 Solution

Regard $V(x)$ as an electron potential energy, and take $G=2\pi /a$. Each cosine supplies half its coefficient to each of its two Fourier coefficients. The nonzero coefficients areThe Fourier coefficient connecting the two plane waves at $k=\pm \pi /a$ is $V_G$, so the two-wave band gap isAt $\delta =0$ the actual periodic potential has period $a/2$ and no Fourier harmonic at $G$. The crossing at the artificially halved Brillouin zone boundary then has zero band gap. Higher-order weak-potential corrections can change the proportionality coefficient at nonzero $\delta$; the displayed coefficient belongs to the given two-wave approximation.

There is also a small-displacement qualification in the source: the claimed positions of the potential energy minima are correct to first order in $\delta$, not exactly. Put $t=2\pi x/a$. ThenAt $\delta =0.01$ and $x/a=1/4-\delta =0.24$, this equals approximately $-2.802\times 10^{-3}$, rather than zero. For $V_0>0$, expanding about $x/a=1/4$ and $3/4$ gives minima at $1/4-\delta +O({\delta }^3)$ and $3/4+\delta +O({\delta }^3)$. This does not alter the leading Fourier coefficient or the linear opening of the band gap.