in many ways like a Jost function): $${\label}{2.40} 
j_b(z) = a(z) q_{b-2}(z) + b(z) p_{b-1}(z)$$

[L2.4]{} $j_b$ is nonvanishing on the interior of the bands.

[*Remark.*]{} By $j_b(x)$ for $x$ real, we mean with $a$ defined via $\lim_{{\varepsilon}\downarrow 0} a(x+i{\varepsilon})$ since $P_\pm$ are only defined off ${{\mathbb{C}}}\backslash\text{bands}$.

As already mentioned, the boundary values obey $${\label}{2.41} 
\lim_{{\varepsilon}\downarrow 0}\, P_+ (x+i{\varepsilon}) = \lim_{{\varepsilon}\downarrow 0}\, P_- (x-i{\varepsilon})$$ (by the two-sheeted nature of $P_+$ and $P_-$). Thus, by and , $$\begin{aligned}
a(x+i0) &= -a (x-i0) {\label}{2.42} \\
b(x+i0) &= 1-b(x-i0) {\label}{2.43} \end{aligned}$$ Moreover, since $T_p$ and $\Gamma_\pm$ are real on ${{\mathbb{R}}}\backslash\text{bands}$, $a(z)$ and $b(z)$ are real on ${{\mathbb{R}}}\backslash\text{bands}$ (by ). Thus $$\begin{aligned}
a(x+i0) &= {\overline}{a(x-i0)}  {\label}{2.44} \\
b(x+i0) &= {\overline}{b(x-i0)} {\label}{2.45} \end{aligned}$$ The last four equations imply for $x$ in the bands $$\begin{aligned}
\operatorname{Re}(a(x+i0)) &= 0 {\label}{2.46} \\ 
\operatorname{Re}(b(x+i0)) &= \tfrac12  {\label}{2.47} \end{aligned}$$

$p$ and $q$ are real on ${{\mathbb{R}}}$, so $${\label}{2.48} 
\operatorname{Re}(j_b(x)) = \tfrac12\,  p_{b-1}(x)$$ Thus, if $j_b(x_0) =0$ on the bands, $p_{b-1}(x_0) =0$.

As we have seen, $a(z)=0$ only at the Dirichlet points and so not in the bands. If $p_{b-1}(x_0)=0=j_b(x_0)$, then since $a(x_0)\neq 0$, we also have $q_{b-2}(x_0) 
=0$. By , if $p_{b-1}(x_0) =q_{b-2}(x_0)$, then $\det (T_b (x_0))=0$, which is false. 