Negative Gibbs-volume capacity in a finite regulated shell

Write $e=-E$ and $\delta =e{r}_0\mathrm{/}A$. For $f(s)=s^{1\mathrm{/}2}(1-s{)}^{3\mathrm{/}2}$ and $h\mathrm{\ne }0$ with $\delta +h\in (0,1)$, the oriented-integral identity gives $(J(\delta +h)-J(\delta ))\mathrm{/}h=-(1\mathrm{/}h){\int }_{\delta }^{\delta +h}f(s)ds$; continuity makes this tend to $-f(\delta )$. At $\delta =1\mathrm{/}16$, restrict to $h<0$, which corresponds to right increments in $E$ within the branch. Hence $J^{\mathrm{\prime }}(\delta )=-{\delta }^{1\mathrm{/}2}(1-\delta {)}^{3\mathrm{/}2}$, with the one-sided interpretation at the endpoint. Differentiating $\log \mathrm{\Omega }=C_1-\frac{3}{2}\log e+\log J(\delta )$ with respect to $E=-e$ gives $d{S}_G\mathrm{/}dE=k_B(\frac{3}{2}+z)\mathrm{/}e=k_B\mathcal{B}\mathrm{/}e$. Its reciprocal is the stated positive $T_G$.

On $[1\mathrm{/}4,1\mathrm{/}2]\subseteq [\delta ,1]$, $s^{1\mathrm{/}2}\ge 1\mathrm{/}2$ and $(1-s{)}^{3\mathrm{/}2}\ge 1\mathrm{/}4$, so $J(\delta )\ge 1\mathrm{/}32$. Also ${\delta }^{3\mathrm{/}2}\le 1\mathrm{/}64$ and $(1-\delta {)}^{3\mathrm{/}2}\le 1$, hence $0<z\le 1\mathrm{/}2$.

Differentiating $T_G=e\mathrm{/}[k_B\mathcal{B}(\delta )]$ with respect to $e$, using $d\delta \mathrm{/}de=\delta \mathrm{/}e$, yields $\frac{d{T}_G}{de}=\frac{\mathcal{B}-\delta {\mathcal{B}}^{\mathrm{\prime }}}{k_B{\mathcal{B}}^2}\mathrm{.}$ Since $E=-e$, inversion gives the displayed formula for $C_G$, provided $\mathcal{B}-\delta {\mathcal{B}}^{\mathrm{\prime }}>0$.

Logarithmic differentiation of $z$ gives $\delta z^{\mathrm{\prime }}=\frac{3}{2}z\frac{1-2\delta }{1-\delta }+z^2\le \frac{3}{2}z+z^2,$ because $0<\delta \le 1\mathrm{/}16$ makes $(1-2\delta )\mathrm{/}(1-\delta )\le 1$. Therefore $\mathcal{B}-\delta {\mathcal{B}}^{\mathrm{\prime }}=\frac{3}{2}+z-\delta z^{\mathrm{\prime }}\ge \frac{3}{2}-\frac{z}{2}-z^2\ge 1$ when $0<z\le 1\mathrm{/}2$. Step 2.1 now gives $C_G<0$ throughout the branch.