Proof. Step 1. Starting from \[\begin{align} (\mathrm{e}^{-\mathrm{i}\lambda x}\psi)'= \mathrm{i}e^{-\mathrm{i}\lambda x}qC_+\overline{q}\psi, \end{align}\] using that \(qC_+\overline{q}\psi\in L^1(\mathbb{R})\) and \(\psi\rightarrow 0\) at infinity, we get \[\begin{align} \psi(x)= \mathbb{1}_{\{x\leq 0\}}\mathrm{i}\int^x_{-\infty}\mathrm{e}^{\mathrm{i}\lambda (x-y)}qC_+\overline{q}\psi \, \mathrm{d}y-\mathbb{1}_{\{x>0\}}\mathrm{i}\int_{x}^\infty \mathrm{e}^{\mathrm{i}\lambda (x-y)}qC_+\overline{q}\psi\, \mathrm{d}y. \end{align}\]

For \(R>0\), taking \[\begin{align} H(R):=\int_{|y|>R}|qg|\, \mathrm{d}y, \qquad g:=C_+\overline{q}\psi, \end{align}\] it follows that \[\begin{align} \label{eqn:psiH} |\psi(x)|\leq H(|x|). \end{align}\tag{6}\]

In the following we denote \[\begin{align} Q(R):=\|q\|_{L^2(|y|>R)}, \quad F(R):=\|\psi\|_{L^2(|y|<R)}. \end{align}\] We begin by estimating \(H(R)\), splitting \(g=g_{0,R}+g_{1,R}\) with \[\begin{align} g_{0,R}:=C_+\mathbb{1}_{\{|y|>R/2\}}\overline{q}\psi,\qquad g_{1,R}:=C_+\mathbb{1}_{\{y<R/2\}}\overline{q}\psi. \end{align}\] For the first part, using the \(L^2\) boundedness of \(C_+\), we have \[\begin{align} \|q g_{0,R}\|_{L^1(|y|>R)}\leq Q(R)\|q \psi\|_{L^2(|y|>R/2)}&\leq Q(R)Q(R/2)\|\psi\|_{L^\infty(|y|>R/2)}\\&\leq Q(R/2)^2 H(R/2), \end{align}\] where we have used 6 and the monotonicity of \(Q\) and \(H\) in the last step.

For the second part we use the property of \(C_+\) that for \(f\in L^2( \mathbb{R})\) and a.e. \(x\in\mathbb{R}\) we have \[\begin{align} (C_+f)(x)= \lim_{\varepsilon \rightarrow 0_+}\frac{1}{2\pi \mathrm{i}}\int_{\mathbb{R}}\frac{f(y)}{y-x-\mathrm{i}\varepsilon}\, \mathrm{d}y. \end{align}\] Thus, for \(|x|>R\), since \(|y|<R/2<|x|/2\) and so \(|y-x|>|x|/2\), we get \[\begin{align} g_{1,R}(x)&=\frac{1}{2\pi \mathrm{i}}\int_{|y|<R/2}\frac{q(y)\psi(y)}{y-x}\, \mathrm{d}y \end{align}\] and \[\begin{align} |g_{1,R}(x)|&\leq \frac{1}{\pi|x|}\int_{|y|<R/2}|q(y)\psi(y)|\, \mathrm{d}y\leq \frac{\|q\|_{L^2(\mathbb{R})}}{\pi|x|}F(R/2). \end{align}\] Introducing \[\begin{align} A(R):=\int_{|y|>R}\frac{|q(y)|}{|y|}\, \mathrm{d}y, \end{align}\] we deduce that \[\begin{align} \|qg_{1,R}\|_{L^1(|y|>R)}\leq C F(R/2)A(R) \end{align}\] with \(C:=\pi^{-1}\|q\|_{L^2(\mathbb{R})}\).

Putting the estimates on \(qg_{0,R}\) and \(qg_{1,R}\) together, we have \[\begin{align} \label{eqn:Hbound} H(R) & \leq Q(R) Q(R/2) H(R/2) + C F(R/2) A(R) \notag \\ & \leq Q(R/2)^2 H(R/2)+CF(R/2)A(R). \end{align}\tag{7}\] To show that \(\psi\in L^2(\mathbb{R})\) we will use that, in view of 6 , for any \(R_0>0\), \[\begin{align} \label{eqn:psiHL2} \int_\mathbb{R}|\psi(x)|^2\, \mathrm{d}x\leq 2R_0\|\psi\|_{L^\infty(\mathbb{R})}^2+2\int_{R_0}^\infty H(R)^2\, \mathrm{d}R, \end{align}\tag{8}\] so that the statement follows if we can show the integral on the right is finite.

For any \(0<R_0<L\), it follows from 7 that \[\begin{align} \label{eqn:H2Int} \int_{R_0}^LH(R)^2\, \mathrm{d}R\leq 2\int_{R_0}^L Q(R/2)^4 H(R/2)^2\, \mathrm{d}R+2C^2\int_{R_0}^LF(R/2)^2A(R)^2\, \mathrm{d}R. \end{align}\tag{9}\] Since \(Q(R)\rightarrow 0\) as \(R\rightarrow \infty\), we can choose \(R_0\) such that for any \(R\geq R_0\) we have \[\begin{align} 2\int_{R_0}^LQ(R/2)^4 H(R/2)^2\, \mathrm{d}R \leq C_{R_0} +\frac{1}{2}\int_{R_0}^LH(R)^2\, \mathrm{d}R. \end{align}\] where \[C_{R_0} := 2\int_{R_0}^{2R_0} Q(R/2)^4 H(R/2)^2\, \mathrm{d}R \,.\] Note that \(C_{R_0}<\infty\) since \(Q\) and \(H\) are bounded.

As a consequence, 9 implies \[\begin{align} \int_{R_0}^LH(R)^2\, \mathrm{d}R \leq 2 C_{R_0} + 4 C^2 \int_{R_0}^LF(R/2)^2A(R)^2\, \mathrm{d}R. \end{align}\]

Using 6 , and \[\begin{align} F(R/2)^2& \leq F(R_0)^2+ 2 \int_{R_0}^R H(T)^2\, \mathrm{d}T, \end{align}\] we get \[\begin{align} \int_{R_0}^LH(R)^2\, \mathrm{d}R & \leq 2C_{R_0} + 4C^2 F(R_0)^2 \int_{R_0}^L A(R)^2\, \mathrm{d}R \\ & \quad + 8 C^2 \int_{R_0}^L A(R)^2 \left(\int_{R_0}^R H(T)^2\, \mathrm{d}T\right)\, \mathrm{d}R. \end{align}\] Applying Grönwall’s inequality gives \[\begin{align} \int_{R_0}^L H(R)^2 \, \mathrm{d}R\leq \left( 2C_{R_0} + 4C^2 F(R_0)^2 \int_{R_0}^L A(R)^2\, \mathrm{d}R \right) e^{8 C^2 \int_{R_0}^L A(R)^2\, \mathrm{d}R}. \end{align}\] If \(A\in L^2(R_0,\infty)\) then taking \(L\rightarrow\infty\) in the above gives \(H\in L^2(R_0,\infty)\) and hence \(\psi\in L^2(\mathbb{R})\) by 8 .

Since \[\begin{align} \|A\|_{L^2(R_0,\infty)}&=\left(\int_{R_0}^\infty\left(\int_{|x|>R}\frac{|q(x)|}{|x|}\, \mathrm{d}x\right)^2\, \mathrm{d}R \right)^{1/2}\\ &=\left(\int_{R_0}^\infty\left(\int_{|y|>1}|q(R y)|\frac{\, \mathrm{d}y}{|y|}\right)^2\, \mathrm{d}R \right)^{1/2}\\ &= \int_{|y|>1}\left(\int_{R_0}^\infty|q(R y)|^2\, \mathrm{d}R\right)^{1/2}\frac{\, \mathrm{d}y}{|y|} \end{align}\] where we have used Minkowski in the last step. Rescaling in the integrand gives \[\begin{align} \|A\|_{L^2(R_0,\infty)}&=\int_{1}^\infty\left(\int_{R_0}^\infty|q(R y)|^2\, \mathrm{d}R\right)^{1/2}\frac{\, \mathrm{d}y}{|y|}+\int_{1}^\infty\left(\int_{R_0}^\infty|q(-R y)|^2\, \mathrm{d}R\right)^{1/2}\frac{\, \mathrm{d}y}{|y|}\\ &\leq \int_{1}^\infty\left(\int_{R_0}^\infty|q(R)|^2\frac{\, \mathrm{d}R}{|y|}\right)^{1/2}\frac{\, \mathrm{d}y}{|y|}+\int_{1}^\infty\left(\int_{R_0}^\infty|q(-R)|^2\frac{\, \mathrm{d}R}{|y|} \right)^{1/2}\frac{\, \mathrm{d}y}{|y|}\\ &\leq \|q\|_{L^2(\mathbb{R})} \int_{1}^\infty\frac{\, \mathrm{d}y}{y^{3/2}}\leq 2\|q\|_{L^2(\mathbb{R})}. \end{align}\] This concludes the proof of \(\psi\in L^2(\mathbb{R})\).

Step 2. To show the asymptotic decay we note that \(F(R/2)\leq \|\psi\|_{L^2(\mathbb{R})}\), hence from 7 we have \[\begin{align} H(R)\leq Q(R)Q(R/2)H(R/2)+C \|\psi\|_{L^2(\mathbb{R})} A(R) \,. \end{align}\] As \[\begin{align} A(R)=\int_{|y|>R}\frac{|q(y)|}{|y|}\, \mathrm{d}y\leq (R/2)^{-1/2}Q(R), \end{align}\] we obtain \[Q(R)^{-1} R^{1/2} H(R) \leq \left( 2^{1/2} Q(R/2)^2 \right) Q(R/2)^{-1} (R/2)^{1/2} H(R/2) + C'\] with \(C' := 2^{1/2} C \|\psi\|_{L^2(\mathbb{R})}\).

We can choose \(R_0'\) such that for all \(R\geq R_0'\) we have \(2^{1/2} Q(R/2)^2\leq1/2\). Thus, for all \(R\geq 2R_0'\). \[Q(R)^{-1} R^{1/2} H(R) \leq \frac{1}{2} Q(R/2)^{-1} (R/2)^{1/2} H(R/2) + C' \,.\] Taking \(R_n := 2^n R_0'\) and \(a_n:=Q(R_n)^{-1}R_n^{1/2}H(R_n)\) for \(n\geq 0\) we get \[a_{n+1} \leq \frac{1}{2} a_n + C' \qquad\text{for all}\;n\geq 0 \,.\] It follows that, for any \(N\geq 1\), \[a_N \leq C'\sum_{n=0}^{N-1}2^{-n}+2^{-N}a_0\lesssim 1.\] Thus, we have shown that \[H(R_N) \lesssim Q(R_N) R_N^{-1/2} \,.\] By 6 and monotonicity of \(H\), we conclude that for \(|x|>R_0\) \[|\psi(x)| \leq H(|x|) \lesssim |x|^{-1/2} \|q\|_{L^2(|y|>|x|/2)},\] This completes the proof of the theorem. ◻

Proof. From Theorem 1 we have \(\psi\in L^2(\mathbb{R})\). From the proof we have \[\begin{align} \int_{|x|<L} |\psi(x)|\, \mathrm{d}x\leq 2R_0\|\psi\|_{L^\infty(\mathbb{R})}+2\int_{R_0}^L H(R)\, \mathrm{d}R \end{align}\] and \[\begin{align} \int_{R_0}^L H(R)\, \mathrm{d}R\leq \int_{R_0}^L Q(R/2)^2H(R/2)\, \mathrm{d}R+C \int_{R_0}^L F(R/2) A(R)\, \mathrm{d}R. \end{align}\] Choosing \(R_0\) large enough, we can absorb the first term on the left to get \[\begin{align} \int_{R_0}^L H(R)\, \mathrm{d}R\leq 2 C_{R_0}' + 2C \int_{R_0}^L F(R/2) A(R)\, \mathrm{d}R \end{align}\] with \(C_{R_0} := \int_{R_0}^{2R_0} Q(R/2)^2H(R/2)\, \mathrm{d}R\). Note that \(F(R/2)\leq\|\psi\|_{L^2(\mathbb{R})}\). Thus, by taking \(L\rightarrow \infty\), \(\psi\in L^1(\mathbb{R})\) will follow if \(A\in L^1(R_0,\infty)\).

Using Fubini–Tonelli, it follows that \[\begin{align} \|A\|_{L^1(R_0,\infty)}&=\int_{R_0}^\infty \int_{|x|>R}\frac{|q(x)|}{|x|}\, \mathrm{d}x\, \mathrm{d}R\\ &\leq \int_{|x|>R_0}\frac{|q(x)|}{|x|}\left(\int_{R_0}^{|x|} \, \mathrm{d}R\right)\, \mathrm{d}x\leq \int_{\mathbb{R}} |q(x)|\, \mathrm{d}x. \end{align}\] This shows that \(\psi\in L^1(\mathbb{R})\).

The asymptotic decay follows from an analogous argument as in Step 2 of the proof of Theorem 1, using \[A(R)\leq R^{-1}\|q\|_{L^1(|x|>R)}.\] Indeed, we can find \(R_0'\) such that for \(|x|\geq R_0'\), \[|\psi(x)|\lesssim |x|^{-1}\max\{\|q\|_{L^1(|y|>|x|/2)},\|q\|_{L^2(|y|>|x|/2)}\}.\qedhere\] ◻

Proof. We multiply the equation \[-\mathrm{i}\psi' - q C_+(\overline{q}\psi) = \lambda \psi\] by \(x\overline{\psi}\) and take the imaginary part to find \[- \frac{1}{2} x (|\psi|^2)' - x\mathop{\mathrm{Im}}(\overline{\psi} q C_+(\overline{q} \psi)) = 0 \,.\] Let \(L>0\) be a parameter that will tend to infinity at the end. Integrating the previous identity over \((-L,L)\) yields \[- \frac{1}{2} x |\psi|^2 \Big|_{-L}^L + \frac{1}{2} \int_{-L}^L |\psi|^2\, \mathrm{d}x - \int_{-L}^L x \mathop{\mathrm{Im}}(\overline{\psi} q C_+(\overline{q} \psi))\, \mathrm{d}x = 0 \,.\] As \(L\to\infty\), the second term on the left side tends to \(\frac{1}{2}\|\psi\|_2^2\) and, by 5 , the first term vanishes. We write the third term as \[\begin{align} \int_{-L}^L x \mathop{\mathrm{Im}}(\overline{\psi} q C_+(\overline{q} \psi))\, \mathrm{d}x & = \int_{-L}^L x \mathop{\mathrm{Im}}(\overline{\psi} q C_+(\mathbb{1}_{\{|y|< L\}} \overline{q} \psi))\, \mathrm{d}x \\ & \quad + \int_{-L}^L x \mathop{\mathrm{Im}}(\overline{\psi} q C_+(\mathbb{1}_{\{|y|\geq L\}} \overline{q} \psi))\, \mathrm{d}x \,. \end{align}\] We note that \[\begin{align} \int_{-L}^L x \mathop{\mathrm{Im}}(\overline{\psi} q C_+(\mathbb{1}_{\{|y|\leq L\}} \overline{q} \psi))\, \mathrm{d}x & = \mathop{\mathrm{Im}}\frac{1}{2\pi\mathrm{i}} \iint_{(-L,L)\times(-L,L)} \frac{x}{y-x} \overline{\psi(x)} q(x) \overline{q(y)} \psi(y)\, \mathrm{d}x\, \mathrm{d}y \\ & = \frac{1}{4\pi} \left| \int_{-L}^L \overline{q(x)} \psi(x)\, \mathrm{d}x \right|^2. \end{align}\] As \(L\to\infty\), this converges to \[\frac{1}{4\pi} \left| \int_\mathbb{R}\overline{q(x)} \psi(x)\, \mathrm{d}x \right|^2.\] This shows that the identity in the lemma is equivalent to \[\lim_{L\to\infty} \int_{-L}^L x \mathop{\mathrm{Im}}(\overline{\psi} q C_+(\mathbb{1}_{\{|y|\geq L\}} \overline{q} \psi))\, \mathrm{d}x = 0 \,.\] We decompose \[\int_{-L}^L x \mathop{\mathrm{Im}}(\overline{\psi} q C_+(\mathbb{1}_{\{|y|\geq L\}} \overline{q} \psi))\,dx = I_1 + I_2\] with \[\begin{align} I_1 & := \int_{L/2<|x|<L} x \mathop{\mathrm{Im}}(\overline{\psi} q C_+(\mathbb{1}_{\{|y|\geq L\}} \overline{q} \psi))\, \mathrm{d}x \,,\\ I_2 & := \int_{|x|<L/2} x \mathop{\mathrm{Im}}(\overline{\psi} q C_+(\mathbb{1}_{\{|y| \geq L\}} \overline{q} \psi))\, \mathrm{d}x \,, \end{align}\] and show that both terms tend to zero as \(L\to\infty\).

To bound \(I_1\), we use the \(L^2(\mathbb{R})\) boundedness of \(C_+\) to see that \[\begin{align} |I_1| & \leq \|\mathbb{1}_{\{L/2<|x|<L\}} x \psi q\|_{L^2(\mathbb{R})}\|\mathbb{1}_{\{|y| \geq L\}}\overline{q}\psi\|_{L^2(\mathbb{R})}\\ &\leq L \|\psi\|_{L^\infty(|x|>L/2)}^2\|q\|_{L^2(\mathbb{R})}^2 \,. \end{align}\] The right side tends to zero from 5 .

For \(I_2\) we use that for \(|x|<L/2\) and \(|y|\geq L\) we have \(|y-x|\geq L/2\), so that \(|x/(y-x)|\leq 1\) and \[\begin{align} |I_2| &\leq \frac{1}{2\pi} \int_{|y|>L}\int_{|x|<L/2} |\overline{\psi(x)} q(x)| |\overline{q(y)} \psi(y)|\, \mathrm{d}x\, \mathrm{d}y\\ &= \frac{1}{2\pi}\int_{|x|<L/2}|\psi(x)q(x)|\, \mathrm{d}x\int_{|y|>L}|\psi(y)q(y)|\, \mathrm{d}y. \end{align}\] Since \(q\psi\in L^1(\mathbb{R})\), one factor is bounded while the other goes to zero. ◻

Proof. Step 1. The fact that convolution with \(G_k\) defines a bounded operator from \(L^1_+\) to \(L^\infty\) follows immediately from the fact that \(G_k\in L^\infty(\mathbb{R})\) for all \(k\in\widehat{\mathbb{C}}\). The fact that for \(k\in\mathbb{C}\setminus\mathbb{R}\) this coincides with \(R_0(k)\) on \(L^2_+\cap L^1_+\) follows from 13 .

Step 2. Let us show the following fact: If \(k\in\widehat{\mathbb{C}}\) and \((k_n)\subset\widehat{\mathbb{C}}\) satisfy \(k_n\to k\) in \(\widehat{\mathbb{C}}\), then for any \(f\in L^1_+(\mathbb{R})\) we have \(R_0(k_n)f\to R_0(k)f\) pointwise and boundedly. Once this is shown, it follows by dominated convergence that \((R_0(k_n)f,g)\to (R_0(k)f,g)\) for any \(g\in L^1_+(\mathbb{R})\).

We write \(k=\lambda+\mu\mathrm{i}\) and \(k_n = \lambda_n+\mu_n\mathrm{i}\). We may assume that \(\mu=\mathop{\mathrm{Im}}k\geq 0\), the opposite case being similar. When \(\lambda>0\), we know that \(\mu_n\geq 0\) for all sufficiently large \(n\). When \(\lambda\leq 0\) we make the additional assumption that \(\mu_n\geq 0\) for all sufficiently large \(n\). This does not represent any loss of generality, because we can handle the case where \(\mu_n<0\) with the case where \(\mu\leq 0\) using 15 .

As a consequence, we have \[(R_0(k_n)f - R_0(k)f)(x) = \mathrm{i}\int_{-\infty}^x \mathrm{e}^{\mathrm{i}\lambda(x-y)} \mathrm{e}^{-\mu(x-y)} \left( \mathrm{e}^{\mathrm{i}(\lambda_n-\lambda)(x-y)} \mathrm{e}^{-(\mu_n-\mu)(x-y)} - 1 \right) f(y)\, \mathrm{d}y \,.\] When \(\mu>0\), we bound \[\left| \mathrm{e}^{\mathrm{i}(\lambda_n-\lambda)(x-y)} \mathrm{e}^{-(\mu_n-\mu)(x-y)} - 1 \right| \leq 1+ \mathrm{e}^{|\mu_n-\mu|(x-y)} \leq 2 \mathrm{e}^{|\mu_n-\mu|(x-y)} \,.\] In particular, when \(n\) is so large that \(|\mu_n-\mu|\leq\mu\), then \[| (R_0(k_n)f - R_0(k)f)(x) | \leq 2 \|f\|_1 \,.\] This bound and pointwise convergence of the integrand implies that \((R_0(k_n)f - R_0(k)f)(x)\to 0\) for each \(x\in\mathbb{R}\).

Now let \(\mu=0\). Using \(\mu_n\geq 0\) we can bound \[\left| \mathrm{e}^{\mathrm{i}(\lambda_n-\lambda)(x-y)} \mathrm{e}^{-\mu_n(x-y)} - 1 \right| \leq 1+ \mathrm{e}^{-\mu_n(x-y)} \leq 2 \,.\] Using this bound we can argue as before.

Step 3. Let us note that the operator \(C_+\overline{q}R_0(k)qC_+\) is bounded on \(L^2_+\). Indeed, \(C_+\) is a bounded operator on \(L^2\), multiplication by \(q\in L^2_+\) is bounded from \(L^2_+\) to \(L^1_+\), by part (a) \(R_0(k)\) is bounded from \(L^1_+\) to \(L^\infty\) and multiplication by \(q\) is bounded from \(L^\infty\) to \(L^2\).

In particular, we have for all \(k\in \widehat{\mathbb{C}}\) and \(\varphi\in L^2_+\) \[\begin{align} C_+\overline{q}R_0(k)qC_+\varphi=C_+\overline{q}G_k\ast q\varphi\,. \end{align}\] For the Hilbert–Schmidt norm, we obtain \[\begin{align} \|C_+\overline{q}R_0(k)qC_+\|_{\mathfrak{S}^2(L^2_+)}^2 & \leq \|\overline{q}G_k\ast (q\,\cdot)\|_{\mathfrak{S}^2(L^2)}^2 \\ &\leq \int_{-\infty}^\infty\int_{-\infty}^{\infty} {\left\vert q(x)\right\vert}^2{\left\vert G_k(x-y)\right\vert}^2{\left\vert q(y)\right\vert}^2\, \mathrm{d}x\, \mathrm{d}y \\ & \leq {\left\|G_k\right\|_{}}_{L^\infty}^2{\left\|q\right\|_{}}_2^4 \\ & \leq {\left\|q\right\|_{}}_2^4 \,. \end{align}\] Note, in particular, that this Hilbert–Schmidt norm is uniformly bounded in \(k\).

Step 4. Let us show continuity of \(k\mapsto C_+ \overline{q}R_0(k)q C_+\) at a point \(k=\lambda\pm\mathrm{i}0\). (Continuity at other points will follow from analyticity established in the next step.) Let \(h\in \widehat{\mathbb{C}}\) with \(\pm\mathop{\mathrm{Im}}h\geq 0\), then \[\begin{align} {\left\|C_+(\overline{q}R_0(k+h)q-\overline{q}R_0(k)q)C_+\right\|_{}}_{\mathfrak{S}^2}^2&\leq \|\overline{q}G_{k+h}\ast (q\,\cdot)-\overline{q}G_{k}\ast(q\,\cdot)\|_{\mathfrak{S}_2(L^2)}^2. \end{align}\] As \(h\to 0\), the pointwise convergence of \(G_{k+h}\) to \(G_k\) together with dominated convergence implies that the right side tends to zero. This proves the asserted continuity at \(\lambda>0\). It also proves continuity at \(\lambda\leq 0\) by the same argument as in Step 2.

Step 5. Finally, let us prove the analyticity of \(k\mapsto C_+\overline{q}R_0(k)qC_+\) in \(\mathbb{C}\setminus\mathbb{R}\), for the sake of concreteness in the upper half-plane. For \(\mathop{\mathrm{Im}}k>0\) and \(h\in\mathbb{C}\) with \({\left\vert h\right\vert}\) small we note that for each \(x\in \mathbb{R}\) \[\begin{align} D_hG_k(x)\mathrel{\vcenter{:}}= \frac{1}{h}(G_{k+h}(x)-G_k(x))=\mathrm{i}\mathrm{e}^{\mathrm{i}kx}\mathbb{1}_{\mathbb{R}_+}(x) \frac{1}{h} (\mathrm{e}^{\mathrm{i}hx}-1) \end{align}\] converges pointwise to \(-x \mathrm{e}^{\mathrm{i}kx}\mathbb{1}_{\mathbb{R}_+}(x)\) as \({\left\vert h\right\vert}\rightarrow 0\). Moreover, using Taylor expansion, \[\begin{align} {\left\vert D_hG_k(x) + x \mathbb{1}_{\mathbb{R}_+}(x)\mathrm{e}^{\mathrm{i}kx}\right\vert} & = \mathbb{1}_{\mathbb{R}_+}(x)\mathrm{e}^{ -\mathop{\mathrm{Im}}k x}{\left\vert\frac{\mathrm{e}^{\mathrm{i}hx}-1}{h}-\mathrm{i}x\right\vert} \\ & \leq \frac{1}{2} \mathrm{e}^{|\mathop{\mathrm{Im}}(hx)|} \mathbb{1}_{\mathbb{R}_+}(x){\left\vert h\right\vert} x^2\mathrm{e}^{ -\mathop{\mathrm{Im}}k x}, \end{align}\] which is uniformly bounded in \(x\) (if, say, \(|h|\leq \mathop{\mathrm{Im}}k/2\)). Thus, \(D_hG_k\) has a limit in \(L^\infty(\mathbb{R})\) and \(C_+\overline{q}R_0(k)qC_+\) is analytic on \(k\in \mathbb{C}\backslash\mathbb{R}\).

Note that since in the previous step we have established continuity of \(C_+\overline{q}R_0(k)qC_+\) as \(k\) approaches the real axis, we obtain analyticity in all of \(\mathbb{C}\backslash[0,\infty)\). ◻

Proof. We only prove the assertion when \(\mathop{\mathrm{Im}}k\geq 0\), the opposite case being similar.

Step 1. Let \(g\in\ker(1-C_+\overline{q} R_0(k)q C_+)\). When \(k\in\mathbb{C}\setminus\mathbb{R}\) it follows from the precise form of the resolvent identity that \(g=0\). Thus, in what follows we assume that \(k=\lambda+\mathrm{i}0\) with \(\lambda\in\mathbb{R}\). Since \(qC_+g\in L^1_+\), and since \(R_0(k)\) maps \(L^1_+\) into \(L^\infty\), we have \(\psi\in L^\infty\). Moreover, the formula \[\psi(x) = \mathrm{i}\int_{-\infty}^x \mathrm{e}^{\mathrm{i}\lambda(x-y)} q(y) (C_+g)(y)\, \mathrm{d}y\] implies that \(\psi\) is locally absolutely continuous and satisfies the claimed equation. (Here we use \(g=C_+\overline{q}\psi\) by the equation for \(g\).) Since \(q C_+g\in L^1(\mathbb{R})\), the formula for \(\psi\) implies, by dominated convergence, \[\psi(x) \to 0 \qquad\text{as}\;x\to-\infty\] and \[\mathrm{e}^{-\mathrm{i}\lambda x} \psi(x) \to \mathrm{i}\int_\mathbb{R}\mathrm{e}^{-\mathrm{i}\lambda y} q(y) (C_+g)(y)\, \mathrm{d}y = \mathrm{i}\int_\mathbb{R}\mathrm{e}^{-\mathrm{i}\lambda y}q(y) (C_+ \overline{q}\psi)(y)\, \mathrm{d}y \;\text{as}\;x\to+\infty \,.\]

We multiply the equation for \(\psi\) by \(\overline{\psi}\) and integrate it over a bounded interval \((a,b)\). Using \(2\mathop{\mathrm{Re}}\overline{\psi}\psi' = (|\psi|^2)'\), we find \[- \frac{\mathrm{i}}{2} (|\psi(b)|^2 - |\psi(a)|^2) + \mathop{\mathrm{Im}}\int_a^b \overline{\psi} \psi'\, \mathrm{d}x - \int_a^b \overline{\psi} q C_+\overline{q}\psi\, \mathrm{d}x = \lambda \int_a^b |\psi|^2\, \mathrm{d}x \,.\] Taking the imaginary part gives \[- \frac{1}{2} (|\psi(b)|^2 - |\psi(a)|^2) - \mathop{\mathrm{Im}}\int_a^b \overline{\psi} q C_+\overline{q}\psi\, \mathrm{d}x = 0 \,.\] We note that \(\overline{\psi} q C_+\overline{q}\psi\) belongs to \(L^1\). Therefore we can take the limit \(a\to-\infty\). Since we have already shown that \(\psi(x)\to 0\) as \(x\to-\infty\), we infer that \[-\frac{1}{2} |\psi(b)|^2 - \mathop{\mathrm{Im}}\int_{-\infty}^b \overline{\psi} q C_+\overline{q}\psi\, \mathrm{d}x = 0 \,.\] We now take the limit \(b\to\infty\). Since \(C_+\) is selfadjoint, the integral on the left side vanishes as \(b\to\infty\) and we deduce that \(\psi(b)\to 0\) as \(b\to\infty\). This proves the assertion about the limit at \(+\infty\) and, in view of what we proved before about the limit of \(\overline{\mathrm{e}(\lambda)}\psi\), the claimed vanishing of the integral.

Step 2. We now assume that \(\psi\in\mathcal{S}(k)\).

We first consider the case \(\mathop{\mathrm{Im}}k>0\) and aim at showing that \(\psi=0\). As before, we multiply the equation for \(\psi\) by \(\overline{\psi}\) and integrate over \((a,b)\). Taking the imaginary part gives \[-\frac{1}{2} (|\psi(b)|^2 - |\psi(a)|^2) - \mathop{\mathrm{Im}}\int_a^b \overline{\psi} q C_+ \overline{q} \psi\, \mathrm{d}x = \mathop{\mathrm{Im}}k \int_a^b |\psi|^2\, \mathrm{d}x \,.\] Using the assumed boundary conditions for \(\psi\) together with the fact that \(\overline{\psi} q C_+ \overline{q} \psi\) belongs to \(L^1\), we can let \(a\to-\infty\) and \(b\to+\infty\) to deduce that \[0 = \mathop{\mathrm{Im}}k \int_\mathbb{R}|\psi|^2\, \mathrm{d}x \,.\] Since \(\mathop{\mathrm{Im}}k> 0\), we conclude that \(\psi=0\), as claimed.

Now let \(k=\lambda+0\mathrm{i}\) with \(\lambda\in\mathbb{R}\). Note that \(\psi\in L^\infty\) implies \(g\in L^2_+(\mathbb{R})\). Moreover, the equation for \(\psi\) implies that \[(\overline{\mathrm{e}(\lambda)} \psi)'(x) = \overline{\mathrm{e}(\lambda)} (\psi' - \mathrm{i}\lambda \psi) = \mathrm{i}\overline{\mathrm{e}(\lambda)} q C_+ \overline{q} \psi = \mathrm{i}\overline{\mathrm{e}(\lambda)} q g \,.\] Since the right side belongs to \(L^1\) and since \(\mathrm{e}^{-\mathrm{i}\lambda x} \psi(x)\to 0\) as \(x\to-\infty\), we deduce that \[\mathrm{e}^{-\mathrm{i}\lambda x} \psi(x) = \mathrm{i}\int_{-\infty}^x \mathrm{e}^{-\mathrm{i}\lambda y} q(y) g(y)\, \mathrm{d}y \,,\] that is, \[\psi = R_0(\lambda+0\mathrm{i}) q g \,.\] Multiplying this equation by \(\overline{q}\) and applying \(C_+\), we arrive at \(g = C_+\overline{q}R_0(\lambda+0\mathrm{i})qg\), as claimed.

Step 3. Finally, we prove that the maps \(g\mapsto\psi\) in (a) and \(\psi\mapsto g\) in (b) are inverses to each other. If \(g\in\ker(1-C_+\overline{q} R_0(\lambda\pm 0\mathrm{i})q C_+)\) and \(\psi:=R_0(\lambda\pm 0\mathrm{i}) qC_+g\), then clearly \(C_+ \overline{q} \psi = C_+ \overline{q} R_0(\lambda\pm 0\mathrm{i})q C_+ g = g\), proving one direction.

For the other direction, let \(\psi\in S(\lambda\pm 0\mathrm{i})\) and \(g:=C_+\overline{q}\psi\). Then \(\tilde{\psi} := R_0(\lambda\pm 0\mathrm{i})qC_+ g= R_0(\lambda\pm 0\mathrm{i})qC_+\overline{q}\psi\) satisfies \(-\mathrm{i}\tilde{\psi}' - qC_+\overline{q}\psi = \lambda\tilde{\psi}\) and \(\tilde{\psi}(x)\to 0\) as \(x\to\mp\infty\). Thus, \(\chi:=\tilde{\psi}-\psi\) satisfies \(-\mathrm{i}\chi' = \lambda\chi\) and \(\chi(x)\to 0\) as \(x\to\mp\infty\). This implies \(\chi=0\), that is \(\tilde{\psi}=\psi\), as claimed. ◻

Proof. Let \(\lambda\in\mathcal{N}_+\) and \(0\neq g\in\ker (1-C_+\overline{q}R_0(\lambda+ 0\mathrm{i})qC_+)\). Then, by part (a) of Lemma 4 the function \(\psi:=R_0(\lambda+0\mathrm{i})qC_+g\) belongs to \(\mathcal{S}(\lambda+0\mathrm{i})\) and satisfies \(\psi(x)\to 0\) as \(|x|\to\infty\). According to Theorem 1, this implies \(\psi\in L^2(\mathbb{R})\). Since, again by Lemma 4, \(\psi\neq 0\), we conclude that \(\lambda\in \sigma_{\rm p}(L_q)\). The proof for \(\lambda\in\mathcal{N}_-\) is similar.

Conversely, if \(\lambda\in\sigma_{\rm p}(L_q)\cap\overline{\mathbb{R}_+}\), then there is a \(\psi\neq 0\) satisfying \(-i\psi' - qC_+ \overline{q}\psi = \lambda\psi\). Moreover, from [3] we know that \(\psi\in\mathop{\mathrm{dom}}L_q = H^1_+(\mathbb{R})\mathrel{\vcenter{:}}= H^1(\mathbb{R})\cap L^2_+(\mathbb{R})\). Since all functions in \(H^1_+(\mathbb{R})\) belong to \(L^\infty\) and tend to zero at both infinities, we are in the situation of part (b) of Lemma 4 and infer that \(C_+\overline{q} \psi\) belongs to both \(\ker (1-C_+\overline{q}R_0(\lambda\pm 0\mathrm{i})qC_+)\). Since \(\psi\neq 0\), the equation for \(\psi\) together with the boundary conditions imply that \(C_+\overline{q}\psi\neq 0\), so \(\lambda\in\mathcal{N}_+\cap\mathcal{N}_-\). ◻

Proof. Step 1. For all \(k\in\widehat{\mathbb{C}}\setminus(\sigma_{\mathrm p}(L_{q}))^\diamond\) we define \(R(k)\) as a bounded operator from \(L^1_+\) to \(L^\infty\) through the resolvent identity 12 . Indeed, the first term, \(R_0(k)\), is bounded from \(L^1_+\) to \(L^\infty\) for all \(k\in\widehat{\mathbb{C}}\) by Lemma 3. For the second term we note that, by Corollary 1, for \(k\in \widehat{\mathbb{C}}\setminus(\sigma_{\mathrm p}(L_{q}))^\diamond\) the operator \(1- C_+ \overline{q}R_0(k) qC_+\) is boundedly invertible on \(L^2_+\). Moreover, \(C_+\overline{q} R_0(k)\) maps \(L^1_+\) to \(L^2_+\) and, by duality, \(R_0(k)qC_+\) maps \(L^2_+\) to \(L^\infty\). This proves boundedness of \(R(k)\) from \(L^1_+\) to \(L^\infty\) for all \(k\in\widehat{\mathbb{C}}\setminus(\sigma_{\mathrm p}(L_{q}))^\diamond\).

Step 2. To prove the claimed continuity, we write, using 12 , \[{\left\langle R(k)f,g\right\rangle} _{} = {\left\langle R_0(k)f,g\right\rangle} _{} - \langle(1-C_+ \overline{q} R_0(k)q C_+)^{-1} C_+\overline{q}R_0(k)f,C_+\overline{q} R_0(\overline{k})g\rangle \,.\] According to Lemma 3, the first term on the right side is continuous. The fact that was shown in Step 2 of the proof of Lemma 3 also implies, by dominated convergence, that \(k\mapsto \overline{q} R_0(k)f\) is continuous in \(\widehat{\mathbb{C}}\) with values in \(L^2(\mathbb{R})\). Since \((1-C_+ \overline{q} R_0(k)q C_+)^{-1}\) is also continuous by Lemma 3, we obtain the claimed continuity of \({\left\langle R(k)f,g\right\rangle} _{}\). ◻

Proof. We already know that \(R(k)\) is well defined for \(k\in\widehat{\mathbb{C}}\setminus\sigma_\mathrm{p}(L_q)^\diamond\). It is easy to see that \(R(k)f\) in case (a) and \(R(\lambda\pm 0\mathrm{i})f\) in case (b) satisfies the claimed equation and boundary conditions.

It remains to prove uniqueness. Clearly it suffices to show that the corresponding homogeneous equation, together with the corresponding boundary conditions, has only the trivial solution. This is a consequence of Lemma 4. Indeed, this lemma implies immediately that \(\mathcal{S}(k)=\emptyset\) if \(\mathop{\mathrm{Im}}k\neq 0\). When \(k=\lambda\pm 0\mathrm{i}\), the lemma implies that \(\mathcal{S}(k)\neq\emptyset\) if and only if \(\ker(1-C_+ \overline{q} R_0(k)qC_+)\neq\emptyset\). When \(\lambda<0\) (so that \(R_0(\lambda)\) is a bounded operator on \(L^2_+(\mathbb{R})\)), it is elementary to see that this happens if and only if \(\lambda\in\sigma_\mathrm{p}(L_q)\), which is excluded in part (a). When \(\lambda\geq 0\) this happens, by definition, if and only if \(\lambda\in\sigma_{\mathrm p}(L_{q})\), which is excluded in part (b). Therefore, under the assumptions of the lemma, there is no nontrivial solution of the homogeneous equation, proving uniqueness of the solution of the inhomogeneous equation. ◻

Proof. The function \(\psi:=m_{\rm e}(\lambda\pm 0\mathrm{i})-\mathrm{e}(\lambda)\in L^\infty\) satisfies the equation and asymptotics in part (b) of Lemma 6 with \(f=q C_+ \overline{q}\mathrm{e}(\lambda)\in L^1_+\) and is therefore unique. ◻

Proof. To see the continuity, we insert the definition of \(m_{\rm e}(\lambda+0\mathrm{i})\) into the definition of \(\Gamma\) and find \[\Gamma(\lambda) = 1 + \mathrm{i}\langle C_+ \overline{q} \mathrm{e}(\lambda), \overline{q} \mathrm{e}(\lambda)\rangle + \mathrm{i}\langle R(\lambda+0\mathrm{i})q C_+ \overline{q} \mathrm{e}(\lambda), q C_+ \overline{q} \mathrm{e}(\lambda) \rangle \,.\] Since \(\lambda\mapsto\overline{q} \mathrm{e}(\lambda)\) is continuous with values in \(L^2\) (by dominated convergence), the second term on the right side is continuous and \(\lambda\mapsto q C_+ \overline{q} \mathrm{e}(\lambda)\) is continuous with values in \(L^1\). Since \(R(\lambda+0\mathrm{i})\) is uniformly bounded on compact subsets of \([0,\infty)\setminus\sigma_\mathrm{p}(L_q)\), we can deduce the continuity of the third term from the continuity statement of Lemma 5.

The proof of 19 uses the following, valid for \(\lambda>0\), \[\begin{align} \label{eqn:Ginprodfg} \langle G_{\lambda\pm 0\mathrm{i}}\ast f,g\rangle -\langle f,G_{\lambda\pm 0\mathrm{i}}\ast g\rangle=\pm \mathrm{i}\langle f,\mathrm{e}(\lambda)\rangle \overline{\langle g,\mathrm{e}(\lambda)\rangle }, \end{align}\tag{21}\] which can be seen easily from \[\begin{align} \label{eqn:Gpmdiff} G_{\lambda+0\mathrm{i}}(x)-G_{\lambda-0\mathrm{i}}(x)=\mathrm{i}\, \mathrm{e}(x,\lambda); \end{align}\tag{22}\] see also [30].

Using this and the definition of \(m_\mathrm{e}\), we compute \[\begin{align} \mathrm{i}{\left\vert{\left\langle qC_+\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i}),\mathrm{e}(\lambda)\right\rangle} _{}\right\vert}^2=&{\left\langle G_{\lambda+0\mathrm{i}}\ast qC_+\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i}),qC_+\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i})\right\rangle} _{}\\&-{\left\langle qC_+\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i}),G_{\lambda+0\mathrm{i}}\ast qC_+\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i})\right\rangle} _{}\\ =&{\left\langle m_\mathrm{e}(\lambda+0\mathrm{i})-\mathrm{e}(\lambda),qC_+\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i})\right\rangle} _{}\\&-{\left\langle qC_+\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i}),m_\mathrm{e}(\lambda+0\mathrm{i})-\mathrm{e}(\lambda)\right\rangle} _{}\\ =&2\mathrm{i}\mathop{\mathrm{Im}}{\left\langle qC_+\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i}),\mathrm{e}(\lambda)\right\rangle} _{}, \end{align}\] where we used that \(qC_+\overline{q}\) is self-adjoint in the last step. Using that \(\Gamma(\lambda)=1+\mathrm{i}{\left\langle qC_+\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i}),\mathrm{e}(\lambda)\right\rangle} _{}\), we find \[\begin{align} -2\mathrm{i}\mathop{\mathrm{Re}}(1-\Gamma(\lambda))+\mathrm{i}{\left\vert 1-\Gamma(\lambda)\right\vert}^2=0 \,, \end{align}\] from which it follows that \(|\Gamma(\lambda)|=1\).

To show 20 , we start from the integral equation for \(m_\mathrm{e}(\lambda+0\mathrm{i})\) and apply 22 , \[\begin{align} m_\mathrm{e}(\lambda+0\mathrm{i})&=\mathrm{e}(\lambda)+G_{\lambda+0\mathrm{i}}\ast(qC_+\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i}))\\ &=\mathrm{e}(\lambda)+(\mathrm{i}\, \mathrm{e}(\lambda)+G_{\lambda-0\mathrm{i}})\ast(qC_+\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i}))\\ &=\Gamma(\lambda)\mathrm{e}(\lambda)+R_0(\lambda-0\mathrm{i})qC_+\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i}) \,. \end{align}\] By rearranging and solving for \(m_\mathrm{e}(\lambda+0\mathrm{i})\), we find the desired identity. ◻

Proof. Step 1. Let \(\lambda\in[0,\infty)\setminus\sigma_\mathrm{p}(L_q)\) and \(f\in L^1_+(\mathbb{R})\). We will show that \[\begin{align} R(\lambda+0\mathrm{i})f-R(\lambda-0\mathrm{i})f=\mathrm{i}\sqrt{2\pi} \, (\Phi f)(\lambda) \, m_\mathrm{e}(\lambda-0\mathrm{i}). \end{align}\]

Indeed, the functions \(R(\lambda\pm0\mathrm{i}) f\) satisfy the integral equations \[R(\lambda\pm0\mathrm{i}) f = G_{\lambda\pm0\mathrm{i}}* f + G_{\lambda\pm0\mathrm{i}}* (qC_+ \overline{q} R(\lambda\pm0\mathrm{i}) f) \,.\] Subtracting these two equations and using 22 , we find \[\begin{align} R(\lambda+0\mathrm{i}) f - R(\lambda-0\mathrm{i}) f & = (G_{\lambda+0\mathrm{i}} - G_{\lambda-0\mathrm{i}})* f + (G_{\lambda+0\mathrm{i}} - G_{\lambda-0\mathrm{i}})* (qC_+ \overline{q} R(\lambda+0\mathrm{i}) f) \\ & \quad + G_{\lambda-0\mathrm{i}}*(qC_+ \overline{q} (R(\lambda+0\mathrm{i})f - R(\lambda-0\mathrm{i}) f)) \\ & = c(\lambda) \mathrm{e}(\lambda) + G_{\lambda-0\mathrm{i}}*(qC_+ \overline{q} (R(\lambda+0\mathrm{i})f - R(\lambda-0\mathrm{i}) f)) \end{align}\] with \[c(\lambda) = \mathrm{i}\int_\mathbb{R}\left( f(x) + qC_+\overline{q} R(\lambda+0\mathrm{i}) f \right) \mathrm{e}^{-\mathrm{i}\lambda x}\,\, \mathrm{d}x \,.\] By the definition of \(m_\mathrm{e}(\lambda\pm\mathrm{i}0)\) in 17 , it follows that \[R(\lambda+0\mathrm{i}) f - R(\lambda-0\mathrm{i}) f = c(\lambda)\, m_\mathrm{e}(\lambda-0\mathrm{i}) \,.\] This will give the claimed identity, provided we can prove that \[c(\lambda) = \mathrm{i}\sqrt{2\pi} \, (\Phi f)(\lambda) \,.\]

Observe that, using the equations for \(R(\lambda\pm 0\mathrm{i})f\) and \(m_\mathrm{e}(\lambda-0\mathrm{i})\), \[\begin{align} \mathrm{i}{\left\langle qC_+\overline{q}R(\lambda+0\mathrm{i})f,\mathrm{e}(\lambda)\right\rangle} _{} \, &\overline{{\left\langle qC_+\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i}),\mathrm{e}(\lambda)\right\rangle} _{}}\\=&{\left\langle G_{\lambda+0\mathrm{i}}\ast qC_+\overline{q}R(\lambda+0\mathrm{i})f,qC_+\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i})\right\rangle} _{}\\&-{\left\langle qC_+\overline{q}R(\lambda+0\mathrm{i})f,G_{\lambda+0\mathrm{i}}\ast qC_+\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i})\right\rangle} _{}\\ =&{\left\langle R(\lambda+0\mathrm{i})f-G_{\lambda+0\mathrm{i}}\ast f,qC_+\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i})\right\rangle} _{}\\&-{\left\langle qC_+\overline{q}R(\lambda+0\mathrm{i})f,m_\mathrm{e}(\lambda+0\mathrm{i})-\mathrm{e}(\lambda)\right\rangle} _{}\\ =&-{\left\langle G_{\lambda+0\mathrm{i}}\ast f,qC_+\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i})\right\rangle} _{}+{\left\langle qC_+\overline{q}R(\lambda+0\mathrm{i})f,\mathrm{e}(\lambda)\right\rangle} _{}, \end{align}\] where we used the self-adjointness of \(qC_+\overline{q}\) in the last step. Rearranging gives \[\begin{align} \label{eqn:sctreleqn1} \langle G_{\lambda+0\mathrm{i}}\ast f, q C_+ \overline{q} m_\mathrm{e}(\lambda+0\mathrm{i})\rangle- \overline{\Gamma(\lambda)}\langle q C_+ \overline{q}R(\lambda+0\mathrm{i})f,\mathrm{e}(\lambda)\rangle =0. \end{align}\tag{29}\] Meanwhile, using 21 again and the equation for \(m_\mathrm{e}\), \[\begin{align} \mathrm{i}\langle f, \mathrm{e}(\lambda)\rangle \,& \overline{\langle qC_+\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i}),\mathrm{e}(\lambda) \rangle} \\ = & \langle G_{\lambda+0\mathrm{i}}\ast f, q C_+ \overline{q} m_\mathrm{e}(\lambda+0\mathrm{i})\rangle - \langle f, G_{\lambda+0\mathrm{i}}\ast qC_+\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i}) \\ = & \langle G_{\lambda+0\mathrm{i}}\ast f, q C_+ \overline{q} m_\mathrm{e}(\lambda+0\mathrm{i})\rangle - \langle f,m_\mathrm{e}(\lambda+0\mathrm{i})\rangle + \langle f,\mathrm{e}(\lambda)\rangle \,. \end{align}\] Using 20 we can rewrite this as \[\begin{align} \label{eqn:sctreleqn2} \langle G_{\lambda+0\mathrm{i}}\ast f, q C_+ \overline{q} m_\mathrm{e}(\lambda+0\mathrm{i})\rangle=-\overline{\Gamma(\lambda)}{\left\langle f,\mathrm{e}(\lambda)\right\rangle} _{}+\overline{\Gamma(\lambda)}\langle f,m_\mathrm{e}(\lambda-0\mathrm{i})\rangle. \end{align}\tag{30}\] Putting 29 and 30 together, gives \[\begin{align} -\overline{\Gamma(\lambda)}\langle f,\mathrm{e}(\lambda)\rangle +\overline{\Gamma(\lambda)}\langle f,m_\mathrm{e}(\lambda-0\mathrm{i})\rangle-\overline{\Gamma(\lambda)} \langle q C_+ \overline{q} R(\lambda+0\mathrm{i})f,\mathrm{e}(\lambda)\rangle =0 \end{align}\] and thus, since \({\left\vert\Gamma(\lambda)\right\vert}=1\neq 0\) by 19 ,we have \[\begin{align} \langle f,m_\mathrm{e}(\lambda-0\mathrm{i})\rangle- \langle qC_+ \overline{q} R(\lambda+0\mathrm{i})f+f,\mathrm{e}(\lambda)\rangle=0, \end{align}\] or equivalently, using 24 , \[\begin{align} \sqrt{2\pi}\Phi(\lambda)+\mathrm{i}c(\lambda)=0, \end{align}\] which is the claimed identity.

Step 2. Let \(\Lambda\subset[0,\infty)\) be a bounded interval with \(\overline{\Lambda}\cap \sigma_\mathrm{p}(L_q)=\emptyset\) and let \(f,g\in L^2_{+}(\mathbb{R})\cap L^1_+(\mathbb{R})\). According to Lemma 5, \({\left\langle R(k)f,g\right\rangle} _{}\) is continuous with respect to \(k\). We obtain, using Step 1, that \[\frac{1}{2\pi\mathrm{i}} \left( {\left\langle R(\lambda+\mathrm{i}\varepsilon) f,g\right\rangle} _{} - {\left\langle R(\lambda-\mathrm{i}\varepsilon) f,g\right\rangle} _{} \right) \to (\Phi f)(\lambda)\, \overline{(\Phi g)(\lambda)} \,,\] and these asymptotics are uniform for \(\lambda\in\overline{\Lambda}\).

Meanwhile, by Stone’s formula we have \[\frac{1}{2\pi\mathrm{i}} \int_\Lambda \left( {\left\langle R(\lambda+\mathrm{i}\varepsilon) f,g\right\rangle} _{} - {\left\langle R(\lambda-\mathrm{i}\varepsilon) f,g\right\rangle} _{} \right)\, \mathrm{d}\lambda \to \frac{1}{2} \left( {\left\langle\mathbb{1}_{\overline{\Lambda}}(L_q) f, g\right\rangle} _{} + \langle{\mathbb{1}_{\overset\circ\Lambda}(L_q) f, g}\rangle \right).\] Since the endpoints of \(\Lambda\) are, by assumption, not in \(\sigma_\mathrm{p}(L_q)\) and therefore by Corollary 1 not eigenvalues of \(L_q\), the right side is equal to \({\left\langle\mathbb{1}_{\Lambda}(L_q) f, g\right\rangle} _{}\).

Noting that \({\left\langle R(\lambda\pm\mathrm{i}\varepsilon) f,g\right\rangle} _{}\) is uniformly bounded in \((\lambda,\varepsilon)\in\Lambda\times(0,1]\), we obtain the claimed formula by dominated convergence. ◻

Proof of Theorem 2. As we have mentioned before, the set \(\sigma_\mathrm{p}(L_q)\) is closed and therefore the open set \(\mathbb{R}_+\setminus\sigma_\mathrm{p}(L_q)\) is the countable union of disjoint open intervals, \(\mathbb{R}_+\setminus\sigma_\mathrm{p}(L_q)=\bigcup_n I_n\). Fix one of these intervals \(I_n\) and let \(K\) be a compact subinterval of \(I_n\). For each \(f\in L^2_+(\mathbb{R})\cap L^1_+(\mathbb{R})\), the function \(|\Phi f|^2\) is bounded on \(K\) and therefore, by Lemma 9, the spectral measure \(\Lambda\mapsto (\mathbb{1}_{\Lambda}(L_q)f,f)\) is absolutely continuous on \(K\), that is, \(\mathbb{1}_K(L_q) f\) belongs to the absolutely continuous subspace of \(L_q\). Since \(L^2_+(\mathbb{R})\cap L^1(\mathbb{R})\) is dense in \(L^2_+(\mathbb{R})\), we deduce that \(\mathop{\mathrm{ran}}\mathbb{1}_K(L_q)\) is contained in the absolutely continuous subspace of \(L_q\), that is, the spectum of \(L_q\) is purely absolutely continuous on \(K\). Since \(K\) is an arbitrary compact subinterval of \(I_n\), it follows that the spectrum of \(L_q\) is purely absolutely continuous on \(I_n\). This implies that the spectrum is purely absolutely continuous on \(\bigcup_n I_n =\mathbb{R}_+\setminus\sigma_\mathrm{p}(L_q)\), as claimed. ◻

Proof of Theorem 3. Step 1. We show that \(\Phi\) can be extended to a bounded operator from \(L^2_+(\mathbb{R})\) to \(L^2(\mathbb{R}_+)\) that satisfies the first equality in 25 .

Let \(f\in L^2_+(\mathbb{R})\cap L^1_+(\mathbb{R})\). We write \(\mathbb{R}_+\setminus\sigma_\mathrm{p}(L_q)\) as a countable union of disjoint open intervals. Approximating each one of these intervals by compact intervals and using monotone convergence, we deduce from Lemma 9 that \[{\left\langle\mathbb{1}_{\mathbb{R}_+\setminus\sigma_\mathrm{p}(L_q)} f,f\right\rangle} _{} = \int_{\mathbb{R}_+} |(\Phi f)(\lambda)|^2\,\, \mathrm{d}\lambda \,.\] On the right side we write \(\mathbb{R}_+\) as integration domain rather than \(\mathbb{R}_+\setminus\sigma_\mathrm{p}(L_q)\), the value of \((\Phi f)(\lambda)\) for \(\lambda\) in the null set \(\sigma_\mathrm{p}(L_q)\) being irrelevant. Note that since \[{\left\langle\mathbb{1}_{\mathbb{R}_+\setminus\sigma_\mathrm{p}(L_q)} f,f\right\rangle} _{}\leq \|f\|_2^2 \,,\] we infer that \(\Phi f\) is square-integrable. Moreover, by Theorem 2 we have \(\mathbb{1}_{\mathbb{R}_+\setminus\sigma_\mathrm{p}(L_q)} = P_\mathrm{ac}(L_q)\), so \[{\left\langle P_\mathrm{ac}(L_q) f,f\right\rangle} _{} = \int_{\mathbb{R}_+} |(\Phi f)(\lambda)|^2\,\, \mathrm{d}\lambda \,.\] By density of \(L^2_+(\mathbb{R})\cap L^1(\mathbb{R})\) in \(L^2_+(\mathbb{R})\), we infer that \(\Phi\) extends to a bounded linear operator from \(L^2_+(\mathbb{R})\) to \(L^2(\mathbb{R}_+)\) satisfying the first equality in 25 .

Step 2. We shall prove ‘one half’ of 26 and 27 . Specifically, we shall show the inclusion \(\subset\) in 26 and the equality in 27 for \(f\in\mathop{\mathrm{dom}}L_q\).

First, for \(f\in\mathcal{S}_+(\mathbb{R})\mathrel{\vcenter{:}}= \mathcal{S}(\mathbb{R})\cap L^2_+(\mathbb{R})\) and \(\lambda\in\mathbb{R}_+\setminus\sigma_\mathrm{p}(L_q)\) we can use the fact that \(m_\mathrm{e}(\lambda-0\mathrm{i})\) satisfies the equation \(L_q m_\mathrm{e}(\lambda-0\mathrm{i}) = \lambda m_\mathrm{e}(\lambda-0\mathrm{i})\) in the sense of tempered distributions to deduce that \[(\Phi L_q f)(\lambda) = \lambda (\Phi f)(\lambda) \,.\] Since the operator norm of \(L_q\) is equivalent to the \(H^1(\mathbb{R})\)-norm and since \(\mathcal{S}_+(\mathbb{R})\) is dense in \(H^1_+(\mathbb{R})\), we easily deduce that, if \(f\in\mathop{\mathrm{dom}}L_q\), then \(\int_{\mathbb{R}_+} \lambda^2 |(\Phi f)(\lambda)|^2\,\, \mathrm{d}\lambda<\infty\) and \((\Phi L_qf)(\lambda) = \lambda (\Phi f)(\lambda)\) for almost all \(\lambda\in\mathbb{R}_+\).

Step 3. Let us show the second equality in 25 . Taking into account the first equality there, which we have already proved, it suffices to show that \(\mathop{\mathrm{ran}}\Phi\) is dense in \(L^2(\mathbb{R}_+)\). To do this, suppose that there is a \(h\in L^2(\mathbb{R}_+)\) that is orthogonal to \(\mathop{\mathrm{ran}}\Phi\). Thus \[\begin{align} {\left\langle\Phi \mathbb{1}_\Lambda(L_{q}) f,h\right\rangle} _{}=0 \end{align}\] for every bounded interval \(\Lambda\subset\mathbb{R}_+\) with \(\overline{\Lambda}\cap\sigma_\mathrm{p}(L_q)=\emptyset\) and every \(f\in L^2_+(\mathbb{R})\cap L^1(\mathbb{R})\). According to Step 2 and Lemma 10 below this means that for all such \(\Lambda\) and \(f\) we have \[\begin{align} 0=\int_\Lambda \int_{-\infty}^\infty f(y) \overline{m_\mathrm{e}(y,\lambda-0\mathrm{i})}\, \mathrm{d}y\overline{h(\lambda)}\, \mathrm{d}\lambda =\int_{-\infty}^\infty f(y)\overline{\int_{\Lambda} h(\lambda)m_\mathrm{e}(y,\lambda-0\mathrm{i})\, \mathrm{d}\lambda }\, \mathrm{d}y \,. \end{align}\] The interchange of integrals is allowed since \(m_\mathrm{e}(\lambda-0\mathrm{i})\) is bounded for \(\lambda\) in the intervals \(\Lambda\) under consideration.

Since \(f\) is arbitrary, we deduce that for all such \(\Lambda\) we have for almost every \(x\in\mathbb{R}\) \[\begin{align} \int_{\Lambda} h(\lambda)m_\mathrm{e}(x,\lambda-0\mathrm{i})\, \mathrm{d}x=0 \,. \end{align}\] Restricting to \(\Lambda\) with rational endpoints, say, we may assume that the full measure set of \(x\)’s is independent of \(\Lambda\). Fixing now \(x\) in this full measure set we obtain, by shrinking \(\Lambda\) and applying the Lebesgue differentiation theorem, that \[h(\lambda)m_\mathrm{e}(x,\lambda-0\mathrm{i})=0 \qquad\text{for a.e.}\;\lambda\in\mathbb{R}_+ \,.\] We also used the fact that \(\sigma_\mathrm{p}(L_q)\) is a null set.

We shall use the asymptotics of \(m_\mathrm{e}(x,\lambda-0\mathrm{i})\) as \(x\to\mp\infty\), see Lemma 7. From the integral equation 17 we easily see that these asymptotics are uniform for \(\lambda\) in compact subsets of \(\mathbb{R}_+\setminus\sigma_\mathrm{p}(L_q)\). Thus, fixing such a subset \(K\), we find \(x\) large enough and belonging to our full measure set such that \(m_\mathrm{e}(x,\lambda-0\mathrm{i})\neq 0\) for all \(\lambda\in K\). We deduce that \(h(\lambda)=0\) for a.e. \(\lambda\in K\). Since \(K\) is arbitrary, we find \(h\equiv 0\). Thus, \(\mathop{\mathrm{ran}}\Phi\) is dense in \(L^2(\mathbb{R}_+)\), as we wanted to prove.

Step 4. We finally turn to the proof of the ‘other half’ of 26 and 27 . Specifically, we shall show the inclusion \(\supset\) in 26 .

To do so, let \(f\in L^2_+(\mathbb{R})\) with \(\lambda\Phi f\in L^2(\mathbb{R}_+)\) and set \(g:=\Phi^*(\lambda\Phi f)\in L^2(\mathbb{R}_+)\). Then for any \(h\in L^2_+(\mathbb{R})\) \[\langle g, h \rangle = \int_{\mathbb{R}_+} \lambda (\Phi f)(\lambda) \overline{(\Phi h)(\lambda)}\,\, \mathrm{d}\lambda \,.\] Under the stronger assumption \(h\in\mathop{\mathrm{dom}}L_q\) we have by the first equality in 25 \[\begin{align} \langle f, L_q h\rangle & = \int_{\mathbb{R}_+} (\Phi f)(\lambda) \overline{(\Phi L_q h)(\lambda)}\,\, \mathrm{d}\lambda + \langle f,(1-P_\mathrm{ac}(L_q))L_q h \rangle \\ & = \int_{\mathbb{R}_+} \lambda (\Phi f)(\lambda) \overline{(\Phi h)(\lambda)}\,\, \mathrm{d}\lambda + \langle f,(1-P_\mathrm{ac}(L_q)) L_q h \rangle \,. \end{align}\] Here in the last equality we used Step 2.

In Theorem 5 below we will show that the set \(\sigma_\mathrm{p}(L_q)\) is bounded. As a consequence, we have \(\widetilde{g}:= L_q(1-P_\mathrm{ac}(L_q))f\in L^2_+(\mathbb{R})\).

To summarise, we have shown that for \(h\in\mathop{\mathrm{dom}}L_q\) we have \[\langle f, L_q h\rangle = \langle g + \widetilde{g}, h \rangle \,.\] This proves that \(f\in\mathop{\mathrm{dom}}L_q^*\) and, since \(L_q\) is self-adjoint, that \(f\in\mathop{\mathrm{dom}}L_q\). This concludes the proof. ◻

Proof. Let \(g\in \mathcal{H}\) and \(z\in\mathbb{C}_+\). Then \((A-z)^{-1} g\in\mathop{\mathrm{dom}}A\) and therefore, by assumption 31 with \(f=(A-z)^{-1} g\), we have \[C g = (B-z) C (A-z)^{-1} g \,.\] Applying the bounded operator \((B-z)^{-1}\) we obtain the operator identity \[(B-z)^{-1} C = C (A-z)^{-1} \,.\] By the spectral theorem, this implies, for any \(f\in\mathcal{H}\), \(\varphi\in\mathcal{G}\) \[\int_\mathbb{R}\frac{d(\mathbb{1}_{(-\infty,\lambda)}(B)Cf,\varphi)}{\lambda-z} = \int_\mathbb{R}\frac{d(\mathbb{1}_{(-\infty,\lambda)}(A)f,C^*\varphi)}{\lambda-z} \,.\] Thus, the Stieltjes transform of the finite, signed measure \[d\left( (\mathbb{1}_{(-\infty,\lambda)}(B)Cf,\varphi) - (\mathbb{1}_{(-\infty,\lambda)}(A)f,C^*\varphi)\right)\] vanishes on \(\mathbb{C}_+\). This implies that the measure vanishes, that is, for any Borel set \(\Lambda\subset\mathbb{R}\) we have \[(\mathbb{1}_{\Lambda}(B)Cf,\varphi) - (\mathbb{1}_{\Lambda}(A)f,C^*\varphi) = 0 \,.\] Since \(f\) and \(\varphi\) are arbitrary, this implies 32 . ◻

Proof. The continuity of \(\beta\) follows writing \[\begin{align} \beta(\lambda) = \mathrm{i}{\left\langle q,\mathrm{e}(\lambda)\right\rangle} _{} +{\left\langle R(\lambda+0\mathrm{i})q, q C_+\overline{q}\mathrm{e}(\lambda)\right\rangle} _{} \end{align}\] and arguing similarly to Lemma 8, in particular, using the continuity of \(\lambda\mapsto\overline{q}\mathrm{e}(\lambda)\) in \(L^2\) and the continuity of Fourier transforms.

For 36 , we use the integral equation 33 for \(m_0\) and apply the relation 22 to find \[\begin{align} m_0(\lambda+0\mathrm{i})-m_0(\lambda-0\mathrm{i}) & = \left(G_{\lambda+0\mathrm{i}}-G_{\lambda-0\mathrm{i}}\right)\ast q+(G_{\lambda+0\mathrm{i}}-G_{\lambda-0\mathrm{i}})\ast (qC_+\overline{q}m_0(\lambda+0\mathrm{i}))\\ & \quad +G_{\lambda-0\mathrm{i}}\ast(qC_+\overline{q}(m_0(\lambda+0\mathrm{i})-m_0(\lambda-0\mathrm{i})))\\ & = \mathrm{i}\, \mathrm{e}(\lambda)\ast q+\mathrm{i}\mathrm{e}(\lambda)\ast (qC_+\overline{q}m_0(\lambda+0\mathrm{i}))\\ & \quad +G_{\lambda-0\mathrm{i}}\ast(qC_+\overline{q}(m_0(\lambda+0\mathrm{i})-m_0(\lambda-0\mathrm{i}))) \,. \end{align}\] The identity follows by rearranging for \(m_0(\lambda+0\mathrm{i})-m_0(\lambda-0\mathrm{i})\) and comparing with the definitions of \(m_\mathrm{e}(\lambda-0\mathrm{i})\) and \(\beta(\lambda)\).

The statement 37 follows directly from Step 1 of the proof of Lemma 9, using that \(\beta(\lambda)=c(\lambda)\) with \(f=q\).

For 38 , using 21 and the integral equation 33 for \(m_0\) we compute \[\begin{align} \mathrm{i}{\left\vert\beta(\lambda)\right\vert}^2=&\mathrm{i}{\left\vert{\left\langle qC_+\overline{q}m_0(\lambda+0\mathrm{i})+q,\mathrm{e}(\lambda)\right\rangle} _{}\right\vert}^2\\=&{\left\langle G_{\lambda+0\mathrm{i}}\ast (qC_+\overline{q}m_0(\lambda+0\mathrm{i})+q),qC_+\overline{q}m_0(\lambda+0\mathrm{i})+q\right\rangle} _{}\\&-{\left\langle qC_+\overline{q}m_0(\lambda+0\mathrm{i})+q,G_{\lambda+0\mathrm{i}}\ast (qC_+\overline{q}m_0(\lambda+0\mathrm{i})+q)\right\rangle} _{}\\ =&{\left\langle m_0(\lambda+0\mathrm{i}),qC_+\overline{q}m_0(\lambda+0\mathrm{i})+q\right\rangle} _{}-{\left\langle qC_+\overline{q}m_0(\lambda+0\mathrm{i})+q,m_0(\lambda+0\mathrm{i})\right\rangle} _{} \\=&{\left\langle m_0(\lambda+0\mathrm{i}),q\right\rangle} _{}-{\left\langle q,m_0(\lambda+0\mathrm{i})\right\rangle} _{}=2\mathrm{i}\mathop{\mathrm{Im}}{\left\langle m_0(\lambda+0\mathrm{i}),q\right\rangle} _{}, \end{align}\] where we used the self-adjointness of \(qC_+\overline{q}\). This establishes 38 . ◻

Proof. The proof is based on the operator identity \[\label{eqn:Chprop} \overline{\mathrm{e}(\lambda)} C_+ \mathrm{e}(\lambda) = C_+ + C_\lambda\tag{41}\] with \(C_\lambda\mathrel{\vcenter{:}}=\mathcal{F}^{-1}\mathbb{1}_{(-\lambda,0)}\mathcal{F}\) for \(\lambda\geq 0\), where \(\mathcal{F}\) is the Fourier transform, \[(\mathcal{F}f)(\xi) = \frac{1}{\sqrt{2\pi}} \int_\mathbb{R}\mathrm{e}^{-\mathrm{i}\xi x} f(x)\, \mathrm{d}x \,.\]

Step 1. We show that for every \(\varphi\in L^\infty\) the \(L^1\)-valued function \(\lambda\mapsto qC_\lambda \overline{q}\varphi\) is differentiable on \([0,\infty)\) with \[\partial_\lambda q C_\lambda\overline{q}\varphi= \frac{1}{2\pi} \int_\mathbb{R}\mathrm{e}^{\mathrm{i}\lambda y} \overline{q(y)}\varphi(y) \, \mathrm{d}y \;\overline{\mathrm{e}(\lambda)} q \,.\]

To see this, let \(\lambda\geq 0\) and let \(h\in\mathbb{R}\) with \(\lambda+h\geq 0\). Then \[h^{-1} q(x)((C_{\lambda+h} - C_\lambda)\overline{q}\varphi)(x) = \frac{1}{\sqrt{2\pi} h} q(x) \int_{-\lambda-h}^{-\lambda} \mathrm{e}^{\mathrm{i}\xi x} \mathcal{F}(\overline{q}\varphi)(\xi) \, \mathrm{d}\xi \,.\] Since \(q\in L^1\) and \(\varphi\in L^\infty\) the quantity \(h^{-1} \int_{-\lambda-h}^{-\lambda} \mathrm{e}^{\mathrm{i}\xi x} \mathcal{F}(\overline{q}\varphi)(\xi) \, \mathrm{d}\xi\) is uniformly bounded with respect to \(x\) and \(h\) and converges, for each \(x\in\mathbb{R}\), to \(\mathrm{e}^{-\mathrm{i}\lambda x} \mathcal{F}(\overline{q}\varphi)(-\lambda)\) as \(h\to 0\). Using again \(q\in L^1\) and dominated convergence we see that as \(h\to 0\) we have \[h^{-1} q(C_{\lambda+h} - C_\lambda)\overline{q}\varphi\to \frac{1}{\sqrt{2\pi}}\mathcal{F}(\overline{q}\varphi)(-\lambda) \, q \, \overline{\mathrm{e}(\lambda)}\] in \(L^1\). This proves the claimed differentiability.

Step 2. For \(\lambda\in[0,\infty)\), we note that setting, for \(\varphi\in L^\infty\), \[\widetilde{T}_{\lambda\pm\mathrm{i}0}\varphi:= \overline{\mathrm{e}(\lambda)} \, R_0(\lambda\pm0\mathrm{i}) qC_+\overline{q} \mathrm{e}(\lambda)\varphi\] defines a bounded operator on \(L^\infty\). We shall show that for every \(\varphi\in L^\infty\), the \(L^\infty\)-valued function \(\lambda\mapsto \widetilde{T}_{\lambda\pm\mathrm{i}0}\varphi\) is differentiable on \([0,\infty)\) with \[\partial_\lambda \widetilde{T}_{\lambda\pm\mathrm{i}0}\varphi= \frac{1}{2\pi} \int_\mathbb{R}\mathrm{e}^{\mathrm{i}\lambda y} \overline{q(y)}\varphi(y)\, \mathrm{d}y \, \;\overline{\mathrm{e}(\lambda)} \;R_0(\lambda\pm\mathrm{i}0) q \,.\]

Indeed, it follows from 41 that \[\begin{align} \widetilde{T}_{\lambda\pm\mathrm{i}0}\varphi&=\mathrm{i}\int_{\mp\infty}^x\mathrm{e}^{-\mathrm{i}\lambda y}qC_+\mathrm{e}(\lambda)\overline{q}\varphi\, \mathrm{d}y=\mathrm{i}\int_{\mp\infty}^xqC_+\overline{q}\varphi\, \mathrm{d}y+\mathrm{i}\int_{\mp\infty}^xqC_\lambda \overline{q}\varphi\, \mathrm{d}y, \end{align}\] and therefore, if \(h\in\mathbb{R}\) is so small that \(\lambda+h\geq 0\), then \[\begin{align} h^{-1}(\widetilde{T}_{\lambda+h\pm\mathrm{i}0}- \widetilde{T}_{\lambda\pm\mathrm{i}0})\varphi(x)= \mathrm{i}h^{-1}\int_{\mp\infty}^x q(C_{\lambda+h}-C_\lambda)\overline{q}\varphi\, \mathrm{d}y\,. \end{align}\] By Step 1 we see that as \(h\to 0\) we have, uniformly with respect to \(x\), \[\begin{align} h^{-1}(\widetilde{T}_{\lambda+h\pm\mathrm{i}0}-\widetilde{T}_{\lambda\pm\mathrm{i}0})\varphi(x) & \to \frac{1}{2\pi} \int_\mathbb{R}\mathrm{e}^{\mathrm{i}\lambda y'} \overline{q(y')}\varphi(y') \, \mathrm{d}y' \;\mathrm{i}\int_{\mp\infty}^x q\mathrm{e}^{-\mathrm{i}y\lambda }\, \mathrm{d}y \\ & \quad = \frac{1}{2\pi} \int_\mathbb{R}\mathrm{e}^{\mathrm{i}\lambda y'} \overline{q(y')}\varphi(y') \, \mathrm{d}y' \;\overline{\mathrm{e}(\lambda)} \;R_0(\lambda\pm\mathrm{i}0) q \,. \end{align}\] This proves the claimed differentiability.

Step 3. Let us show the differentiability of \(\lambda\mapsto\overline{\mathrm{e}(\lambda)}m_\mathrm{e}(\lambda\pm 0\mathrm{i})\) and the claimed formula for their derivatives.

It follows from the resolvent identity 12 that for all \(\lambda\in[0,\infty)\setminus\sigma_\mathrm{p}(L_q)\) \[1+R(\lambda\pm 0\mathrm{i}) q C_+ \overline{q} = (1- R_0(\lambda\pm 0\mathrm{i}) q C_+ \overline{q})^{-1}\] and, consequently, \[\label{eq:diffemeproof} \overline{\mathrm{e}(\lambda)} \left( 1+R(\lambda\pm 0\mathrm{i}) q C_+ \overline{q} \right) \mathrm{e}(\lambda) = (1- \widetilde{T}_{\lambda\pm 0\mathrm{i}})^{-1} \,.\tag{42}\] Thus, by definition of \(m_\mathrm{e}(\lambda\pm 0\mathrm{i})\) we have \[\label{eq:diffemeproof2} \overline{\mathrm{e}(\lambda)}m_\mathrm{e}(\lambda\pm 0\mathrm{i}) = (1- \widetilde{T}_{\lambda\pm 0\mathrm{i}})^{-1} 1 \,.\tag{43}\] Fix \(\lambda\in\mathbb{R}_+\setminus\sigma_\mathrm{p}(L_q)\). Note that by 42 the operators \((1- \widetilde{T}_{\lambda+h\pm 0\mathrm{i}})^{-1}\) exist for all sufficiently small \(h\in\mathbb{R}\). Moreover, it follows from Step 2 that \[\| \widetilde{T}_{\lambda+h\pm\mathrm{i}0}- \widetilde{T}_{\lambda\pm\mathrm{i}0} \|_{L^\infty\to L^\infty} \leq (2\pi)^{-1} \|q\|_1^2 |h| \,,\] so, in particular, \(\widetilde{T}_{\lambda+h\pm\mathrm{i}0} \to \widetilde{T}_{\lambda\pm\mathrm{i}0}\) in norm as \(h\to 0\). Since \[\begin{align} (1- \widetilde{T}_{\lambda+h\pm 0\mathrm{i}})^{-1} - (1- \widetilde{T}_{\lambda\pm 0\mathrm{i}})^{-1} & = - \left( 1 - \left(1-(1-\widetilde{T}_{\lambda\pm 0\mathrm{i}})^{-1}(\widetilde{T}_{\lambda+h\pm 0\mathrm{i}} - \widetilde{T}_{\lambda\pm 0\mathrm{i}}) \right)^{-1} \right) \\ & \quad \times (1- \widetilde{T}_{\lambda\pm 0\mathrm{i}})^{-1} \end{align}\] we see that \((1- \widetilde{T}_{\lambda+h\pm 0\mathrm{i}})^{-1} \to (1- \widetilde{T}_{\lambda\pm 0\mathrm{i}})^{-1}\) in norm as \(h\to 0\).

Equation 43 implies \[\begin{align} & h^{-1} \left( \overline{\mathrm{e}(\lambda+h)}m_\mathrm{e}(\lambda+h\pm 0\mathrm{i}) - \overline{\mathrm{e}(\lambda)}m_\mathrm{e}(\lambda\pm 0\mathrm{i}) \right) \\ & =(1- \widetilde{T}_{\lambda+h\pm 0\mathrm{i}})^{-1} h^{-1} \left( \widetilde{T}_{\lambda+h\pm\mathrm{i}0}- \widetilde{T}_{\lambda\pm\mathrm{i}0} \right) (1- \widetilde{T}_{\lambda\pm 0\mathrm{i}})^{-1} 1 \,. \end{align}\] According to Step 2, we have \[h^{-1} \left( \widetilde{T}_{\lambda+h\pm\mathrm{i}0}- \widetilde{T}_{\lambda\pm\mathrm{i}0} \right) (1- \widetilde{T}_{\lambda\pm0\mathrm{i}})^{-1} 1 \to \frac{1}{2\pi} \int_\mathbb{R}\mathrm{e}^{\mathrm{i}\lambda y} \overline{q(y)} (1- \widetilde{T}_{\lambda\pm 0\mathrm{i}})^{-1} 1 \, \mathrm{d}y \;\overline{\mathrm{e}(\lambda)} \;R_0(\lambda\pm \mathrm{i}0) q \,.\] Since \((1- \widetilde{T}_{\lambda+h\pm 0\mathrm{i}})^{-1} \to (1- \widetilde{T}_{\lambda\pm 0\mathrm{i}})^{-1}\) in norm, we conclude that \[\begin{align} & h^{-1} \left( \overline{e(\lambda+h)}m_\mathrm{e}(\lambda+h\pm 0\mathrm{i}) - \overline{e(\lambda)}m_\mathrm{e}(\lambda\pm 0\mathrm{i}) \right) \\ & \to \frac{1}{2\pi} \int_\mathbb{R}\mathrm{e}^{\mathrm{i}\lambda y} \overline{q(y)} (1- \widetilde{T}_{\lambda\pm 0\mathrm{i}})^{-1} 1 \, \mathrm{d}y \;(1- \widetilde{T}_{\lambda\pm 0\mathrm{i}})^{-1} \;\overline{\mathrm{e}(\lambda)} \;R_0(\lambda\pm\mathrm{i}0) q \,. \end{align}\] According to 42 and the resolvent identity, we have \[\begin{align} (1- \widetilde{T}_{\lambda\pm 0\mathrm{i}})^{-1} \;\overline{\mathrm{e}(\lambda)} \;R_0(\lambda\pm\mathrm{i}0) q & = \overline{\mathrm{e}(\lambda)} \, (1+R(\lambda\pm 0\mathrm{i}) q C_+\overline{q}) R_0(\lambda\pm 0\mathrm{i}) q \\&= \overline{\mathrm{e}(\lambda)}\, R(\lambda\pm 0\mathrm{i}) q \\ &= \overline{\mathrm{e}(\lambda)}\, m_0(\lambda\pm 0\mathrm{i}) \,. \end{align}\] Similarly, one finds \[(1- \widetilde{T}_{\lambda\pm 0\mathrm{i}})^{-1} 1 = \overline{\mathrm{e}(\lambda)} \, (1+R(\lambda\pm 0\mathrm{i}) q C_+\overline{q}) \mathrm{e}(\lambda) = \overline{\mathrm{e}(\lambda)} \, m_\mathrm{e}(\lambda\pm 0\mathrm{i})\] and therefore, using 37 and 20 , \[\int_\mathbb{R}\mathrm{e}^{\mathrm{i}\lambda y} \overline{q(y)} (1- \widetilde{T}_{\lambda\pm 0\mathrm{i}})^{-1} 1 \, \mathrm{d}y = \begin{cases} \mathrm{i}\, \overline{\beta(\lambda)} \, \Gamma(\lambda) & \text{for the upper sign} \,,\\ \mathrm{i}\, \overline{\beta(\lambda)} & \text{for the lower sign} \,. \end{cases}\] This proves the claimed differentiability.

Step 4. We shall show that for any \(\varphi\in L^\infty\), the \(L^1\)-valued map \(\lambda\mapsto \overline{\mathrm{e}(\lambda)}q C_+\overline{q}\mathrm{e}(\lambda)\varphi\) is differentiable on \([0,\infty)\) with \[\partial_\lambda \left( \overline{\mathrm{e}(\lambda)}q C_+\overline{q}\mathrm{e}(\lambda)\varphi\right) = \frac{1}{2\pi} \int_\mathbb{R}\mathrm{e}^{-\mathrm{i}\lambda y} \overline{q(y)} \varphi(y)\, \mathrm{d}y \;\overline{\mathrm{e}(\lambda)} q \,.\]

Indeed, in view of 41 we have \[\overline{\mathrm{e}(\lambda)}q C_+\overline{q}\mathrm{e}(\lambda)\varphi= qC_+\overline{q}\varphi+ q C_\lambda \overline{q}\varphi\,,\] so the assertion follows immediately from Step 1.

Step 5. Let us show the differentiability of \(\Gamma\).

From the definition 18 of \(\Gamma\) and the commutator identity 41 we get \[\begin{align} \Gamma(\lambda) & = 1 + \mathrm{i}\, \langle \overline{\mathrm{e}(\lambda)} m_\mathrm{e}(\lambda+0\mathrm{i}), \overline{\mathrm{e}(\lambda)} q C_+ \overline{q} \mathrm{e}(\lambda) \rangle \,. \end{align}\] The second term on the right side is differentiable by Steps 2 and 4, together with the fact that the \(L^\infty\) norm of \(\overline{\mathrm{e}(\lambda)} m_\mathrm{e}(\lambda+0\mathrm{i})\) and the \(L^1\) norm of \(\overline{\mathrm{e}(\lambda)} q C_+ \overline{q} \mathrm{e}(\lambda)\) are bounded on compact subsets of \(\mathbb{R}_+\setminus\sigma_\mathrm{p}(L_q)\). Using the formulas for the derivatives from Steps 2 and 4 we obtain \[\begin{align} \partial_\lambda \Gamma(\lambda) & = -\frac{\overline{\beta(\lambda)}\,\Gamma(\lambda)}{2\pi} \, \langle \overline{\mathrm{e}(\lambda)} m_0(\lambda+0\mathrm{i}), \overline{\mathrm{e}(\lambda)} q C_+ \overline{q} \mathrm{e}(\lambda) \rangle \\ & \quad + \frac{\mathrm{i}}{2\pi} \int_\mathbb{R}\mathrm{e}^{-\mathrm{i}\lambda y} q(y) \, \mathrm{d}y \,\langle \overline{\mathrm{e}(\lambda)} m_\mathrm{e}(\lambda+0\mathrm{i}), \overline{\mathrm{e}(\lambda)} q \rangle \\ & = -\frac{\overline{\beta(\lambda)}\,\Gamma(\lambda)}{2\pi} \left( \langle m_0(\lambda+0\mathrm{i}), q C_+ \overline{q} \mathrm{e}(\lambda) \rangle + \int_\mathbb{R}\mathrm{e}^{-\mathrm{i}\lambda y} q(y) \, \mathrm{d}y \right), \end{align}\] where, in the last step, we used 20 and 37 . The claimed identity now follows from the definition 18 of \(\Gamma\). This completes the proof of the lemma. ◻

Proof. Step 1. Let \(\varepsilon>0\). We show that there is a \(\kappa_\varepsilon<\infty\) such that if \(k\in\mathbb{C}\) satisfies \(\min\{|\mathop{\mathrm{Im}}k|,-\mathop{\mathrm{Re}}k\} \geq \kappa_\varepsilon\), then \[\begin{align} \label{eq:tkinvertproof} \|T_k\|_{L^\infty\rightarrow L^\infty}\leq \varepsilon. \end{align}\tag{58}\] In particular, choosing \(\varepsilon<1\) we see that \(1-T_k\) is invertible and its inverse is given by a convergent Neumann series.

Thus, fix \(\varepsilon>0\) and suppose \(\mathop{\mathrm{Im}}k\geq 0\), the case \(\mathop{\mathrm{Im}}k\leq 0\) being similar. Then \[\begin{align} {\left\vert T_k\varphi(x)\right\vert}= {\left\vert \int_{-\infty}^{x}\mathrm{e}^{\mathrm{i}k (x-y)}q(y)[C_+\overline{q}\varphi](y)\, \mathrm{d}y\right\vert}= {\left\vert \int_0^{\infty}\mathrm{e}^{\mathrm{i}k y}q(x-y)[C_+\overline{q}\varphi](x-y)\, \mathrm{d}y\right\vert}. \end{align}\] Since \(q\in L^2\), there is a \(\delta>0\) such that \[\sup_x\left(\int_{x-\delta}^{x}{\left\vert q(y)\right\vert}^2\, \mathrm{d}y\right)^{1/2} \leq \frac{\varepsilon}{2} \|q\|_2^{-1} \,.\] We then split the integral as \[\begin{align} {\left\vert T_k\varphi(x)\right\vert}&\leq \int_0^{\delta}{\left\vert q(x-y)[C_+\overline{q}\varphi](x-y)\right\vert}\, \mathrm{d}y+ \int_\delta^{\infty} \mathrm{e}^{- \mathop{\mathrm{Im}}k y} |q(x-y)[C_+\overline{q}\varphi](x-y)| \, \mathrm{d}y \\ &\leq \left( \int_0^\delta |q(x-y)|^2 \, \mathrm{d}y \right)^{1/2} \|q \varphi\|_2 + \mathrm{e}^{- (\mathop{\mathrm{Im}}k) \delta} \|q\|_2 \|q \varphi\|_2 \\ & \leq \frac{\varepsilon}{2}{\left\|\varphi\right\|_{}}_{\infty}+\mathrm{e}^{- (\mathop{\mathrm{Im}}k) \delta} \|q\|_2^2 \|\varphi\|_\infty \,. \end{align}\] Hence, we can choose \(\mathop{\mathrm{Im}}k\) large enough so that 58 holds.

Next, suppose \(\mathop{\mathrm{Re}}k<0\). Then we estimate \[\begin{align} {\left\vert T_k\varphi(x)\right\vert}&\leq {\left\vert\int_{0}^\infty \overline{\mathcal{F}\left(\overline{\mathrm{e}^{\mathrm{i}k(x-\cdot )}q}\mathbb{1}_{<x}\right)(\xi)}\mathcal{F}(\overline{q}\varphi)(\xi)\, \mathrm{d}\xi\right\vert}\\ &\leq {\left\|\overline{q}\varphi\right\|_{}}_2 \left( \frac{1}{2\pi} \int_{0}^\infty{\left\vert\int^x_{-\infty}\overline{\mathrm{e}^{\mathrm{i}k(x-y)}q(y)}\mathrm{e}^{-\mathrm{i}\xi y}\, \mathrm{d}y\right\vert}^2\, \mathrm{d}\xi\right)^{1/2}\\ &={\left\|\overline{q}\varphi\right\|_{}}_2 \left( \frac{1}{2\pi} \int_{-\mathop{\mathrm{Re}}k}^\infty{\left\vert\int^x_{-\infty}\mathrm{e}^{-\mathop{\mathrm{Im}}k(x-y)}q(y)\mathrm{e}^{\mathrm{i}\xi y}\, \mathrm{d}y\right\vert}^2\, \mathrm{d}\xi\right)^{1/2}. \end{align}\] It follows from Lemma 14 below (with \(\sigma =\mathop{\mathrm{Im}}k\) and \(\Xi=-\mathop{\mathrm{Re}}k\)) that the second term tends to zero as \(\mathop{\mathrm{Re}}k\to-\infty\), uniformly in \(\mathop{\mathrm{Im}}k\geq 0\) and \(x\in\mathbb{R}\). Thus, bounding \({\left\|\overline{q}\varphi\right\|_{}}_2\leq {\left\|q\right\|_{}}_2{\left\|\varphi\right\|_{}}_\infty\), we arrive again at 58 for \(-\mathop{\mathrm{Re}}k\) sufficiently large.

Step 2. We now turn to the second claim. We split \(T_k=S_k-T^-_k\) with \(T^-_k\) as defined in 53 and with \[\begin{align} (S_k\varphi)(x) \mathrel{\vcenter{:}}= \mathrm{i}\int_{\mp\infty}^x \mathrm{e}^{\mathrm{i}k(x-y)}{\left\vert q(y)\right\vert}^2 \varphi(y) \, \mathrm{d}y \,. \end{align}\] Arguing as in [30] we find that \(1-S_k\) is invertible with \[(1-S_k)^{-1} = 1+\widetilde{S}_k \,.\]

We aim to show that for any \(\varepsilon>0\) there is a \(\kappa_\varepsilon<\infty\) such that if \(k\in\widehat{\mathbb{C}}\) satisfies \(\min\{|\mathop{\mathrm{Im}}k|, \mathop{\mathrm{Re}}k\} \geq\kappa_\varepsilon\), then \[\begin{align} \label{eq:tktildedecay} \|T^-_k\|_{L^\infty\rightarrow L^\infty} \leq \varepsilon. \end{align}\tag{59}\]

Once we have shown this, recalling also 55 , we obtain the invertibility of \((1-T_k)^{-1}\) via \[\begin{align} (1-T_k)^{-1}=(1-S_k+T^-_k)^{-1} =(1+(1-S_k)^{-1}T^-_k)^{-1}(1-S_k)^{-1} \end{align}\] and, at the same time, we obtain a convergent Neumann expansion for \((1-T_k)^{-1}\) by expanding \((1+(1-S_k)^{-1} T^-_k)^{-1} = (1+(1+\widetilde{S}_k)T^-_k)^{-1}\) into a Neumann series.

To prove 59 , we fix \(\varepsilon>0\) and suppose \(\mathop{\mathrm{Im}}k\geq 0\), the case \(\mathop{\mathrm{Im}}k\leq 0\) being similar. Then repeating the argument from Step 1 above we can show that 59 holds when \(\mathop{\mathrm{Im}}k\) is large enough. Next, suppose \(\mathop{\mathrm{Re}}k>0\). Then we estimate \[\begin{align} |T^-_k\varphi(x)|&\leq {\left\vert\int_{-\infty}^0 \overline{\mathcal{F}\left(\overline{\mathrm{e}^{\mathrm{i}k(x-\cdot )}q}\mathbb{1}_{<x}\right)(\xi)}\mathcal{F}(\overline{q}\varphi)(\xi)\, \mathrm{d}\xi\right\vert}\\ &\leq {\left\|\overline{q}\varphi\right\|_{}}_2 \left( \frac{1}{2\pi} \int_{-\infty}^0 {\left\vert\int^x_{-\infty} \overline{\mathrm{e}^{\mathrm{i}k (x-y)}q(y)} \mathrm{e}^{-\mathrm{i}\xi y}\, \mathrm{d}y\right\vert}^2\, \mathrm{d}\xi\right)^{1/2}\\ &\leq {\left\|\overline{q}\varphi\right\|_{}}_2\left( \frac{1}{2\pi} \int_{\mathop{\mathrm{Re}}k}^\infty {\left\vert\int^x_{-\infty}\mathrm{e}^{- \mathop{\mathrm{Im}}k (x-y)}q(y)\mathrm{e}^{-\mathrm{i}\xi y}\, \mathrm{d}y\right\vert}^2\, \mathrm{d}\xi\right)^{1/2}. \end{align}\] As before, by Lemma 14 we arrive at 59 . This concludes the proof. ◻

Proof. We first prove the assertion under the assumption that \(f\in C^1_c(\mathbb{R})\). In this case we can integrate by parts and find \[\int_{-\infty}^x \mathrm{e}^{\mathrm{i}\xi(x-y)} \mathrm{e}^{-\sigma(x-y)} f(y)\, \mathrm{d}y = \frac{1}{-\mathrm{i}\xi+\sigma} f(x) - \frac{1}{-\mathrm{i}\xi+\sigma} \int_{-\infty}^x \mathrm{e}^{\mathrm{i}\xi(x-y)} \mathrm{e}^{-\sigma(x-y)} f'(y)\, \mathrm{d}y \,.\] Thus, \[\left| \int_{-\infty}^x \mathrm{e}^{\mathrm{i}\xi(x-y)} \mathrm{e}^{-\sigma(x-y)} f(y)\, \mathrm{d}y \right| \leq \frac{1}{\sqrt{\xi^2+\sigma^2}} \left( \|f\|_\infty + \|f'\|_1 \right) \leq \frac{1}{|\xi|} \left( \|f\|_\infty + \|f'\|_1 \right),\] from which both assertions follow easily.

Now for general \(f\in L^p(\mathbb{R})\), \(p=1,2\), and \(\varepsilon>0\) we choose \(\tilde{f}\in C^1_c(\mathbb{R})\) with \(\|\tilde{f} - f\|_p \leq \varepsilon\). We bound \[\left| \int_{-\infty}^x \mathrm{e}^{\mathrm{i}\xi(x-y)} \mathrm{e}^{-\sigma(x-y)} (\tilde{f}(y)-f(y)) \, \mathrm{d}y \right| \leq \|\tilde{f} - f\|_1 \leq \varepsilon\,,\] so the first assertion for \(f\) follows from that for \(\tilde{f}\). Meanwhile, by Plancherel’s theorem \[\begin{align} & \int_\Xi^\infty \left| \int_{-\infty}^x \mathrm{e}^{\mathrm{i}\xi(x-y)} \mathrm{e}^{-\sigma(x-y)} (\tilde{f}(y) - f(y)) \, \mathrm{d}y \right|^2 \, \mathrm{d}\xi \\ & \quad \leq \int_\mathbb{R}\left| \int_{-\infty}^x \mathrm{e}^{\mathrm{i}\xi(x-y)} \mathrm{e}^{-\sigma(x-y)} (\tilde{f}(y) - f(y)) \, \mathrm{d}y \right|^2 \, \mathrm{d}\xi \\ & \quad = 2\pi \int_{-\infty}^x \left| \mathrm{e}^{-\sigma(x-y)} (\tilde{f}(y)- f(y)) \right|^2 \, \mathrm{d}y \\ & \quad \leq 2\pi \| \tilde{f} - f\|_2^2 \leq 2\pi\varepsilon^2 \,. \end{align}\] Thus, once again the assertion for \(f\) follows from that for \(\tilde{f}\). ◻

Proof of Theorem 5. We begin with the proof of 47 . Applying the Neumann expansion 57 to the integral equation 51 gives \[\begin{align} m_\mathrm{e}(\lambda+0\mathrm{i})=(1+\widetilde{S}_{\lambda+0\mathrm{i}})\mathrm{e}(\lambda)+o_{\lambda\rightarrow \infty}(1) \end{align}\] with the error term in \(L^\infty\). Using 54 we can compute the leading term, \[\begin{align} \label{eq:tildeskone} (1+\widetilde{S}_{\lambda\pm0\mathrm{i}})\mathrm{e}(\lambda)&=\mathrm{e}^{\mathrm{i}\lambda x}+\mathrm{i}\mathrm{e}^{\mathrm{i}\lambda x}\int_{\mp \infty}^x \mathrm{e}^{\mathrm{i}\int_y^x{\left\vert q(t)\right\vert}^2\, \mathrm{d}t}{\left\vert q(y)\right\vert}^2\, \mathrm{d}y \notag \\ &=\mathrm{e}^{\mathrm{i}\lambda x}-\mathrm{e}^{\mathrm{i}\lambda x}\mathrm{e}^{\mathrm{i}\int_{-\infty}^x{\left\vert q(t)\right\vert}^2\, \mathrm{d}t}\int_{\mp \infty}^x \partial_y\left(\mathrm{e}^{-\mathrm{i}\int_{-\infty}^y{\left\vert q(t)\right\vert}^2\, \mathrm{d}t}\right)\, \mathrm{d}y \notag\\ &=\mathrm{e}^{\mathrm{i}\lambda x}\mathrm{e}^{\mathrm{i}\int_{\mp\infty}^x{\left\vert q(t)\right\vert}^2\, \mathrm{d}t}, \end{align}\tag{60}\] which proves the asymptotics 47 .

To establish 48 , we write \[\begin{align} \Gamma(\lambda)&=1+\mathrm{i}\int_{-\infty}^\infty \mathrm{e}^{-\mathrm{i}\lambda x}qC_+\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i})\, \mathrm{d}x\\ &=1+\mathrm{i}\int_{-\infty}^\infty \mathrm{e}^{-\mathrm{i}\lambda x}{\left\vert q\right\vert}^2m_\mathrm{e}(\lambda+0\mathrm{i})\, \mathrm{d}x-\mathrm{i}\int_{-\infty}^\infty \mathrm{e}^{-\mathrm{i}\lambda x}qC_{-}\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i})\, \mathrm{d}x. \end{align}\] For the first term, we substitute the asymptotics for \(m_\mathrm{e}\) from 47 , giving \[\begin{align} 1+\mathrm{i}\int_{-\infty}^\infty \mathrm{e}^{-\mathrm{i}\lambda x}{\left\vert q\right\vert}^2m_\mathrm{e}(\lambda+0\mathrm{i})\, \mathrm{d}x&=1+\mathrm{i}\int_{-\infty}^\infty {\left\vert q(x)\right\vert}^2\mathrm{e}^{\mathrm{i}\int_{-\infty}^x{\left\vert q(t)\right\vert}^2\, \mathrm{d}t}\, \mathrm{d}x+o_{\lambda\rightarrow \infty}(1)\\ &=1+ \int_{-\infty}^\infty\partial_x\left(\mathrm{e}^{\mathrm{i}\int_{-\infty}^x{\left\vert q(t)\right\vert}^2\, \mathrm{d}t}\right)\, \mathrm{d}x+o_{\lambda\rightarrow \infty}(1)\\ &=\mathrm{e}^{\mathrm{i}\int_\mathbb{R}{\left\vert q(t)\right\vert}^2\, \mathrm{d}t}+o_{\lambda\rightarrow \infty}(1). \end{align}\] For the second term, we estimate, using Lemma 13, \[\begin{align} {\left\vert\mathrm{i}\int_{-\infty}^\infty \mathrm{e}^{-\mathrm{i}\lambda x}qC_{-}\overline{q}m_\mathrm{e}(\lambda+0\mathrm{i})\, \mathrm{d}x\right\vert}\leq \|T^-_{\lambda+0\mathrm{i}}m_\mathrm{e}(\lambda+0\mathrm{i})\|_{\infty}=o_{\lambda\rightarrow \infty}(1) \end{align}\] which completes the proof of 48 .

In the remainder of the proof we assume, in addition, \(q\in L^1\). To prove 49 we assume \(\mathop{\mathrm{Im}}k\geq 0\), the opposite case being similar. We claim that \[\begin{align} \label{eqn:Gkinftyzero} \lim_{{\left\vert k\right\vert}\rightarrow\infty} {\left\|G_k\ast q\right\|_{}}_\infty=0 \,. \end{align}\tag{61}\] Indeed, when \(\mathop{\mathrm{Im}}k\to\infty\) this is easy (see the proof of Lemma 13) and when \(|\mathop{\mathrm{Re}}k|\to\infty\) this follows from the first part of Lemma 14 (with \(\sigma = \mathop{\mathrm{Im}}k\) and \(\xi=\mathop{\mathrm{Re}}k\)).

In view of the integral equation 52 for \(m_0(k)\) we find \[\| m_0(k)\|_\infty \leq \|(1-T_k)^{-1}\|_{L^\infty\to L^\infty} \|G_k\ast q\|_\infty \,.\] According to Lemma 13 \(\|(1-T_k)^{-1}\|_{L^\infty\to L^\infty}\) is uniformly bounded as \(|k|\to\infty\), so 49 follows from 61 .

Finally, to prove 50 , we observe \[\begin{align} {\left\vert\beta(\lambda)\right\vert}={\left\vert \int_{-\infty}^\infty q(C_+\overline{q}m_0(\lambda+0\mathrm{i})+1)\mathrm{e}^{-\mathrm{i}\lambda x}\right\vert}\leq {\left\|q\right\|_{}}_2^2{\left\|m_0( \lambda+0\mathrm{i})\right\|_{}}_\infty+ \sqrt{2\pi}\, {\left\vert\mathcal{F}(q)(\lambda)\right\vert}, \end{align}\] and the claim follows from Riemann–Lebesgue and the asymptotics 49 for \(m_0\). ◻

Proof. For any \(k\in \mathbb{C}\backslash\mathbb{R}\) and \(r>0\) sufficiently large, from the Neumann expansion 56 , we have \[\begin{align} \label{eqn:m0asymprkexpan} m_0(rk)=G_{rk}\ast q+\sum_{n=1}^\infty T_{rk}^nG_{rk}\ast q. \end{align}\tag{63}\] Then we use that \[\begin{align} K_r(x)\mathrel{\vcenter{:}}= -rk G_k(rx)=-rk G_{rk}(x) \end{align}\] defines an approximation of the identity, see 14 . In particular, we have pointwise convergence \[\begin{align} \lim_{r\rightarrow \infty}\left[K_r\ast q\right](x)= q(x). \end{align}\]

Since \(q\) is continuous, standard results imply convergence in \(L^\infty(\mathbb{R})\). That is \[\begin{align} \lim_{r\rightarrow \infty}{\left\|-rkG_{rk}\ast q-q\right\|_{}}_\infty=0, \end{align}\] which, together with the Neumann expansion 63 , gives the desired result. ◻

Proof. Step 1. Given that \(q\in\mathcal{S}_+(\mathbb{R})\), it follows by induction that each \(c_n\) is well defined and belongs to \(\mathcal{S}_+(\mathbb{R})\). This uses the fact that \(C_+\) maps \(\mathcal{S}(\mathbb{R})\) to \(C^\infty(\mathbb{R})\cap L^2_+(\mathbb{R})\), so that \(qC_+\overline{q}c_n \in \mathcal{S}(\mathbb{R})\).

Step 2. Fix \(M\in \mathbb{N}\) and, for \(k\in \widehat{\mathbb{C}}\setminus\{0\}\), define the remainder term \[\begin{align} r_M(x,k)\mathrel{\vcenter{:}}= m_0(x,k)-\sum_{n=1}^M \frac{c_n(x)}{k^n}. \end{align}\] Using the differential equation for \(m_0\), we compute \[\begin{align} (-\mathrm{i}\partial_x-k-qC_+\overline{q}) r_M(k)&= (-\mathrm{i}\partial_x-k-qC_+\overline{q})m_0(k)-\sum_{n=1}^M\frac{1}{k^n}(-\mathrm{i}\partial_x-k-qC_+\overline{q})c_n\\ &=q+\sum_{n=1}^M\frac{c_n}{k^{n-1}}-\sum_{n=1}^M\frac{(-\mathrm{i}\partial_x-qC_+\overline{q})c_n}{k^n}\\ &=q+\sum_{n=1}^M\frac{c_n}{k^{n-1}}-\sum_{n=1}^M\frac{c_{n+1}}{k^n}=-\frac{c_{M+1}}{k^M}, \end{align}\] using the recurrence relation defining \(c_n\). Let us assume \(k\in\widehat{\mathbb{C}}\setminus\sigma_\mathrm{p}(L_{q})^\diamond\). Since \(r_M(k)\) belongs to \(L^\infty(\mathbb{R})\) and decays rapidly at infinity and since \(c_{M+1}\in L^1_+(\mathbb{R})\), it follows from Lemma 6 that \[r_M(k) = - \frac{1}{k^M}\, R(k) c_{M+1}\] and therefore, by the resolvent identity 12 , \[\begin{align} r_M(k)=-\frac{1}{k^M}G_{k}\ast c_{M+1}+R_0(k) q C_+ \overline{q} r_M(k) \,. \end{align}\] In terms of the operator \(T_k\), this can be written as \[\begin{align} r_M(k)=-\frac{1}{k^{M}}(1-T_k)^{-1}(G_{k}\ast c_{M+1}). \end{align}\] By Theorem 5 the assumption \(k\in\widehat{\mathbb{C}}\setminus\sigma_\mathrm{p}(L_{q})^\diamond\) is satisfied for all sufficiently large \(|k|\) and, repeating the argument for proving 49 , it follows that \[\begin{align} {\left\|r_M(k)\right\|_{}}_\infty=o_{{\left\vert k\right\vert}\rightarrow\infty}({\left\vert k\right\vert}^{-M}) \,. \end{align}\] This gives the desired asymptotic expansion 64 for \(m_0\).

Step 3. To obtain the asymptotic behaviour of \(\beta(\lambda)\), we substitute the expansion 64 for \(m_0\) into its definition. We find \[\begin{align} {\left\vert\beta(\lambda)\right\vert}&\leq \sqrt{2\pi}\, {\left\vert\widehat{q}(\lambda)\right\vert}+\sum_{n=1}^{M-1} \frac{1}{\lambda^n} {\left\vert\int_\mathbb{R}\mathrm{e}^{-\mathrm{i}\lambda y} qC_+\overline{q}c_n\, \mathrm{d}y\right\vert}+\mathcal{O}_{\lambda\rightarrow\infty}(\lambda^{-M}). \end{align}\] Using the recurrence relation again, this yields \[\begin{align} \frac{{\left\vert\beta(\lambda)\right\vert}}{\sqrt{2\pi}}&\leq \, {\left\vert\widehat{q}(\lambda)\right\vert} + \sum_{n=1}^{M-1} \frac{1}{\lambda^n} \left( \lambda {\left\vert\widehat{ c_n}(\lambda)\right\vert}+{\left\vert\widehat{c_{n+1}}(\lambda)\right\vert}\right)+\mathcal{O}_{\lambda\rightarrow\infty}(\lambda^{-M})=\mathcal{O}_{\lambda\rightarrow\infty}(\lambda^{-M}), \end{align}\] since each \(c_n\) belongs to \(\mathcal{S}(\mathbb{R})\). This establishes 67 .

Step 4. To obtain the asymptotic behaviour 65 of \(m_\mathrm{e}(\lambda\pm0\mathrm{i})\), we use the Neumann series expansion 57 , \[\begin{align} m_\mathrm{e}(\lambda\pm0\mathrm{i})- (1+\widetilde{S}_{\lambda\pm0\mathrm{i}}) \mathrm{e}(\lambda) =\sum_{n=1}^\infty (-(1+\widetilde{S}_{\lambda\pm0\mathrm{i}}) T^-_{\lambda\pm0\mathrm{i}})^n (1+\widetilde{S}_{\lambda\pm0\mathrm{i}})\mathrm{e}(\lambda), \end{align}\] and we recall from 60 that \[\begin{align} (1+\widetilde{S}_{\lambda\pm0\mathrm{i}})\mathrm{e}(\lambda)=\mathrm{e}^{\mathrm{i}\lambda x}\mathrm{e}^{\mathrm{i}\int_{\mp\infty}^x{\left\vert q(t)\right\vert}^2\, \mathrm{d}t}. \end{align}\] To estimate the first iterate, we bound \[\begin{align} |T^-_{\lambda\pm0\mathrm{i}}\varphi(x)|&= {\left\vert\int_{\mp\infty}^x \mathrm{e}^{-\mathrm{i}\lambda y}qC_{-}\overline{q}\varphi\, \mathrm{d}y\right\vert} \leq{\left\|q\right\|_{}}_2 \left(\int_{-\infty}^\infty {\left\vert C_{-}\overline{q}\varphi\right\vert}^2\, \mathrm{d}y\right)^{1/2}\\ &\leq{\left\|q\right\|_{}}_2 \left( \frac{1}{2\pi} \int_{-\infty}^0{\left\vert\int^\infty_{-\infty}\mathrm{e}^{-\mathrm{i}y \xi}\overline{q(y)}\varphi(y)\, \mathrm{d}y\right\vert}^2\, \mathrm{d}\xi\right)^{1/2}. \end{align}\] Applying this to \(\varphi=(1+\widetilde{S}_{\lambda\pm0\mathrm{i}})\mathrm{e}(\lambda)\), we obtain \[\begin{align} \|T^-_{\lambda\pm0\mathrm{i}}(1+\widetilde{S}_{\lambda\pm0\mathrm{i}})\mathrm{e}(\lambda)\|&\leq {\left\|q\right\|_{}}_2 \left( \frac{1}{2\pi} \int_{-\infty}^0{\left\vert\int^\infty_{-\infty}\mathrm{e}^{\mathrm{i}y(\lambda- \xi)}\mathrm{e}^{\mathrm{i}\int_{\mp\infty}^y{\left\vert q(t)\right\vert}^2\, \mathrm{d}t}\overline{q(y)}\, \mathrm{d}y\right\vert}^2\, \mathrm{d}\xi\right)^{1/2}\\ &={\left\|q\right\|_{}}_2 \left(\int_{\lambda}^\infty {\left\vert\mathcal{F}(\mathrm{e}^{-\mathrm{i}\int_{\mp\infty}^y{\left\vert q(t)\right\vert}^2\, \mathrm{d}t} q)(\xi)\right\vert}^2\, \mathrm{d}\xi\right)^{1/2}. \end{align}\] Since \(q(x) \mathrm{e}^{-\mathrm{i}\int_{\pm\infty}^x{\left\vert q(t)\right\vert}^2\, \mathrm{d}t}\) is a Schwartz function, the tail integral decays rapidly as \(\lambda\rightarrow\infty\) and we have \[\|T^-_{\lambda\pm0\mathrm{i}}(1+\widetilde{S}_{\lambda\pm0\mathrm{i}})\mathrm{e}(\lambda)\| \leq C_M \lambda^{-M}\] for any \(M\). This bound, together with \[\begin{align} \| m_\mathrm{e}(\lambda+0\mathrm{i}) - (1+\widetilde{S}_{\lambda\pm0\mathrm{i}})e(\lambda)\|_\infty & \leq \sum_{n=1}^\infty \| (1+\widetilde{S}_{\lambda\pm0\mathrm{i}}) T^-_{\lambda\pm0\mathrm{i}} \|_{L^\infty\to L^\infty} ^{n-1} \\ & \quad\times \| 1+\widetilde{S}_{\lambda\pm0\mathrm{i}} \|_{L^\infty\to L^\infty} \|T^-_{\lambda\pm0\mathrm{i}}(1+\widetilde{S}_{\lambda\pm0\mathrm{i}})\mathrm{e}(\lambda)\|_\infty \end{align}\] and the norm bounds for \(\widetilde{S}_{\lambda\pm0\mathrm{i}}\) in 55 and for \(T^-_{\lambda\pm0\mathrm{i}}\) in Lemma 13, establish the asymptotic behaviour of \(m_\mathrm{e}(\lambda\pm0\mathrm{i})\) in 65 .

Step 5. Finally, the asymptotics 66 for \(\Gamma(\lambda)\) follow by substituting the expansion 65 for \(m_\mathrm{e}(\lambda+0\mathrm{i})\) into the definition of \(\Gamma(\lambda)\) and using the first-order term identified in Theorem 5. ◻

Proof. Eigenfunctions belong to the operator domain \(H^1_+(\mathbb{R})\) [3], which is contained in \(L^\infty(\mathbb{R})\). This, together with Lemma 1, gives \[\begin{align} \label{eqn:assump2} P_{\mathrm{p}}(L_{q})L^2_+(\mathbb{R})\subset L^1_+(\mathbb{R})\cap L^\infty(\mathbb{R}). \end{align}\tag{69}\] Therefore, the projections \[\begin{align} P_jf\mathrel{\vcenter{:}}=\frac{{\left\langle f,\varphi_j\right\rangle} _{}}{{\left\|\varphi_j\right\|_{}}^2_2} \, \varphi_j \end{align}\] define bounded operators on \(L^\infty(\mathbb{R})\). Set \(Q_j\mathrel{\vcenter{:}}= 1-P_j\) and write \[\begin{align} \tag{70} m_0(k)&=P_jm_0(k)+Q_jm_0(k) \intertext{and} \tag{71} \overline{\mathrm{e}(\lambda)}m_\mathrm{e}(\lambda\pm 0\mathrm{i})&=\overline{\mathrm{e}(\lambda)}P_jm_\mathrm{e}(\lambda\pm 0\mathrm{i})+\overline{\mathrm{e}(\lambda)}Q_jm_\mathrm{e}(\lambda\pm 0\mathrm{i}) \,. \end{align}\] The index \(j\) is fixed and, in the setting of part (b) we assume \(\lambda_j\geq 0\).

We shall show in Step 1 that \[\label{eq:embeddedexpanproof1} P_jm_0(k) = -\mathrm{i}\frac{\varphi_j}{k-\lambda_j} \qquad\text{and}\qquad P_j m_\mathrm{e}(\lambda\pm0\mathrm{i}) = 0\tag{72}\] in a neighbourhood of \(\lambda_j\pm0\mathrm{i}\) in \(\widehat{\mathbb{C}}\) and of \(\lambda_j\) in \([0,\infty)\), respectively. In Step 2 we show that \[\label{eq:embeddedexpanproof2} Q_jm_0(k) \;\text{is}\;(L^\infty,w^*)\text{-continuous} \qquad\text{and}\qquad \overline{\mathrm{e}(\lambda)}Q_j m_\mathrm{e}(\lambda\pm0\mathrm{i}) \;\text{is}\;L^\infty\text{-continuous}\tag{73}\] in a neighbourhood of \(\lambda_j\pm0\mathrm{i}\) in \(\widehat{\mathbb{C}}\) and of \(\lambda_j\) in \([0,\infty)\), respectively.

Once these two properties are shown, we immediately obtain the assertions of the theorem by 70 and 71 , except for the asymptotics of \(\overline{\mathrm{e}(\lambda)}m_\mathrm{e}(\lambda\pm0\mathrm{i})\) as \(\lambda\to 0_+\), which are dealt with in Step 3.

Step 1. We shall show that \[\begin{align} \label{eqn:m0meinnerphij} \quad {\left\langle m_0(k),\varphi_j\right\rangle} _{}=-\frac{{\left\langle q,\varphi_j\right\rangle} _{}}{k-\lambda_j} \quad \text{and}\quad {\left\langle m_\mathrm{e}(\lambda\pm 0\mathrm{i}),\varphi_j\right\rangle} _{}=0 \end{align}\tag{74}\] for all \(k\in\widehat{\mathbb{C}}\setminus\sigma_\mathrm{p}(L_{q})^\diamond\) and \(\lambda\in[0,\infty)\setminus\sigma_\mathrm{p}(L_{q})\), respectively. (We mention that a more intricate argument shows that this holds for all \(k\in \widehat{\mathbb{C}}\setminus\{\lambda_j\}^\diamond\) and \(\lambda\in \mathbb{R}_+\setminus\{\lambda_j\}\), but we will not need this.) Note that 74 proves 72 in view of the normalisation 68 .

For the proof of the identities in 74 , we use the formula \[\begin{align} \label{eqn:Glambdaphij} \varphi_j-R_0(k)qC_+\overline{q}\varphi_j=(\lambda_j-k)G_{k}\ast \varphi_j, \end{align}\tag{75}\] for any \(k\in\widehat{\mathbb{C}}\), which follows by the same reasoning as Lemma 6 since \(\varphi_j\in L^2_+(\mathbb{R})\cap L^1_+(\mathbb{R})\) and \[\begin{align} \label{eqn:Glambdaphij2} (-\mathrm{i}\partial_x-k)\varphi_j=(\lambda_j-k)\varphi_j+qC_+\overline{q}\varphi_j. \end{align}\tag{76}\]

Step 1a. Let us prove the first identity in 74 . In the case with \(k\in \mathbb{C}\backslash\sigma(L_{q})\), this follows immediately since \(m_0(k)=R(k)q\), and therefore \[\begin{align} {\left\langle m_0(k),\varphi_j\right\rangle} _{}=\langle q,R(\overline{k})\varphi_j\rangle=-\frac{{\left\langle q,\varphi_j\right\rangle} _{}}{k-\lambda_j}. \end{align}\] Using the \((L^\infty,w^*)\)-continuity of \(m_0(k)\), which we have observed right after its definition, we obtain the identity for all \(k\in\mathbb{C}\setminus\sigma_\mathrm{p}(L_{q})^\diamond\), as claimed.

Step 1b. Next, we prove the second identity in 74 for \(\lambda\in[0,\infty)\setminus\sigma_\mathrm{p}(L_{q})\). Using the definition of \(m_\mathrm{e}\), we find \[\begin{align} {\left\langle m_\mathrm{e}(\lambda\pm 0\mathrm{i}),\varphi_j\right\rangle} _{} & = {\left\langle\mathrm{e}(\lambda),\varphi_j\right\rangle} _{} + {\left\langle R(\lambda\pm 0\mathrm{i})qC_+\overline{q}\mathrm{e}(\lambda),\varphi_j\right\rangle} _{} \\ & = {\left\langle\mathrm{e}(\lambda),\varphi_j\right\rangle} _{} + {\left\langle\mathrm{e}(\lambda),q C_+ \overline{q} R(\lambda\mp 0\mathrm{i})\varphi_j\right\rangle} _{} \\ & = {\left\langle\mathrm{e}(\lambda),\varphi_j\right\rangle} _{} + \frac{1}{\lambda_j - \lambda} \, {\left\langle\mathrm{e}(\lambda),q C_+ \overline{q} \varphi_j\right\rangle} _{}. \end{align}\] Because of 76 , we have \[\label{eq:efeqev} {\left\langle\mathrm{e}(\lambda),q C_+ \overline{q} \varphi_j\right\rangle} _{} = {\left\langle\mathrm{e}(\lambda), -\mathrm{i}\varphi_j'- \lambda_j\varphi_j\right\rangle} _{} = (\lambda-\lambda_j) {\left\langle\mathrm{e}(\lambda),\varphi_j\right\rangle} _{} \,.\tag{77}\] Combining these two equations, we arrive at the second identity in 74 .

Step 2. We now turn to the proof of 73 . From the integral equation 33 for \(m_0\), the normalisation of \(\varphi_j\), the identity 75 and the first part of 74 we find \[\begin{align} \label{eq:embeddedexpanproof3a} Q_jm_0(k) &=Q_jR_0(k)Q_jq+Q_jR_0(k)qC_+\overline{q}Q_jm_0(k) \end{align}\tag{78}\] for \(k\in \widehat{\mathbb{C}}\). This gives \[\begin{align} \label{eq:embeddedexpanproof3b} C_+\overline{q}Q_jm_0(k) &=C_+\overline{q}Q_jR_0(k)Q_jq+C_+\overline{q}Q_jR_0(k)qC_+\overline{q}Q_jm_0(k). \end{align}\tag{79}\]

Similarly, using integral equation 17 for \(m_\mathrm{e}\) and the second part of 74 we find \[\begin{align} \label{eq:embeddedexpanproof4a} Q_jm_\mathrm{e}(\lambda\pm0\mathrm{i})=Q_j\mathrm{e}(\lambda)+Q_jR_0(\lambda\pm0\mathrm{i})qC_+\overline{q}Q_jm_\mathrm{e}(\lambda\pm0\mathrm{i}) \end{align}\tag{80}\] for \(\lambda\in \mathbb{R}_+\), which gives \[\begin{align} \label{eq:embeddedexpanproof4b} C_+\overline{q} Q_jm_\mathrm{e}(\lambda\pm0\mathrm{i})=C_+\overline{q} Q_j\mathrm{e}(\lambda) + C_+\overline{q} Q_jR_0(\lambda\pm0\mathrm{i})qC_+\overline{q}Q_jm_\mathrm{e}(\lambda\pm0\mathrm{i}) \,. \end{align}\tag{81}\]

Identities 79 and 81 motivate the study of the operators \(1-C_+\overline{q}Q_jR_0(k)qC_+\), acting on \(L^2_+\).

Step 2a. Let us show that \(k\mapsto C_+\overline{q}Q_jR_0(k)qC_+\) is a family of \(L_+^2\)-compact operators a \(\widehat{\mathbb{C}}\)-neighbourhood of \(\lambda_j\pm0\mathrm{i}\).

Indeed, since this is true for \(C_+\overline{q}R_0(k)qC_+\) by Lemma 3, it suffices to prove it for \(C_+\overline{q}P_jR_0(k)qC_+\). Note that for any \(f\in L^2_+\) we have \[\begin{align} {\left\|\varphi_j\right\|_{}}^2_2 \, C_+\overline{q}P_jR_0(k)qC_+f = {\left\langle R_0(k)qC_+f,\varphi_j\right\rangle} _{} C_+\overline{q}\varphi_j = {\left\langle f,C_+\overline{q}R_0(\overline{k})\varphi_j\right\rangle} _{} C_+\overline{q}\varphi_j. \end{align}\] Thus \(C_+\overline{q}P_jR_0(k)qC_+\) is a rank one operator and therefore compact. The continuity in \(L^2_+\)-operator norm follows from the \(L^2_+\)-continuity of \(k\mapsto C_+\overline{q}R_0(\overline{k})\varphi_j\), which follows from the argument used in Step 2 of the proof of Lemma 3. (Indeed, since \(\varphi_j\in L^1_+\), \(R_0(\overline{k_n})\varphi_j\) converges pointwise and boundedly, and therefore, by dominated convergence, \(qR_0(\overline{k_n})\varphi_j\) converges in \(L^2\).)

Step 2b. Let us show that \(\ker(1-C_+\overline{q} Q_j R_0(\lambda_j\pm0\mathrm{i})qC_+)=\{0\}\).

Indeed, suppose that \(g\in L^2_+(\mathbb{R})\) is a solution to \[\begin{align} \label{eqn:Qjcontrassump} (1-C_+\overline{q}Q_jR_0(\lambda_j\pm0\mathrm{i})qC_+)g=0. \end{align}\tag{82}\] Then, \[\begin{align} (1-C_+\overline{q}R_0(\lambda_j\pm0\mathrm{i})qC_+) g&=-C_+\overline{q}P_jR_0(\lambda_j\pm0\mathrm{i})qC_+g\\ &=-{\left\|\varphi_j\right\|_{}}_2^{-2}{\left\langle R_0(\lambda_j\pm 0\mathrm{i})qC_+g,\varphi_j\right\rangle} _{}C_+\overline{q}\varphi_j. \end{align}\] Applying \(1-C_+\overline{q}R_0(\lambda_j\pm0\mathrm{i})qC_+\) to both sides, using the definition of \(P_j\) and using Lemma 4 yields \[\begin{align} \label{eqn:Qjkernel} (1-C_+\overline{q}R_0(\lambda_j\pm0\mathrm{i})qC_+)^2 g=0. \end{align}\tag{83}\]

Since \(C_+\overline{q}R_0(\lambda_j\pm0\mathrm{i})qC_+\) is compact, \(1-C_+\overline{q}R_0(\lambda_j\pm0\mathrm{i})qC_+\) is a Fredholm operator of index zero on \(L^2_+(\mathbb{R})\). Namely, the dimensions of its kernel and cokernel coincide. Consequently, \[\begin{align} \ker\left( (1-C_+\overline{q}R_0(\lambda_j\pm0\mathrm{i})qC_+)^2\right) =\ker \left( 1-C_+\overline{q}R_0(\lambda_j\pm0\mathrm{i})qC_+ \right) =\mathrm{span}\{C_+\overline{q}\varphi_j\} \,, \end{align}\] where we used Lemma 4 and the simplicity of eigenvalues discussed after Lemma 2. Thus, 83 implies that \(g = \alpha\, C_+\overline{q}\varphi_j\) for some \(\alpha\in\mathbb{C}\). Since \(R_0(\lambda_j\pm0\mathrm{i})qC_+\overline{q}\varphi_j=\varphi_j\), it follows from 82 that \[g= C_+ \overline{q}Q_jR_0(\lambda_j\pm0\mathrm{i})qC_+ g = \alpha\, C_+ \overline{q}Q_jR_0(\lambda_j\pm0\mathrm{i})qC_+ \overline{q}\varphi_j = \alpha C_+ \overline{q} Q_j \varphi_j = 0 \,,\] proving triviality of the kernel.

Step 2c. By the Fredholm alternative, it follows from the compactness in Step 2a and the injectivity in Step 2b that the operator \(1-C_+\overline{q} Q_j R_0(k)qC_+\) is boundedly invertible for \(k=\lambda_j\pm0\mathrm{i}\). By the continuity in Step 2a, this holds for all \(k\) in a \(\widehat{\mathbb{C}}\)-neighbourhood of \(\lambda_j\pm 0\).

Therefore, solving equations 79 and 81 and inserting the resulting expressions into 78 and 80 , we obtain \[\begin{align} \quad Q_jm_0(k)&=Q_jR_0(k)Q_jq+Q_jR_0(k)qC_+ (1-C_+\overline{q}Q_jR_0(k)qC_+)^{-1} C_+\overline{q} Q_j R_0(k)Q_jq \end{align}\] and \[\begin{align} Q_jm_\mathrm{e}(\lambda\pm0\mathrm{i}) &=Q_j\mathrm{e}(\lambda)+Q_jR_0(\lambda\pm 0\mathrm{i}) qC_+ (1-C_+\overline{q}Q_jR_0(\lambda\pm0\mathrm{i})qC_+)^{-1} C_+\overline{q}Q_j\mathrm{e}(\lambda) \,. \end{align}\]

The fact that \(k\mapsto\langle Q_jm_0(k), f\rangle\) is continuous in a \(\widehat{\mathbb{C}}\)-neighbourhood of \(\lambda_j\pm0\mathrm{i}\) for any \(f\in L^1_+\) now follows by the same reasoning as Lemma 5, using, in particular, the continuity in operator norm of \(k\mapsto (1-C_+\overline{q}Q_jR_0(k)qC_+)^{-1}\).

To prove that \(\lambda\mapsto \overline{\mathrm{e}(\lambda)} Q_jm_\mathrm{e}(\lambda\pm0\mathrm{i})\) is continuous in \(L^\infty\), we write \[\label{eq:eqmecont} \begin{align} \overline{\mathrm{e}(\lambda)} Q_jm_\mathrm{e}(\lambda\pm0\mathrm{i}) & = 1 - \|\varphi_j\|_2^{-2} \langle \mathrm{e}(\lambda),\varphi_j \rangle \, \overline{\mathrm{e}(\lambda)}\varphi_j \\ & \quad + \overline{\mathrm{e}(\lambda)} R_0(\lambda\pm0\mathrm{i}) F(\lambda) - \|\varphi_j\|_2^{-2} \langle R_0(\lambda\pm 0\mathrm{i}) F(\lambda),\varphi_j\rangle \, \overline{\mathrm{e}(\lambda)}\varphi_j \end{align}\tag{84}\] with \[F(\lambda) := qC_+ (1-C_+\overline{q}Q_jR_0(\lambda\pm0\mathrm{i})qC_+)^{-1} C_+\overline{q}Q_j\mathrm{e}(\lambda) \,.\]

We begin by noting that \(\lambda\mapsto \langle \mathrm{e}(\lambda),\varphi_j\rangle\) is continuous since \(\varphi_j\in L^1\). Moreover, \(\lambda\mapsto\overline{q}\mathrm{e}(\lambda)\) is \(L^2\)-continuous by dominated convergence, so \[\overline{q}Q_j \mathrm{e}(\lambda) = \overline{q}\mathrm{e}(\lambda) - \|\varphi_j\|_2^{-2} \langle \mathrm{e}(\lambda),\varphi_j \rangle \overline{q}\varphi_j\] is also \(L^2\)-continuous. It follows from Steps 2a and 2b that the inverse operator in the definition of \(F\) is continuous as a bounded operator on \(L^2\) and therefore \(F\) is \(L^1\)-continuous. The \((L^\infty,w^*)\)-continuity of \(\lambda\mapsto R_0(\lambda\mp0\mathrm{i})\varphi_j\) together with the strong continuity of \(F\) imply continuity of \(\lambda\mapsto \langle R_0(\lambda\pm 0\mathrm{i}) F(\lambda),\varphi_j\rangle\).

A combination of these facts shows that the second and fourth term on the right side of 84 are \(L^\infty\)-continuous, provided we can show that \(\lambda\mapsto\overline{\mathrm{e}(\lambda)}\varphi_j\) is \(L^\infty\)-continuous. To prove this, consider a sequence \((\lambda_n)\subset\mathbb{R}\) with \(\lambda_n\to\lambda\) and let \(\varepsilon>0\). Since \(\varphi_j\) tends to zero at infinity (by Lemma 4), there is an \(R>0\) such that \(|\varphi_j(x)|\leq\varepsilon/2\) if \(|x|\geq R\). Thus, \(|(\overline{\mathrm{e}(\lambda_n)}-\overline{\mathrm{e}(\lambda)})\varphi_j(x)|\leq \varepsilon\) if \(|x|\geq R\). Meanwhile, for \(|x|\leq R\) we have \(|(\overline{\mathrm{e}(\lambda_n)}-\overline{\mathrm{e}(\lambda)})\varphi_j(x)|\leq |x||\lambda_n-\lambda| \|\varphi_j\|_\infty \leq R |\lambda_n - \lambda| \|\varphi_j\|_\infty\) and this is \(\leq \varepsilon\) provided \(n\) is large enough, thus proving the claimed continuity.

It remains to prove continuity of the third term on the right side of 84 . For this purpose we note that \(\lambda\mapsto\overline{\mathrm{e}(\lambda)}F(\lambda)\) is \(L^1\)-continuous. This follows by bounding \[\| \overline{\mathrm{e}(\lambda_n)}F(\lambda_n) - \overline{\mathrm{e}(\lambda)} F(\lambda) \|_1 \leq \| \overline{\mathrm{e}(\lambda_n)}(F(\lambda_n)-F(\lambda))\|_1 + \|(\overline{\mathrm{e}(\lambda_n)}-\overline{\mathrm{e}(\lambda)})F(\lambda)\|_1\] and noting that the first term goes to zero by \(L^1\)-continuity of \(F\) and the second one by dominated convergence. Noting that \[(\overline{\mathrm{e}(\lambda)} R_0(\lambda\pm0\mathrm{i}) F(\lambda))(x) = \mathrm{i}\int_{\mp\infty}^x (\overline{\mathrm{e}(\lambda)}F(\lambda))(y) \, \mathrm{d}y \,,\] we deduce the \(L^\infty\)-continuity of this term, thereby completing the proof of the \(L^\infty\) continuity of \(\overline{\mathrm{e}(\lambda)} Q_jm_\mathrm{e}(\lambda\pm0\mathrm{i})\).

Step 3. To finish the proof of the lemma it remains to study the behaviour of \(\overline{\mathrm{e}(\lambda)}m_\mathrm{e}(\lambda\pm0\mathrm{i})\) as \(\lambda\to 0_+\).

The claimed result is straightforward when \(0\) is not an eigenvalue of \(L_{q}\) by the continuity of \(\overline{\mathrm{e}(\lambda)}m_\mathrm{e}(\lambda\pm0\mathrm{i})\) at \(0\), which we have already shown, and the fact that \(m_\mathrm{e}(0\pm0\mathrm{i})=1\), see 23 .

When \(0\) is an eigenvalue of \(L_q\), let \(P\) denote the projection onto the corresponding eigenfunction and set \(Q :=1-P\). The decomposition 71 , the vanishing in 72 and the formula for \(m_\mathrm{e}(\lambda\pm0\mathrm{i})\) in Step 2c give \[\begin{align} \overline{\mathrm{e}(\lambda)}m_{\mathrm{e}}(\lambda\pm0\mathrm{i})-1=&-\overline{\mathrm{e}(\lambda)}P\mathrm{e}(\lambda)\\&+\overline{\mathrm{e}(\lambda)}Q R_0(k)qC_+(1-C_+\overline{q}QR_0(\lambda\pm0\mathrm{i})qC_+)^{-1}C_+\overline{q}Q\mathrm{e}(\lambda) \end{align}\] for all \(\lambda\) in a right-neighbourhood of \(0\). We have also seen that \((1-C_+\overline{q}QR_0(\lambda\pm0\mathrm{i})qC_+)^{-1}\) is bounded and continuous for \(\lambda\) in a right-neighbourhood of \(0\). The result then follows since \(\overline{\mathrm{e}(\lambda)}P\mathrm{e}(\lambda)\to 0\) in \(L^\infty\) and \(C_+\overline{q}Q\mathrm{e}(\lambda)\to 0\) in \(L^2\). The latter limits follow since \(\varphi\in L^1_+\) implies \(\langle \mathrm{e}(\lambda),\varphi\rangle \to 0\) and \(C_+\overline{q}=0\). This completes the proof. ◻

Proof. As in the proof of Lemma 12 we write \[\Gamma(\lambda) = 1 + \mathrm{i}\langle \overline{\mathrm{e}(\lambda)} m_\mathrm{e}(\lambda+0\mathrm{i}), \overline{\mathrm{e}(\lambda)} qC_+\overline{q}\mathrm{e}(\lambda) \rangle \,.\] By Theorem 7, \(\lambda\mapsto \overline{\mathrm{e}(\lambda)} m_\mathrm{e}(\lambda+0\mathrm{i})\) extends \(L^\infty\)-continuously to all of \([0,\infty)\) and, by Step 4 in the proof of Lemma 12, \(\lambda\mapsto \overline{\mathrm{e}(\lambda)} qC_+\overline{q}\mathrm{e}(\lambda)\) is \(L^1\)-continuous on \([0,\infty)\). This defines a continuous extension of \(\Gamma\) to all of \([0,\infty)\). As \(\lambda\to 0\) we have \(\overline{\mathrm{e}(\lambda)} m_\mathrm{e}(\lambda+0\mathrm{i})\to 1\) in \(L^\infty\) and \(\overline{\mathrm{e}(\lambda)} qC_+\overline{q}\mathrm{e}(\lambda)\to 0\) in \(L^1\) (since \(C_+\overline{q}=0\)), so \(\Gamma(0)=0\).

For \(\beta\), we fix an eigenvalue \(\lambda_j\in[0,\infty)\) and apply the expansion for \(m_0\) about \(\lambda_j\) from Theorem 7. Substituting into the definition of \(\beta\), we obtain \[\begin{align} \beta(\lambda)&=\frac{1}{\lambda-\lambda_j} \int_{-\infty}^\infty \mathrm{e}^{-\mathrm{i}y\lambda}qC_+\overline{q}\varphi_j\, \mathrm{d}y +\mathrm{i}\sqrt{2\pi}\, (\mathcal{F}q)(\lambda)+\mathrm{i}\int_{-\infty}^\infty \mathrm{e}^{-\mathrm{i}y\lambda}qC_+\overline{q}h(\lambda+0\mathrm{i})\, \mathrm{d}y \,. \end{align}\] Here the Fourier transform \(\mathcal{F}q\) is continuous since \(q\in L^1\). Moreover, the last term is continuous since \(\lambda\mapsto q C_+\overline{q}\mathrm{e}(\lambda)\) is continuous in \(L^1\) and since \(\lambda\mapsto h(\lambda\pm 0\mathrm{i})\) is bounded in \(L^\infty\) for \(\lambda\) in a neighbourhood of \(\lambda_j\). The latter is implicit in the proof of Theorem 7.

To handle the first term, we use 77 to write \[\begin{align} \frac{1}{\lambda-\lambda_j} \int_{-\infty}^\infty \mathrm{e}^{-\mathrm{i}y\lambda}qC_+\overline{q}\varphi_j\, \mathrm{d}y =\frac{1}{\lambda-\lambda_j}(\lambda-\lambda_j) \langle \varphi_j,\mathrm{e}(\lambda) \rangle = \sqrt{2\pi}\, (\mathcal{F}\varphi_j)(\lambda) \,. \end{align}\] Since \(\varphi_j\in L^1(\mathbb{R})\), this term is continuous as well. This defines a continuous extension of \(\beta\) to all of \([0,\infty)\).

The limit \(\beta(\lambda)\rightarrow 0\) as \(\lambda\rightarrow 0_+\) follows from the continuity and 39 when \(0\) is not an eigenvalue of \(L_q\). When it is, it follows from the above formula for \(\beta\) with \(\lambda_j=0\), since \(qC_+\overline{q}h,q,\varphi\in L^1(\mathbb{R})\) and their Fourier transforms vanish on negative frequencies.

Finally, for the asymptotics of \(\Gamma(\lambda)\) as \(\lambda\rightarrow 0_+\), we deduce from \(\Gamma(0)=0\) and the differential equation 40 in Lemma 12 that \[\Gamma(\lambda) = \exp\left(-\frac{1}{2\pi\mathrm{i}} \int_0^\lambda |\beta(\lambda')|^2\,\, \mathrm{d}\lambda' \right).\] The stated asymptotic order now follows from that of \(\beta\). ◻

Proof. We recall from Lemma 1 that \(q\in L^2_+\cap L^1_+\) also gives \[\begin{align} \label{eqn:assump3} P_{\mathrm{p}}(L_{q})L^2_+(\mathbb{R})\subset \langle x\rangle^{-1} L^\infty(\mathbb{R}). \end{align}\tag{85}\]

Step 1. Fix \(j\in \{1,\ldots,N\}\). Then by Theorem 7 \[\begin{align} m_0(k)=-\frac{\mathrm{i}\varphi_j}{k-\lambda_j}+ h_j(k) \,, \end{align}\] where \(h_j\) is \((L^\infty,w^*)\) continuous in \(\widehat{\mathbb{C}}\)-neighbourhoods of both \(\lambda_j+ 0\mathrm{i}\) and \(\lambda_j-0\mathrm{i}\).

To refine this expansion, we use the integral equation 33 for \(m_0\), obtaining \[\begin{align}\label{eqn:improvlaurentexp1} h_j(k)&=\frac{\mathrm{i}\varphi_j}{k-\lambda_j}+R_0(k)q+R_0(k)qC_+\overline{q}\left[-\frac{\mathrm{i}\varphi_j}{k-\lambda_j}+h_j(k)\right]\\ &=\frac{\mathrm{i}}{k-\lambda_j}(R_0(\lambda_j\pm0\mathrm{i})-R_0(k))qC_+\overline{q}\varphi_j+R_0(k) q+R_0(k)qC_+\overline{q}h_j(k), \end{align}\tag{86}\] where we used \(\varphi_j=R_0(\lambda_j\pm0\mathrm{i})qC_+\overline{q}\varphi_j\).

We now take \(k=(\lambda_j+h)\pm 0\mathrm{i}\) with \(h\in\mathbb{R}\) and take the limit \(h\to 0\). By Theorem 7 \(h_j(k)\) is \(w^*\)-continuous and by Lemma 3 \(R_0(k)q\) is \(w^*\)-continuous. Moreover, it is easy to see that \(R_0(k)qC_+\overline{q}h_j(k)\) is also \(w^*\)-continuous. Indeed, if \(f\in L^1_+\), then, by Step 2 in the proof of Lemma 3, \(R_0(\overline{k})f\) converges pointwise and boundedly and therefore, by dominated convergence, \(\overline{q}R_0(\overline{k})f\) converges in \(L^2\) and \(qC_+\overline{q}R_0(\overline{k})f\) converges in \(L^1\). Since this convergence is strong, when paired with the \(w^*\)-convergence of \(h_j(k)\) in \(L^\infty\) we obtain convergence of \(\langle h_j(k), qC_+\overline{q}R_0(\overline{k})f\rangle\), as claimed.

In Step 2 we will show that \[\begin{align} & \frac{\mathrm{i}}{k-\lambda_j}(R_0(\lambda_j+0\mathrm{i})-R_0(k))qC_+\overline{q}\varphi_j =-\frac{\mathrm{i}}{h}(R_0(\lambda_j+h+0\mathrm{i})-R_0(\lambda_j+0\mathrm{i}))qC_+\overline{q}\varphi_j\\ & \to -\mathrm{i}\left( \mathrm{i}x R_0(\lambda_j+0\mathrm{i})qC_+\overline{q}\varphi_j - \mathrm{i}R_0(\lambda_j+0\mathrm{i})q C_+ \overline{q} x\varphi_j + (2\pi)^{-1} \langle \varphi_j,q \rangle \, R_0(\lambda_j+0\mathrm{i}) q \right) \\ & \quad = x \varphi_j - R_0(\lambda_j+0\mathrm{i})q C_+ \overline{q} x\varphi_j - R_0(\lambda_j+0\mathrm{i}) q \,. \end{align}\] In the last equation we use \(R_0(\lambda_j+0\mathrm{i})qC_+\overline{q}\varphi_j=\varphi_j\) as well as the normalisation 68 .

The convergence shown in Step 2 holds only when integrated against a test function in \(\langle x \rangle^{-1} L^1\), but since this set is dense in \(L^1\), we deduce in the limit from 86 that \[\begin{align} h_j(\lambda_j\pm 0\mathrm{i})= x \varphi_j-R_0(\lambda_j+0\mathrm{i})qC_+\overline{q}x\varphi_j +R_0(\lambda_j+0\mathrm{i})qC_+\overline{q}h_j(\lambda_j+0\mathrm{i}) \,, \end{align}\] which can be rewritten as \[\begin{align} h_j(\lambda_j\pm0\mathrm{i})-x\varphi_j=R_0(\lambda_j\pm0\mathrm{i})qC_+\overline{q}(h_j(\lambda_j\pm0\mathrm{i})-x\varphi_j) \,. \end{align}\] By 85 , we have \(h_j(\lambda_j\pm0\mathrm{i})-x\varphi_j\in L^\infty(\mathbb{R})\). Thus, Lemma 4 shows that \(h_j(\lambda_j\pm0\mathrm{i})- x\varphi_j\) are eigenfunctions of \(L_{q}\) with corresponding eigenvalue \(\lambda_j\). Therefore, by simplicity (see the discussion after Lemma 2), \[\begin{align} h_j(\lambda_j\pm0\mathrm{i})- x\varphi_j=\gamma_j^\pm\varphi_j \end{align}\] for some constants \(\gamma_j^\pm\in \mathbb{C}\).

It remains to be shown that \(\gamma_j^+=\gamma_j^{-}\). When \(\lambda_j< 0\), this is clear since \(h_j(\lambda_j+0\mathrm{i})=h_j(\lambda_j-0\mathrm{i})\). When \(\lambda_j\geq0\), we observe that \[\begin{align} (\gamma^{+}_j-\gamma^{-}_j)\varphi_j&=h_j(\lambda_j+0\mathrm{i})-h_j(\lambda_j-0\mathrm{i})\\ &=m_0(\lambda_j+0\mathrm{i})-m_0(\lambda_j-0\mathrm{i})=\beta(\lambda_j)m_\mathrm{e}(\lambda_j-0\mathrm{i}), \end{align}\] where the final identity uses 36 from Lemma 11. Note that \((\gamma^{+}_j-\gamma^{-}_j)\varphi_j\) is square-integrable, while, in view of the asymptotics in Lemma 7, \(m_\mathrm{e}(\lambda_j-0\mathrm{i})\) is not square-integrable near \(+\infty\). We deduce that \(\beta(\lambda_j)=0\), and hence \(\gamma^+_j=\gamma^{-}_j\). This concludes the proof, except for the convergence result.

Step 2. Let \(\varphi\in L^1_+\) with \(\langle x\rangle\varphi\in L^\infty\) and let \(\lambda\in\mathbb{R}\). We are going to show that \[\begin{align} & h^{-1} ( R_0(\lambda +h\pm 0\mathrm{i}) - R_0(\lambda\pm 0\mathrm{i})) qC_+\overline{q}\varphi\\ & \to \mathrm{i}x R_0(\lambda\pm 0\mathrm{i}) qC_+\overline{q}\varphi- \mathrm{i}R_0(\lambda\pm 0\mathrm{i}) qC_+\overline{q} x\varphi+ (2\pi)^{-1} \langle\varphi,q\rangle \, R_0(\lambda\pm\mathrm{i}0) q \end{align}\] as \(h\to 0\) with \(h\in\mathbb{R}\). The convergence is understood in a weak sense when integrated again a test function \(f\in L^1_+\) with \(\langle x \rangle f\in L^1\).

We prove this for the upper sign, the opposite case being similar. We begin by writing \[\begin{align} (R_0(\lambda+0\mathrm{i})&qC_+\overline{q}\varphi)(x) \\&=\int_\mathbb{R}\mathrm{e}^{-\mathrm{i}h x}G_{\lambda+h+0\mathrm{i}}(x-y)\mathrm{e}^{\mathrm{i}h y}qC_+\overline{q}\varphi\, \mathrm{d}y\\ &=\int_\mathbb{R}\mathrm{e}^{-\mathrm{i}h x}G_{\lambda+h+0\mathrm{i}}(x-y)qC_+\overline{q}\mathrm{e}(h)\varphi\, \mathrm{d}y - \int_\mathbb{R}G_{\lambda+0\mathrm{i}}(x-y) qC_{h}\overline{q}\varphi\, \mathrm{d}y \end{align}\] with \(C_h := \mathcal{F}^{-1}\mathbb{1}_{(-h,0)}\mathcal{F}\) if \(h\geq 0\) and \(C_{h} := -\mathcal{F}^{-1}\mathbb{1}_{(0,-h)}\mathcal{F}\) if \(h\leq 0\), see 41 . Thus, \[\begin{align} & h^{-1} ( R_0(\lambda +h\pm 0\mathrm{i}) - R_0(\lambda\pm 0\mathrm{i})) qC_+\overline{q}\varphi\\ & = \frac{1}{h}\int_\mathbb{R}G_{\lambda+h+0\mathrm{i}}(x-y)\left(qC_+\overline{q}\varphi-\mathrm{e}^{-\mathrm{i}h x}qC_+\overline{q}\mathrm{e}(h)\varphi\right)\, \mathrm{d}y + \frac{1}{h}\int_\mathbb{R}G_{\lambda+0\mathrm{i}}(x-y)qC_{h} \overline{q}\varphi\, \mathrm{d}y\\ & \eqqcolon (1)+(2) \,. \end{align}\]

For the term \((1)\), we write \[\begin{align} (1) & =\mathrm{e}^{-\mathrm{i}hx}\int_\mathbb{R}G_{\lambda+h+0\mathrm{i}}(x-y)\left(\frac{\mathrm{e}^{\mathrm{i}hx}-1}{h}qC_+\overline{q}\varphi+qC_+\overline{q}\frac{1-\mathrm{e}(h)}{h}\varphi\right)\, \mathrm{d}y \\ & = \int_\mathbb{R}\frac{1-\mathrm{e}^{-\mathrm{i}hx}}{h} G_{\lambda+h+0\mathrm{i}}(x-y)qC_+\overline{q}\varphi\, \mathrm{d}y + \int_\mathbb{R}G_{\lambda+0\mathrm{i}}(x-y) \overline{\mathrm{e}(h)} qC_+\overline{q}\frac{1-\mathrm{e}(h)}{h}\varphi\, \mathrm{d}y \,. \end{align}\] Thus, if \(f\) is a test function as specified, then \[\langle (1),f \rangle = \langle R_0(\lambda+h+0\mathrm{i}) qC_+\overline{q}\varphi, \frac{1-\mathrm{e}(h)}{h} f \rangle + \langle \overline{\mathrm{e}(h)} q C_+ \overline{q} \frac{1-\mathrm{e}(h)}{h} \varphi, R_0(\lambda-0\mathrm{i})f \rangle \,.\] By dominated convergence, the extra assumption on \(f\) implies that \(h^{-1}(1-\mathrm{e}(h))f\to -\mathrm{i}xf\) in \(L^1\). This strong convergence coupled with the \((L^\infty,w^*)\)-convergence of \(R_0(\lambda+h+0\mathrm{i}) qC_+\overline{q}\varphi\) (Lemma 3) shows that \[\langle R_0(\lambda+h+0\mathrm{i}) qC_+\overline{q}\varphi, \frac{1-\mathrm{e}(h)}{h} f \rangle \to \mathrm{i}\;\langle R_0(\lambda+0\mathrm{i}) qC_+\overline{q}\varphi, x f \rangle\] Similarly, the extra boundedness assumption on \(\varphi\) and dominated convergence implies \(h^{-1}\overline{q}(1-\mathrm{e}(h))\varphi\to -\mathrm{i}\overline{q} x\varphi\) in \(L^2\), and, by dominated convergence, \(\overline{\mathrm{e}(h)} q\to q\) in \(L^2\). This shows that \[\langle \overline{\mathrm{e}(h)} q C_+ \overline{q} \frac{1-\mathrm{e}(h)}{h} \varphi, R_0(\lambda-0\mathrm{i})f \rangle \to -\mathrm{i}\, \langle q C_+ \overline{q} x \varphi, R_0(\lambda-0\mathrm{i})f \rangle \,.\] This gives the limit of term \((1)\).

For the term \((2)\) with \(h\geq 0\) we can apply Step 1 in the proof of Lemma 12, noting that \(C_0=0\), and obtain \[(2) \to \frac{1}{2\pi} \int_\mathbb{R}\varphi(y') \overline{q(y')} \, \mathrm{d}y' \; \int_\mathbb{R}G_{\lambda+0\mathrm{i}}(x-y) q(y)\, \mathrm{d}y = (2\pi)^{-1} \langle \varphi,q \rangle \;R_0(\lambda+0\mathrm{i}) q \,.\] The same holds for \(h\leq 0\) by essentially the same argument. This gives the limit of term \((2)\) and thereby concludes the proof of the lemma. ◻

Proof of Theorem 10. It suffices to take \(\Lambda=\mathbb{R}\) in the first equality in Lemma 17. ◻

Proof of Lemma 17. Let \(\nu\) be the spectral measure of \(q\) with respect to \(L_{q}\), that is, \(\nu(\Lambda):=\langle \mathbb{1}_\Lambda(L_q) q,q\rangle\) for any Borel set \(\Lambda\subset\mathbb{R}\). It follows from Corollary 2 and Theorem 3 that, for any Borel set \(\Lambda\subset\mathbb{R}\), \[\begin{align} \label{eq:nu} \nu(\Lambda) & = \int_{\Lambda\cap\mathbb{R}_+} |(\Phi q)(\lambda)|^2\,\, \mathrm{d}\lambda + \sum_{\lambda_j\in\Lambda} \frac{|\langle q,\varphi_j\rangle|^2}{\|\varphi_j\|^2} \notag \\ & = \frac{1}{2\pi} \int_{\Lambda\cap\mathbb{R}_+} |\beta(\lambda)|^2\,\, \mathrm{d}\lambda + 2\pi \, \#\{ j:\;\lambda_j\in\Lambda\} \,, \end{align}\tag{90}\] where we used the definition of \(\beta\), as well as Lemma 2. This proves the first assertion in the lemma. Note that this formula means that the absolutely continuous part of \(\nu\) has density \((2\pi)^{-1}|\beta|^2\) and its singular part consists in \(2\pi\) times a delta function at the eigenvalues.

Let \(k\in\mathbb{C}\setminus\sigma(L_{q})\) and recall that \(m_0(k)= R(k) q\). This implies that \[\label{eq:cauchymeasure} \int_{-\infty}^{\infty} \overline{q(x)} m_0(x,k)\,\, \mathrm{d}x = \langle R(k) q,q \rangle = \int_\mathbb{R}\frac{\, \mathrm{d}\nu(\lambda)}{\lambda-k} \,,\tag{91}\] where we used the spectral theorem in the last step.

According to our formula for \(\nu\), we can rewrite this as \[\int_{-\infty}^{\infty} \overline{q(x)} m_0(x,k) \, \mathrm{d}x = \frac{1}{2\pi} \int_0^\infty \frac{|\beta(\lambda')|^2}{\lambda'-k}\,\, \mathrm{d}\lambda' + 2\pi \sum_{j=1}^N \frac{1}{\lambda_j - k} \,,\] which proves the second assertion in the lemma.

To prove the third assertion we want to take \(k=\lambda-\mathrm{i}\varepsilon\) with \(\lambda\in\mathbb{R}\setminus\sigma_\mathrm p(L_{q})\) and \(\varepsilon>0\) and let \(\varepsilon\to 0\). The left side has a limit according to the continuity statement made right after the definition of \(m_0(k)\), and the second term on the right side clearly has a limit. Thus, the first term on the right side has a limit as well and we have \[\int_{-\infty}^{\infty} \overline{q(x)} m_0(x,\lambda-0\mathrm{i}) \, \mathrm{d}x = \lim_{\varepsilon\to 0^+} \frac{1}{2\pi} \int_0^\infty \frac{|\beta(\lambda')|^2}{\lambda'-\lambda + \mathrm{i}\varepsilon}\,\, \mathrm{d}\lambda' + 2\pi \sum_{j=1}^N \frac{1}{\lambda_j - \lambda} \,.\] By the Sokhotski–Plemelj formula (see, e.g., [35]) it follows that for almost every \(\lambda\in\mathbb{R}\) we have \[\lim_{\varepsilon\to 0^+} \int_0^\infty \frac{|\beta(\lambda')|^2}{\lambda'-\lambda + \mathrm{i}\varepsilon}\,\, \mathrm{d}\lambda' = -\mathrm{i}\pi\,|\beta(\lambda)|^2 + \mathrm{p.v.} \int_0^\infty \frac{|\beta(\lambda')|^2}{\lambda'-\lambda}\,\, \mathrm{d}\lambda' = -2\pi\mathrm{i}\, C_-(|\beta|^2)(\lambda) \,.\] It follows, in particular, that the principal value integral exists for a.e. \(\lambda\in\mathbb{R}\). (This follows also standard harmonic analysis results concerning the Hilbert transform of an \(L^1\)-function.) This shows that the formula in the lemma holds for almost every \(\lambda\in\mathbb{R}\). ◻

Proof. We will show the formula for \(k=\kappa<\inf\sigma (L_{q})\) sufficiently negative, the extension to all \(k\in \mathbb{C}\backslash\sigma(L_{q})\) will follow by analytic continuation.

From the resolvent formula it follows that \(\kappa\in\mathbb{R}\setminus\sigma(L_{q})\) if and only if the operator \(1-C_+\overline{q}R_0(\kappa)qC_+\) is invertible and, in this case, \[\begin{align} C_+\overline{q}R(\kappa)=(1-C_+\overline{q}R_0(\kappa)q C_+)^{-1}C_+\overline{q}R_0(\kappa) \,. \end{align}\] It follows that \[R(\kappa)-R_0(\kappa) = R_0(\kappa)qC_+\overline{q}R(\kappa) = R_0(\kappa)q C_+ (1-C_+\overline{q}R_0(\kappa)q C_+)^{-1}C_+\overline{q}R_0(\kappa) \,.\] According to [3] we have \(\| C_+\overline{q}R_0(\kappa)q C_+\|<1\) for all sufficiently negative \(\kappa\) and then we can expand \((1-C_+\overline{q}R_0(\kappa)q C_+)^{-1}\) in a norm-convergent Neumann series. From now on we assume that \(\kappa\) is sufficiently negative. Taking into account that \(C_+\overline{q} R_0(\kappa)\) is Hilbert–Schmidt [3], we obtain \[\begin{align} \mathop{\mathrm{Tr}}(R(\kappa)-R_0(\kappa))&=\mathop{\mathrm{Tr}}(R_0(\kappa)q C_+(1-C_+\overline{q}R_0(\kappa)qC_+)^{-1}C_+\overline{q}R_0(\kappa))\\ &=\mathop{\mathrm{Tr}}((1-C_+\overline{q}R_0(\kappa)q)^{-1}C_+\overline{q}R_0(\kappa)^2qC_+)\\ & = \sum_{n=0}^\infty \mathop{\mathrm{Tr}}((C_+\overline{q}R_0(\kappa)q C_+)^n C_+\overline{q}R_0(\kappa)^2qC_+). \end{align}\] For each \(n\in\mathbb{N}_0\) we shall show that \[\label{eq:detgoal} \mathop{\mathrm{Tr}}((C_+\overline{q}R_0(\kappa)q C_+)^n C_+\overline{q}R_0(\kappa)^2qC_+) = \frac{1}{2\pi} \, \langle (R_0(\kappa) q C_+ \overline{q})^n R_0(\kappa)q,q \rangle \,.\tag{92}\] Once we have shown this, we can sum with respect to \(n\in\mathbb{N}_0\) and use the Neumann series \(\sum_{n=0}^\infty (R_0(\kappa) q C_+ \overline{q})^n R_0(\kappa) = R(\kappa)\) to obtain the claimed bound.

Let us turn to the proof of 92 . The case \(n=0\) is easy. Indeed, we have \[\mathop{\mathrm{Tr}}(R_0(\kappa)qC_+\overline{q}R_0(\kappa))=\frac{1}{2\pi}\int_0^\infty \frac{|\widehat{q}(\xi)|^2}{\xi+\kappa} \, \mathrm{d}\xi = \frac{1}{2\pi}{\left\langle R_0(\kappa)q,q\right\rangle} _{} \,,\] where the left equality follows from [3] and the right equality follows by Plancherel’s theorem. For \(n\geq 1\) we write \[\mathop{\mathrm{Tr}}((C_+\overline{q}R_0(\kappa)q C_+)^n C_+\overline{q}R_0(\kappa)^2qC_+) = \frac{1}{n+1}\partial_\kappa(\mathop{\mathrm{Tr}}((C_+\overline{q}R_0(\kappa)q C_+)^{n+1})) \,.\] We abbreviate \(m=n+1\) and compute \[\begin{align} &\frac{1}{m}\mathop{\mathrm{Tr}}((C_+\overline{q}R_0(\kappa)q C_+)^{m}) \\&=\frac{1}{m(2\pi)^{m}}\int_{[0,\infty)^{m}}\int_{[0,\infty)^{m}}\frac{\widehat{q}(\xi_1-\eta_1)\widehat{\overline{q}}(\eta_1-\xi_2)}{\xi_1-\kappa}\cdots \frac{\widehat{q}(\xi_{m}-\eta_{m})\widehat{\overline{q}}(\eta_{m}-\xi_1)}{\xi_{m}-\kappa}\, \mathrm{d}\eta\, \mathrm{d}\xi\\ &=\frac{1}{(2\pi)^{m}}\int_{[0,\infty)^{m}}\int_{\omega_{m}}\frac{\widehat{q}(\xi_1-\eta_1)\widehat{\overline{q}}(\eta_1-\xi_2)}{\xi_1-\kappa}\cdots \frac{\widehat{q}(\xi_{m}-\eta_{m})\widehat{\overline{q}}(\eta_{m}-\xi_1)}{\xi_{m}-\kappa}\, \mathrm{d}\eta\, \mathrm{d}\xi \,, \end{align}\] where \(\omega_{m} :=\{\eta\in [0,\infty)^{m}\colon \eta_{m} \leq \min\{\eta_1,\ldots,\eta_{m-1}\}\}\), and the last step follows by cyclicity. Taking \(\nu_j\mathrel{\vcenter{:}}=\xi_j-\eta_{m}\) for \(j=1, \ldots,m\), we have \[\begin{align} \ldots &=\frac{1}{(2\pi)^{m}}\int_{\omega_{m}}\int_{[-\eta_{m},\infty)^{m}}\frac{\widehat{q}(\nu_1+\eta_{m}-\eta_1)\widehat{\overline{q}}(\eta_1-\eta_{m}-\nu_2)}{\nu_1+\eta_{m}-\kappa}\cdots \frac{\widehat{q}(\nu_{m})\widehat{\overline{q}}(-\nu_1)}{\nu_{m}+\eta_{m}-\kappa} \, \mathrm{d}\nu \, \mathrm{d}\eta \,. \end{align}\] Finally, with \(\zeta_j\mathrel{\vcenter{:}}=\eta_j-\eta_m\geq0\) for \(j=1,\ldots,m-1\) and \(\lambda:=\kappa -\eta_m\leq \kappa\) and noting that \(\widehat{q}(\nu)=0\) for a.e. \(\nu<0\), we have \[\begin{align} \ldots &=\frac{1}{(2\pi)^m}\int_{-\infty}^\kappa \int_{[0,\infty)^{m-1}} \int_{[0,\infty)^{m}} \overline{\widehat{q}(\nu_1)}\frac{\widehat{q}(\nu_1-\zeta_1)\widehat{\overline{q}}(\zeta_1-\nu_2)}{\nu_1-\lambda} \cdots \frac{\widehat{q}(\nu_m)}{\nu_m-\lambda}\, \mathrm{d}\nu\, \mathrm{d}\zeta\, \mathrm{d}\lambda \\ &=\frac{1}{2\pi}\int_{-\infty}^\kappa \int_0^\infty \overline{\widehat{q}(\nu_1)} \mathcal{F}\left(\left[ R_0(\lambda)qC_+\overline{q}\right]^{m-1} R_0(\lambda)q\right)(\nu_1)\, \mathrm{d}\nu_1\, \mathrm{d}\lambda \\ &=\frac{1}{2\pi}\int_{-\infty}^{\kappa} \langle \left[ R_0(\lambda)qC_+\overline{q}\right]^{m-1}R_0(\lambda)q, q \rangle \, \mathrm{d}\lambda, \end{align}\] where we used Plancherel in the last step. Taking the derivative in \(\kappa\), we arrive at the claimed equality 92 . This completes the proof. ◻

Proof of Theorem 11. According to [3], the difference \((L_{q}-k)^{-1}-(L_{0}-k)^{-1}\) is trace class all sufficiently negative \(k\in\mathbb{R}\) and consequently for all \(k\in\mathbb{C}\setminus\sigma(L_{q})\). Therefore by [34], there is a function \(\xi\in L^1(\mathbb{R};(1+|\lambda|)^{-2})\), vanishing on \((-\infty,\inf\sigma(L_{q}))\), such that for any \(f\in\mathcal{C}\) the operator \(f(L_{q})-f(L_{0})\) is trace class and \[\label{eq:krein} \mathop{\mathrm{Tr}}\left( f(L_{q})-f(L_{0}) \right) = \int_\mathbb{R}\xi(\lambda) f'(\lambda)\, \mathrm{d}\lambda \,.\tag{93}\] In particular, this gives \[\mathop{\mathrm{Tr}}\left( (L_{q}-k)^{-1} - (L_{0}-k)^{-1} \right) = - \int_\mathbb{R}\frac{\xi(\lambda)}{(\lambda-k)^2}\, \mathrm{d}\lambda \,.\] Meanwhile, combining Lemma 18 with 91 , we find \[\mathop{\mathrm{Tr}}\left( (L_{q}-k)^{-1} - (L_{0}-k)^{-1} \right) = \int_\mathbb{R}\frac{d\nu(\lambda)}{\lambda-k}\, \mathrm{d}\lambda\] with the measure \(\nu\) from 90 . By uniqueness of the Cauchy–Stieltjes transform and the fact that \(\xi\) vanishes near \(-\infty\), a comparison of the previous two relations shows that \(\xi(\lambda) = \nu((-\infty,\lambda))\) for a.e. \(\lambda\in\mathbb{R}\), that is, \[\xi(\lambda) = - \frac{1}{(2\pi)^2} \int_0^\lambda |\beta(\lambda')|^2\, \mathrm{d}\lambda' - \#\{ j:\;\lambda_j<\lambda\} \,.\] Inserting this formula into 93 and integrating by parts we obtain the assertion of the theorem. Let us justify this integration by parts. From the definition of the class \(\mathcal{C}\) we see that \(f(\lambda) = c\lambda^{-1} + O(\lambda^{-1-\varepsilon})\) for some constant \(c\). Thus the integration by parts requires \(M^{-1} \int_0^M |\beta(\lambda')|^2\, \mathrm{d}\lambda'\to 0\) as \(M\to\infty\), which follows from the square integrability of \(\beta\), see Remark 4. ◻

Proof. Let us fix \(M\in \mathbb{N}\). We want to take the limit as \(s\rightarrow \infty\) with \(k=\mathrm{i}s\) in the identity 89 from Lemma 17. We consider the three terms separately.

Contribution from \(q\). It follows from Theorem 6 that \[\begin{align} \int_{-\infty}^\infty \overline{q(x)} m_0(x,\mathrm{i}s)\, \mathrm{d}x = \sum_{n=1}^{M+1} \frac{1}{(\mathrm{i}s)^n} \int_{-\infty}^\infty \overline{q(x)} \, c_{n}(x) \, \mathrm{d}x + o_{s\rightarrow\infty}(s^{-M-1}). \end{align}\]

Contribution from the eigenvalues. Since \(N\mathrel{\vcenter{:}}= N(L_{q})<\infty\), we clearly have \[\sum_{j=1}^N \frac{1}{\lambda_j-\mathrm{i}s} = - \sum_{j=1}^N \frac{1}{\mathrm{i}s}\frac{1}{1-\lambda_j/\mathrm{i}s} = - \sum_{n=0}^M \frac{1}{(\mathrm{i}s)^{n+1}}\sum_{j=1}^N\lambda_j^{n}+o_{s\rightarrow\infty}(s^{-M-1}).\]

Contribution from \(\beta\). Similarly, for the term with \(\beta\), we find \[\begin{align} \int_0^\infty \frac{|\beta(\lambda)|^2}{\lambda-\mathrm{i}s}\, \mathrm{d}\lambda=-\sum_{n=0}^M\frac{1}{(\mathrm{i}s)^{n+1}}\int_0^\infty|\beta(\lambda)|^2 \lambda^n\, \mathrm{d}\lambda+o_{s\rightarrow\infty}(s^{-M-1}), \end{align}\] where we have used the series expansion above together with the rapid decay of \(\beta\) from Theorem 6. More precisely, we use \((1-z)^{-1} - \sum_{n=0}^M z^n = (1-z)^{-1} z^{M+1}\) with \(z=\mathrm{i}s/\lambda\) and control the remainder using dominated convergence.

Putting everything together. Inserting the contributions from \(q\), from the eigenvalues and from \(\beta\) into 89 , we obtain \[\begin{align} \sum_{n=1}^{M+1} \frac{1}{(\mathrm{i}s)^n} \int_{-\infty}^\infty \overline{q(x)} \, c_{n}(x) \, \mathrm{d}x & = - \frac{1}{2\pi}\sum_{n=0}^{M}\frac{1}{(\mathrm{i}s)^{n+1}}\int_0^\infty {\left\vert\beta(\lambda)\right\vert}^2 \lambda^n\, \mathrm{d}\lambda - 2\pi \sum_{n=0}^M \frac{1}{(\mathrm{i}s)^{n+1}}\sum_{j=1}^N\lambda_j^{n} \\ & \quad +o_{s\rightarrow\infty}(s^{-M-1}). \end{align}\] The trace formulas are derived through an iterative process, repeatedly multiplying through by \(\mathrm{i}s\) and taking the limit \(s\rightarrow \infty\). Since \(M\) can be chosen arbitrarily large, we obtain the claimed infinite family of trace formulas. ◻