= Solution
Choose any basis of two independent local solutions and collect their coefficients into a column $c_n$ immediately to the right of the $n$th cell boundary. The \b[<Floquet matrix>] is the one-period matrix $F(E)$ defined by
$$
c_{n+1}=F(E)c_n.
$$
Equivalently, it advances the Cauchy data $(\psi,\psi')^T$ by one period. The <Wronskian> of two solutions of the stationary <Schrödinger equation> is constant between the delta functions, and the derivative jump at each delta function also has determinant one. Consequently
$$
\boxed{\det F=1.}
$$
For a real potential the Floquet matrix in the real Cauchy-data basis is real, so its trace is real in every basis. Its <Floquet multipliers> $\rho_\pm$ satisfy
$$
\rho_+\rho_-=1,
\qquad
\rho_++\rho_-=\operatorname{tr}F.
$$
By <Bloch theorem>, an energy is in the interior of an <allowed energy band> when $\rho_\pm=e^{\pm iqa}$ are distinct unit-modulus numbers. Since $\operatorname{tr}F$ is real, this is equivalent to
$$
\boxed{(\operatorname{tr}F)^2<4.}
$$
For the <attractive delta-comb Kronig-Penney model>, integrating the <Schrödinger equation> across a lattice point gives
$$
\psi(na^+)=\psi(na^-),
\qquad
\psi'(na^+)-\psi'(na^-)=-2\lambda\psi(na).
$$
At negative energy write $E=-\hbar^2\mu^2/(2m)$ and use the local basis $e^{\pm\mu(x-na)}$. Free propagation through one cell followed by the derivative jump gives
$$
F_-(\mu)=
\begin{pmatrix}1-\lambda/\mu&-\lambda/\mu\\
\lambda/\mu&1+\lambda/\mu\end{pmatrix}
\begin{pmatrix}e^{\mu a}&0\\0&e^{-\mu a}\end{pmatrix}.
$$
Thus
$$
\det F_-=1,
\qquad
\frac12\operatorname{tr}F_-
=\cosh(\mu a)-\frac\lambda\mu\sinh(\mu a).
$$
The band condition is therefore
$$
-1<\cosh(\mu a)-\frac\lambda\mu\sinh(\mu a)<1.
$$
Using $(\cosh u-1)/\sinh u=\tanh(u/2)$ and $(\cosh u+1)/\sinh u=\coth(u/2)$ gives the requested inequality
$$
\boxed{\lambda\tanh\frac{\mu a}{2}<\mu<\lambda\coth\frac{\mu a}{2}.}
$$
At positive energy write $E=\hbar^2k^2/(2m)$ and use $e^{\pm ik(x-na)}$. The corresponding matrix is
$$
F_+(k)=
\begin{pmatrix}1+i\lambda/k&i\lambda/k\\
-i\lambda/k&1-i\lambda/k\end{pmatrix}
\begin{pmatrix}e^{ika}&0\\0&e^{-ika}\end{pmatrix},
$$
so
$$
\det F_+=1,
\qquad
\frac12\operatorname{tr}F_+
=\cos(ka)-\frac\lambda k\sin(ka).
$$
Hence the positive-energy bands are exactly the values of $k>0$ for which
$$
\boxed{\left|\cos(ka)-\frac\lambda k\sin(ka)\right|<1.}
$$
Finally, the zero-energy limit has $\tfrac12\operatorname{tr}F=1-\lambda a$. If $\lambda a>2$, then $|1-\lambda a|>1$, and therefore \b[zero energy lies in a forbidden gap].
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