The ideal-gas admissible state set is open and convex

On the convex half-space $\rho >0$, set $g(\rho ,m)=\mathrm{\mid }m{\mathrm{\mid }}^2\mathrm{/}(2\rho )$. For any direction $(a,b)\in \mathbb{R}\times {\mathbb{R}}^d$, differentiating twice along $(\rho +ta,m+tb)$ gives $\frac{d^2}{d{t}^2}g(\rho +ta,m+tb){\mathrm{\mid }}_{t=0}=\mathrm{\mid }b-(m\mathrm{/}\rho )a{\mathrm{\mid }}^2\mathrm{/}\rho \ge 0$. Thus $g$ is convex on that half-space.

Let $U_0=({\rho }_0,m_0,E_0)$ and $U_1=({\rho }_1,m_1,E_1)$ lie in $\mathcal{G}$, and let $0\le t\le 1$. Then ${\rho }_t=(1-t){\rho }_0+t{\rho }_1>0$ and convexity from step 1.1 gives $g({\rho }_t,m_t)\le (1-t)g({\rho }_0,m_0)+tg({\rho }_1,m_1)$. The strict endpoint inequalities $E_i>g({\rho }_i,m_i)$ imply $E_t>g({\rho }_t,m_t)$, so $U_t\in \mathcal{G}$.

The function $g$ is continuous where $\rho >0$, so the strict inequalities $\rho >0$ and $E-g(\rho ,m)>0$ define an open set there. Step 2.1 proves convexity, and choosing $t=1\mathrm{/}2$ proves the arithmetic-mean assertion.