90 .

For the prefactor error estimate, we use the above asymptotics for \(K(s)\) to see that \[\begin{align} P(s) &= \frac{f_+f_-}{2\pi}\sqrt{\frac{2k_1k_2}{K(s)}} = \frac{f_+f_-}{2\pi}\sqrt{\frac{2k_1k_2}{e^{f_+s}\left[\frac{f_-^2(k_1+k_2)^2}{2} + \mathcal{O}(e^{-(f_+-f_-)s})\right]}}\notag\\ &= \frac{f_+f_-}{2\pi}\sqrt{\frac{4k_1k_2}{f_-^2(k_1+k_2)^2}}e^{-f_+s/2}\left(1+\mathcal{O}(e^{-(f_+-f_-)s})\right)\notag\\ &= C_P\exp(-E_0 s)\big(1+\text{ERROR}_2(s)\big), \end{align}\] where one checks the \(C_P\) is the same prefactor that appears in the counterterm \[e^{-E_0s}\psi_0(x,\lambda)\psi_0^*(y,\lambda^*).\] This completes the proof of the lem. ◻

Proof. First note that \(\operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}_\infty(x,y,s) \geq c|\lambda|(|x|^2+|y|^2)\). Indeed, \[\begin{align} &\operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}_\infty(x,y,s) = \operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\big[\eta(x_1^2+y_1^2)+\zeta(x_2^2+y_2^2)\big]-\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda \right\}\xi(x_1x_2-y_1y_2)\notag\\ &\geq c''\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}(|x|^2+|y|^2) - |\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}||B|C'(|x_1x_2|+|y_1y_2|)\notag\\ &= c^\sharp(\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda \right\}- C|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}||B|)(|x|^2+|y|^2) \geq c|\lambda|(|x|^2+|y|^2), \end{align}\] provided \(|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}|<C'|\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}|\) and \(|B|\) is small enough (remember \(\eta,\zeta>0\)). Now, by 37, we have \[\begin{align} &\operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}(x,y,s) = \operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}_\infty(x,y,s) + \operatorname{\mathbb{R}\mathbb{e}}\left\{\text{ERROR}\right\}_1(x,y,s,\lambda) \notag\\ &\geq c|\lambda|(|x|^2+|y|^2) - |\text{ERROR}_1(x,y,s,\lambda)| \notag\\ &\geq c|\lambda|(|x|^2+|y|^2) \geq \frac{1}{2}c|\lambda|(|x|^2+|y|^2) + \frac{1}{2}c\exp(c^\sharp s), \end{align}\]

where we used the region 6 hypothesis \(|\lambda|(|x|^2+|y|^2)>\exp(c^\sharp s)\). Picking \(c^\sharp<c\), we have the desired result.

For the prefactor estimate, we have shown in 37 that \[P(s) = C_P\exp(-E_0 s)\big(1+\text{ERROR}_2(s)\big),\;\text{with}\;|\text{ERROR}_2(s)|\leq C''\exp(-c s),\] and it follows that \(|P(s)|\leq Ce^{-E_0 s}\). The proof of the lem is complete. ◻

Proof. We again move to \(w\) coordinates. In this case, we write \[\begin{align} \label{eq:phi95second95thm95w95coords} \phi(w,s) &= \operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda \right\}\bigg[\frac{\lambda}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}}\frac{f_+f_- a_{12}(s)}{2K(s)}w_1^2 +\frac{\lambda}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}} \frac{f_+f_- a_{21}(s)}{2K(s)}w_2^2 \notag\\ &+ \frac{\lambda}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}}\frac{f_+f_- b_{12}(s)}{2K(s)}w_3^2 + \frac{\lambda}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}}\frac{f_+f_- b_{21}(s)}{2K(s)}w_4^2 \notag\\ &+ i\frac{\lambda}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}}\frac{B g_-(s)}{2K(s)}w_1w_4 +i\frac{\lambda}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}}\frac{B g_+(s)}{2K(s)}w_2w_3\bigg]. \end{align}\tag{93}\] We define the three expressions \[\begin{align} \label{eq:phi95195295395second95thm} \phi_{1ij}(s)=\frac{\lambda}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}}\frac{f_+f_- a_{ij}(s)}{2K(s)},\quad \phi_{2ij}(s)=\frac{\lambda}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}}\frac{f_+f_- b_{ij}(s)}{2K(s)},\quad \phi_{3\pm}(s)=\frac{\lambda}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}}\frac{B g_\pm(s)}{2K(s)}, \end{align}\tag{94}\] and we will prove estimates for each of these. Starting with \(\phi_{1ij}(s)\), we have \[\phi_{1ij}(s) = \frac{\lambda}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}}\frac{f_+f_-}{2}\frac{\hat{a}_{ij}(s)}{\hat{K}(s)}\frac{1}{s}=\varphi_{1ij}(s,\lambda,B)\frac{1}{s},\] where \(\hat{a}_{ij}, \hat{K}(s)\) are defined as in 80 . Notice that \(\hat{a}_{ij}(s)/\hat{K}(s)\) uniformly continuous for \((s,B)\) on the compact set \([0,c_1]\times\Lambda\). Indeed, we saw in 8.4.1 that \(\hat{a}_{ij}(s)/\hat{K}(s)\) is uniformly continuous for \(s\in[0,c_1]\). Furthermore, notice that the only dependence on \(B\) is through \(f_\pm=\sqrt{(k_1\pm k_2)^2+4B^2}\). In particular, \(\hat{K}(s)\) is never zero for any \(B\in\Lambda\). Therefore, \(\hat{a}_{ij}(s)/\hat{K}(s)\) is analytic in \(B\) and we get the uniform continuity.

Thus, we get that \(\varphi_{1ij}(s,\lambda,B)\) is uniformly continuous for \((s,\lambda/\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\},B)\in[0,c_1]\times\{\text{The set of } \lambda/\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\;\text{as in } \eqref{eq:lambda95over95re95lambda}\}\times\Lambda\). Indeed, \(f_+f_-\) is analytic in \(B\) and one can write \[\label{eq:lambda95over95re95lambda} \frac{\lambda}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}} = 1 + i\frac{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}},\quad \text{where } \quad \left|\frac{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}}\right|\leq c_\lambda,\tag{95}\] because \(|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}| \leq c_\lambda|\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}|\) for \(\lambda\in\Omega'\), and this is a closed and bounded set.

Next, we see that when \(\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0\), we have for \(s\in(0,c_1)\), \[\label{eq:phi951ij95re} \operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}_{1ij}(s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0} = \phi_{1ij}(s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0} = \frac{f_+f_-}{2}\frac{\hat{a}_{ij}(s)}{\hat{K}(s)}\frac{1}{s} \geq \frac{c}{s} > 0.\tag{96}\] This is because the argument in 8.4.1 shows that \(\hat{a}_{ij}(s)/\hat{K}(s)\geq c\) for \(s\in[0,c_1]\) and \(B\in[-C_B,C_B]\). Therefore, by uniform continuity, there exists a \(\delta>0\) such that whenever \(|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}|<\delta\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\) and \(|\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}|<\delta\) we have \[|\phi_{1ij}(s)-\phi_{1ij}(s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0}| < \frac{c}{2s},\] with \(c\) as in 96 , and therefore \[\begin{align} &\operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}_{1ij}(s) \geq \operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}_{1ij}(s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0} - |\phi_{1ij}(s)-\phi_{1ij}(s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0}| \notag\\ &\geq \frac{c}{s} - \frac{c}{2s} = \frac{c'}{s} > 0. \end{align}\]

We continue to \(\phi_{2ij}(s)\): \[\phi_{2ij}(s) = \frac{\lambda}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}}\frac{f_+f_-}{2}\frac{\hat{b}_{ij}(s)}{\hat{K}(s)}s=\varphi_{2ij}(s,\lambda,B)s,\] where \(\hat{b}_{ij}(s)\) is defined as in 80 . By the same arguments as for \(\phi_{1ij}(s)\), we see that \(\varphi_{2ij}(s,\lambda,B)\) is uniformly continuous for \[(s,\lambda/\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\},B)\in[0,c_1]\times\{\text{The set of } \lambda/\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\;\text{as in } \eqref{eq:lambda95over95re95lambda}\}\times\Lambda\]. Furthermore, when \(\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0\), we have \[\label{eq:phi952ij95re} \operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}_{2ij}(s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0} = \phi_{2ij}(s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0} = \frac{f_+f_-}{2}\frac{\hat{b}_{ij}(s)}{\hat{K}(s)}s \geq c s > 0,\tag{97}\] for \(s\in(0,c_1)\), because \(\hat{b}_{ij}(s)/\hat{K}(s)\geq c\) for \(s\in[0,c_1]\) and \(B\in[-C_B,C_B]\). Therefore, by uniform continuity, there exists a \(\delta>0\) such that whenever \(|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}|<\delta\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\) and \(|\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}|<\delta\) we have \[|\phi_{2ij}(s)-\phi_{2ij}(s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0}| < \frac{c}{2}s,\] with \(c\) as in 97 , and therefore \[\begin{align} &\operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}_{2ij}(s) \geq \operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}_{2ij}(s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0} - |\phi_{2ij}(s)-\phi_{2ij}(s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0}| \notag\\ &\geq c s - \frac{c}{2}s = c's > 0. \end{align}\]

Finally, we consider \(\phi_{3\pm}(s)\): \[\phi_{3\pm}(s) = i\frac{\lambda}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}}\frac{B}{2}\frac{\hat{g}_\pm(s)}{\hat{K}(s)} = \varphi_{3\pm}(s,\lambda,B),\] where \(\hat{g}_\pm(s)\) is defined as in 80 . By the same arguments as for \(\phi_{1\pm}(s)\), we see that \(\varphi_{3\pm}(s,\lambda,B)\) is uniformly continuous for \[(s,\lambda/\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\},B)\in[0,c_1]\times\{\text{The set of } \lambda/\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\;\text{as in } \eqref{eq:lambda95over95re95lambda}\}\times\Lambda\]. Furthermore, when \(\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0\), we have \[\operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}_{3\pm}(s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0} = 0,\] for \(s\in(0,c_1)\), because \(\hat{g}_\pm(s)/\hat{K}(s)\) is bounded for \(s\in[0,c_1]\) and \(B\in[-C_B,C_B]\) and everything is real. Therefore, by uniform continuity, for every \(\varepsilon>0\) there exists a \(\delta>0\) such that whenever \(|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}|<\delta\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\) and \(|\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}|<\delta\) we have \[|\phi_{3\pm}(s)-\phi_{3\pm}(s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0}| < \varepsilon,\] and therefore \[\begin{align} &\operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}_{3\pm}(s) \leq |\phi_{3\pm}(s)| < \varepsilon + |\phi_{3\pm}(s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0}| = \varepsilon. \end{align}\]

Putting everything together, we have \[\begin{align} &\operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}(w,s) \geq \operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\bigg[\frac{c'_1}{s}(w_1^2+w_2^2) + c'_2s(w_3^2+w_4^2) - (\varepsilon|w_1w_4|+\varepsilon|w_2w_3|)\bigg]\notag\\ &\geq c''\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\bigg[\frac{1}{s}(w_1^2+w_2^2)+s(w_3^2+w_4^2)\bigg]\geq c|\lambda|s(|x|^2+|y|^2) + \frac{c|\lambda|}{s}|x-y|^2,\nonumber\\ \end{align}\] provided \(\varepsilon\) is small enough. We now fix \(c_\lambda, c_B\) smaller than the above \(\delta\)’s. In moving from the first to the second line, we used the usual Cauchy-Schwarz application 81 . The proof of the first part of the lem is complete.

For the prefactor estimate, we have \[P(s) = \frac{f_+f_-}{2\pi}\sqrt{\frac{2k_1k_2}{K(s)}} = \frac{f_+f_-}{2\pi}\sqrt{\frac{2k_1k_2}{\hat{K}(s)}}\frac{1}{s} = \hat{p}(s,B)\frac{1}{s}.\] As seen above, \(\hat{K}(s)\) is uniformly continuous for \((s,B)\in[0,c_1]\times\Lambda\). Therefore, \(\hat{p}(s,B)\) is uniformly continuous for \((s,B)\in[0,c_1]\times\Lambda\). Now for \(\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0\), we have \[|\hat{p}(s,B)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0}| = \frac{f_+f_-}{2\pi}\sqrt{\frac{2k_1k_2}{\hat{K}(s)}} \leq C,\] for \(s\in[0,c_1]\) and \(B\in[-C_B,C_B]\) from our work in 8.4.1. Therefore, by uniform continuity, there exists a \(\delta>0\) such that whenever \(|\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}|<\delta\) we have \[|\hat{p}(s,B)-\hat{p}(s,B)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0}| < C,\] and therefore \[|\hat{p}(s,B)| \leq |\hat{p}(s,B)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0}| + |\hat{p}(s,B)-\hat{p}(s,B)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0}| \leq C + C = C',\] for sufficiently small \(c_1, c_\lambda, c_B\) (in particular, we reduce \(c_B\) to be less than \(\delta\)). This completes the proof of the lem. ◻

Proof. Still working in \(w\) coordinates, we first set \(\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0\). Then, from 93 and 94 , we have \[\label{eq:phi95second95thm95w95coords95re} \operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}(w,s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0} = \lambda [ \phi_{112}(s)w_1^2 + \phi_{121}(s)w_2^2 + \phi_{212}(s)w_3^2 + \phi_{221}(s)w_4^2],\tag{98}\] for \(s\in[c_1,C_1]\) and \(B\in[-C_B,C_B]\). From 80 , we see that the coefficients \[\phi_{112},\phi_{121},\phi_{212},\phi_{221}\] are bounded below by a constant \(c\). Therefore, for \(s\in[c_1,C_1]\) and \(B\in[-C_B,C_B]\), we have \[\operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}(w,s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0} \geq c|\lambda|(|w_1|^2+|w_2|^2+|w_3|^2+|w_4|^2) = c|\lambda|(|x|^2+|y|^2).\]

As seen in the proof of the previous 40, combined with the fact that \(s\) in this region is bounded away from \(0\), we have that the \(\phi\)’s appearing in 98 are uniformly continuous for \((s,\lambda/\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\},B)\in[c_1,C_1]\times\{\text{The set of } \lambda/\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\;\text{as in } \eqref{eq:lambda95over95re95lambda}\}\times\Lambda\). Therefore, there exists a \(\delta>0\) such that whenever \(|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}|<\delta\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\) and \(|\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}|<\delta\) we have \[|\phi(w,s)-\phi(w,s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0}| < \frac{c}{2}|\lambda||w|^2,\] which gives \[\begin{align} &\operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}(w,s) \geq \operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}(w,s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0} - |\phi(w,s)-\phi(w,s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0}| \notag\\ &\geq c|\lambda||w|^2 - \frac{c}{2}|\lambda||w|^2 = \frac{c'}{2}|\lambda||w|^2 = c'|\lambda|(|x|^2+|y|^2), \end{align}\] for sufficiently small \(c_\lambda, c_B\) (again, we reduce \(c_\lambda,c_B\) so they are less than \(\delta\)). This completes the proof of the first part of the lem.

For the prefactor estimate, again combined with the fact that \(s\) in this region is bounded away from \(0\), we see that \(P(s)\) is uniformly continuous for \((s,B)\in[c_1,C_1]\times\Lambda\). Indeed, by an argument similar to that in the proof of the 40, \(\hat{K}(s)\) is uniformly continuous for \((s,B)\in[c_1,C_1]\times\Lambda\) and bounded away from zero. More explicitly, we have \[P(s) = \frac{f_+f_-}{2\pi}\sqrt{\frac{2k_1k_2}{K(s)}} = \frac{f_+f_-}{2\pi}\sqrt{\frac{2k_1k_2}{\hat{K}(s)}}\frac{1}{s},\] where \(\hat{K}(s)\) is defined as in 80 . Therefore, \(P(s)\) is uniformly continuous for \((s,B)\in[c_1,C_1]\times\Lambda\) and consequently bounded. This completes the proof of the lem. ◻

Proof. We start by writing again \[\begin{align} \phi(x,y,s) = \lambda &\bigg[\frac{f_+f_- \alpha_{12}(s)}{2K(s)}(x_1^2+y_1^2) +\frac{f_+f_- \beta_{12}(s)}{K(s)}x_1y_1 \notag\\ &+ \frac{f_+f_- \alpha_{21}(s)}{2K(s)}(x_2^2+y_2^2) + \frac{f_+f_- \beta_{21}(s)}{K(s)}x_2y_2\notag\\ &+ i\frac{B \gamma_1(s)}{K(s)}(x_1x_2-y_1y_2) -i\frac{B \gamma_2(s)}{K(s)}(x_1y_2-x_2y_1)\bigg], \end{align}\] and \(\phi_\infty(x,y,s):=\lim_{s\to\infty}\phi(x,y,s)\) as \[\phi_\infty(x,y,s) = \lambda\bigg[ \frac{\eta(B)}{2}(x_1^2+y_1^2) + \frac{\zeta(B)}{2}(x_2^2+y_2^2) - i\frac{\xi(B)}{2}(x_1x_2-y_1y_2) \bigg],\] where \(\eta(B),\zeta(B),\xi(B)\) are defined as in 74 .

With \(\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0\), we have \(f_+>f_->0\). Therefore, by continuity of the square root, we have that for \(|\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}|<\delta\) small enough, \[\begin{align} &\operatorname{\mathbb{R}\mathbb{e}}\left\{f\right\}_\pm > \frac{1}{2}f_\pm|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0} > 0,\quad \text{and likewise}\notag\\ &\operatorname{\mathbb{R}\mathbb{e}}\left\{(\right\}f_+-f_-) > \frac{1}{2}(f_+|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0} - f_-|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0}) > 0. \end{align}\] Thus, for the exponential functions we have \[|e^{\pm f_\pm s}| = e^{\pm\operatorname{\mathbb{R}\mathbb{e}}\left\{f\right\}_\pm s},\] and the same arguments goes exactly as in the proof of 37 but with the change \(f_\pm \to \operatorname{\mathbb{R}\mathbb{e}}\left\{f\right\}_\pm\). In particular, we get the result for \(\text{ERROR}_1(x,y,s,\lambda)\). ◻

Proof. First we show that \(\operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}_\infty(x,y,s) \geq c|\lambda|(|x|^2+|y|^2)\) for some controlled constant \(c>0\). For the case where \(\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0\), the argument in 38 again applies here to show that \[\label{eq:phi95infty95re} \phi_\infty(x,y,s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0} \geq c|\lambda|(|x|^2+|y|^2).\tag{99}\] In particular, \(\eta(B),\zeta(B)\) are real, positive and bounded by \(|B|\), and \(\xi(B)\) is real and bounded by \(|B|\). Also, \(|B||\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}|<c\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\) where \(c\) is a small constant. Now write \(\phi_\infty(x,y,s)\) as \[\phi_\infty(x,y,s) = \frac{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}}{2}\bigg[ \frac{\lambda}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}}\eta(B)(x_1^2+y_1^2) + \frac{\lambda}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}}\zeta(B)(x_2^2+y_2^2) - i\frac{\lambda}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}}\xi(B)(x_1x_2-y_1y_2) \bigg],\] where \(\eta(B),\zeta(B),\xi(B)\) are defined as in 74 and analytic in \(B\) for \(B\in\Lambda\). Therefore, the expressions \(\frac{\lambda}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}}\eta(B), \frac{\lambda}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}}\zeta(B), \frac{\lambda}{\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}}\xi(B)\) are uniformly continuous for \[(\lambda/\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\},B)\in\{\text{The set of } \lambda/\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\;\text{as in } \eqref{eq:lambda95over95re95lambda}\}\times\Lambda.\] Then, there exists a \(\delta>0\) such that whenever \(|\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}|<\delta\operatorname{\mathbb{R}\mathbb{e}}\left\{\lambda\right\}\) and \(|\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}|<\delta\) we have \[|\phi_\infty(x,y,s)-\phi_\infty(x,y,s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0}| < \frac{c|\lambda|}{2}(|x|^2+|y|^2),\] with \(c\) as in 99 . Therefore, \[\begin{align} &\operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}_\infty(x,y,s) \geq \operatorname{\mathbb{R}\mathbb{e}}\left\{\phi\right\}_\infty(x,y,s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0} - |\phi_\infty(x,y,s)-\phi_\infty(x,y,s)|_{\operatorname{\mathbb{I}\mathbb{m}}\left\{\lambda\right\}=\operatorname{\mathbb{I}\mathbb{m}}\left\{B\right\}=0}| \notag\\ &\geq c|\lambda|(|x|^2+|y|^2) - \frac{c|\lambda|}{2}(|x|^2+|y|^2) = \frac{c}{2}|\lambda|(|x|^2+|y|^2), \end{align}\] for sufficiently small \(c_\lambda, c_B\). This completes the proof of the first part of the lem.

For the prefactor estimate, we have shown in 42 that \(|P(s)|= Ce^{-\operatorname{\mathbb{R}\mathbb{e}}\left\{E\right\}_0 s}(1+\text{ERROR}_2)\) with \(|\text{ERROR}_2|\leq C\exp(-cs)\) for \(s>C_1\). Then it clearly follows that \(|P(s)|\leq C'e^{-\operatorname{\mathbb{R}\mathbb{e}}\left\{E\right\}_0 s}\) for some controlled constant \(C'>0\). This completes the proof of the lem. ◻

Proof. We first define \[G_N(x,y,\lambda,\mu) = \int_{1/N}^N \frac{e^{\mu s}}{\lambda}\left[q(x,y,s) - e^{-E_0 s}\psi_0(x, \lambda)\psi_0^*(y, \lambda^*)\right]ds,\] and will show that \(\lim_{N\to\infty}G_N(x,y,\lambda,\mu) = G(x,y,\lambda,\mu)\) uniformly in \((x,y,\lambda,\mu)\) varying in any compact subset of \(\mho\). This will give the desired result, since \(G_N(x,y,\lambda,\mu)\) is analytic in \((\lambda,\mu)\) for fixed \((x,y)\) and continuous in \((x,y,\lambda,\mu)\).

To show the uniform convergence, we need to show that for any \(\epsilon>0\) and any compact subset \(K\subset\mho\), there exists an \(N_0>0\) such that for all \(N>N_0\) and all \((x,y,\lambda,\mu)\in K\) we have \[\begin{align} \label{eq:analyticity95uniform95convergence} |G(x,y,\lambda,\mu) - G_N(x,y,\lambda,\mu)|&\notag\\ &\leq \int_0^{1/N}\Big|e^{\mu s}\left[q(x,y,s) - e^{-E_0 s}\psi_0(x, \lambda)\psi_0^*(y, \lambda^*)\right]\Big| \notag\\ &+ \int_N^\infty \Big|e^{\mu s}\left[q(x,y,s) - e^{-E_0 s}\psi_0(x, \lambda)\psi_0^*(y, \lambda^*)\right]\Big| \notag\\ &< \epsilon. \end{align}\tag{100}\] To prove 100 , we fix \(x_0,y_0,\lambda_0,\mu_0\in\mho\). Then for small \(\delta>0\) depending on \(x_0,y_0,\lambda_0,\mu_0\), we prove both statements: \[\begin{align} &\lim_{\tau\to0}\int_0^\tau \Big|e^{\mu s}\left[q(x,y,s) - e^{-E_0 s}\psi_0(x, \lambda)\psi_0^*(y, \lambda^*)\right]\Big| ds = 0,\quad \text{and}\notag\\ &\lim_{T\to\infty}\int_T^\infty \Big|e^{\mu s}\left[q(x,y,s) - e^{-E_0 s}\psi_0(x, \lambda)\psi_0^*(y, \lambda^*)\right]\Big| ds = 0, \end{align}\] uniformly for \((x,y,\lambda,\mu)\) satisfying \[\label{eq:uniform95convergence95conditions} |x-x_0|,|y-y_0|,|\lambda-\lambda_0|,|\mu-\mu_0|<\delta.\tag{101}\] This will establish and complete the proof of the theorem.

Since \(x_0\neq y_0\) for \((x_0,y_0,\lambda_0,\mu_0)\in\mho\), 101 implies that \[|x-y| > \frac{|x_0-y_0|}{2} > 0, \quad \text{and}\quad |x^2|+|y^2| < 2(|x_0|^2+|y_0|^2),\] provided \(\delta\) is small enough in 101 . Therefore, we can use the first and second main theorems, specifically 35 40, and estimates 82 , 85 , 86 to estimate the first integral: \[\begin{align} &\int_0^\tau \Big|e^{\mu s}\left[q(x,y,s) - e^{-E_0 s}\psi_0(x, \lambda)\psi_0^*(y, \lambda^*)\right]\Big| ds \notag\\ &\leq \int_0^\tau \frac{C}{s}\exp\left(-c|\lambda|\left[s(|x|^2+|y|^2)+\frac{|x-y|^2}{s}\right]\right)ds + \int_0^\tau C\exp\left(-c|\lambda|(|x|^2+|y|^2)\right) ds \notag\\ &\leq \int_0^\tau \frac{C}{s}\exp\left(-c|\lambda|\frac{|x_0-y_0|^2}{s}\right)ds + \int_0^\tau C\exp\left(-c|\lambda|(|x_0|^2+|y_0|^2)\right) ds \to 0 \quad \text{as } \tau\to0, \end{align}\] uniformly for \((x,y,\lambda,\mu)\) satisfying 101 .

For the second integral, we use the analysis of regions 5 and 6 in the proof of the first and second main theorems, 37 38, to prove that \[\int_T^\infty \Big|e^{\mu s}\left[q(x,y,s) - e^{-E_0 s}\psi_0(x, \lambda)\psi_0^*(y, \lambda^*)\right]\Big| ds \to 0 \quad \text{as } T\to\infty,\] uniformly for \((x,y,\lambda,\mu)\) satisfying 101 . This completes the proof of the theorem. ◻

Proof. Through the proof we write \(e_\lambda\) and \(e_\lambda^{\rm MHO}\) for the ground state eigenvalues of \(h_\lambda\) and \(h_\lambda^{\rm MHO}\) respectively.

Let \[\begin{align} v_{\rm err}(x) := v(x) - \left(-1+\frac{1}{2}\left\langle x,\left(\nabla\otimes\nabla^\ast v\right)\left(0\right)x\right\rangle\right)\,. \end{align}\] Then as we remarked, \(\left|v_{\rm err}(x)\right|=O(\left\lVert x\right\rVert^3)\) as \(x\to0\). Hence \[\begin{align} \label{eq:MHO32quasimode} \left\lVert\left(h_\lambda - \left(-\lambda^2+e_\lambda^{\rm MHO}\right)\mathbb{1}\right)\varphi_\lambda^{\rm MHO}\right\rVert = \lambda^2 \left\lVert v_{\rm err}(X)\varphi_\lambda^{\rm MHO}\right\rVert\,. \end{align}\tag{103}\]

We recall the exact ground state of the magnetic harmonic oscillator is of the form \[\begin{align} \label{kbafntcj} \varphi_\lambda^{\rm MHO}(x) := C \sqrt{\lambda}\exp\left(-\lambda \langle x, S x \rangle\right)\qquad(x\in\mathbb{R}^2) \end{align}\tag{104}\] for some \(2\times 2\) matrix \(S>0\), and constant \(C<\infty\), both dependent on the Hessian of \(v\) at the origin. Using this, we can estimate \[\begin{align} \label{eq:potential32remainder32on32MHO32ground32state} \lambda^2 \left\lVert v_{\rm err}(X)\varphi_\lambda^{\rm MHO}\right\rVert \lesssim \lambda^{1/2} + \operatorname{e}^{-c\lambda }\lesssim \lambda^{1/2}\,. \end{align}\tag{105}\]

Now, for any \(\psi\in L^2(\mathbb{R}^2)\) such that \(\psi\perp \varphi_\lambda\), \[\begin{align} \left\lVert\left(h_\lambda - e_\lambda\mathbb{1}\right)\psi\right\rVert \geq \frac{1}{\left\lVert\psi\right\rVert}\langle \psi, \left(h_\lambda-e_\lambda\mathbb{1}\right)\psi \rangle \geq c_{\rm gap} \left\lVert\psi\right\rVert \end{align}\] where, by 9 , \(c_{\rm gap} = \left(e_1^{\rm MHO} - e_0^{\rm MHO}\right)\lambda+O(\lambda^{1/2})\). Thus, if \(P:=\varphi_\lambda\otimes\varphi_\lambda^\ast\), then \[\begin{align} \left\lVert P^\perp\psi\right\rVert \leq \frac{1}{c_{\rm gap}}\left\lVert\left(h_\lambda - e_\lambda\mathbb{1}\right)\psi\right\rVert \qquad(\psi\in L^2(\mathbb{R}^2))\,. \end{align}\] Applying this with \(\varphi_\lambda^{\rm MHO}\) and using 103 , 105 as well as 9 again, we find \[\begin{align} \left\lVert P^\perp\varphi_\lambda^{\rm MHO}\right\rVert \lesssim \lambda^{-1/2}\,. \end{align}\] ◻

Proof. We have \[\begin{align} \left(H_\lambda - \left(-\lambda^2+e_\lambda^{\rm MHO}\right)\mathbb{1}\right)\widehat{R}^{ d}\varphi_\lambda^{\rm MHO} &= \widehat{R}^{ d} \left(h_\lambda - \left(-\lambda^2+e_\lambda^{\rm MHO}\right)\mathbb{1}\right)\varphi_\lambda^{\rm MHO}+\widehat{R}^{ d}\lambda^2v(X-2d)\varphi_\lambda^{\rm MHO} \end{align}\] so by the proof of 105 , we have \[\begin{align} \left\lVert\left(H_\lambda - \left(-\lambda^2+e_\lambda^{\rm MHO}\right)\mathbb{1}\right)\widehat{R}^{ d}\varphi_\lambda^{\rm MHO}\right\rVert \lesssim \lambda^{1/2} + \exp\left(-c\lambda 2d\right) \lesssim \lambda^{1/2}\,. \end{align}\]

Otherwise the proof proceeds as in 105 and we get the same result. ◻