loss of generality that $H_{q_{+}}$ has at most one bound state $-\varkappa ^{2}$. Consider the problem with $$q_{n}(x)=\begin{cases}
q(x)\quad & ,\quad x\geq -n \\ 
0 & ,\quad x<-n\end{cases}.$$

It is well-known that for such initial profiles[^6] $$u_{n}(z,t)=-2\partial _{z}^{2}\log \det \left( 1+\mathbb{M}_{n}(z,t)\right) .
\label{eq5.1}$$By Proposition \[pr4.1\] $$\mathbb{M}(z,t)=\mathbb{M}_{+}(z,t)+\mathbb{A}(z,t)+\delta \mathbb{A}(z,t)$$where $\delta \mathbb{A}\equiv\mathbb{A}_{n}-\mathbb{A}$ is meromorphic in $z $ for any $t>0$ and small in the $\mathfrak{S}_{1}$-norm for $n$ large enough. I.e. $$\left\Vert \mathbb{M}_{n}(z,t)-\mathbb{M}(z,t)\right\Vert _{\mathfrak{S}_{1}}\rightarrow 0,\quad n\rightarrow \infty .  \label{eq5.2.1}$$Therefore, $\mathbb{M}(z,t)\in \mathfrak{S}_{1}$ is proven if we show that $\mathbb{M}_{+}(z,t)\in \mathfrak{S}_{1}$.

Split $$\mathbb{M}_{+}(z,t)=\mathbb{M}_{1}^{+}(z,t)+\mathbb{M}_{2}^{+}(z,t)$$where $\mathbb{M}_{1}^{+}(z,t),\mathbb{M}_{2}^{+}(z,t)$ are the Hankel operators with the kernels $$c_{0}^{2}\zeta _{z,t}(i\varkappa )e^{-\varkappa (x+y)},$$and $$\frac{1}{2\pi }\int_{-\infty }^{\infty }e^{i\lambda (x+y)}\zeta
_{z,t}(\lambda )R_{+}(\lambda )d\lambda$$respectively. Here $c_{0}$ stands for the norming constant associated with the bound state $-\varkappa ^{2}$.

The operator $\mathbb{M}_{1}(z,t)$ is rank 1 and clearly entire in $z$. Thus we only need to properly control $\partial _{z}^{n}\mathbb{M}_{2}^{+}(z,t)$ in the $\mathfrak{S}_{1}$-norm. Evaluate (so far formally) the kernel of $\partial _{z}^{n}\mathbb{M}_{2}^{+}(z,t),n=0,1,2,...,$ by the Green formula applied to the strip $\mathbb{R}\times (0,\varkappa /2)$ and by Proposition \[pr3.2\] ($\lambda =\alpha +i\beta ,\partial _{\overline{\lambda }}=\frac{1}{2}(\partial _{\alpha }+i\partial _{\beta })$) $$\begin{aligned}
\frac{1}{2\pi }\int_{-\infty }^{\infty }e^{i\lambda (x+y)}(2i\lambda
)^{n}\zeta _{z,t}(\lambda )R_{+}(\lambda )d\lambda & =\int_{\mathbb{R}+i\varkappa /2}e^{i\lambda (x+y)}(2i\lambda )^{n}\left( B^{-1}SG\right)
(\lambda )\frac{d\lambda }{2\pi }  \notag \\
