Unique finite gravitational Gibbs variational optimum

For $s>0$, $\varphi (s)=s\log s-s+1$ has derivative $\log s$, is decreasing for $0<s<1$, and increasing for $s>1$. Its continuous value at $s=0$ is $1$, so $\varphi (s)\ge 0$ with equality exactly at $s=1$.

Since $p_{\beta }>0$, set $r=\rho \mathrm{/}{p}_{\beta }$ almost everywhere. The assumptions and $\log p_{\beta }=-\beta H_{N,a}-\log Z_N$ make $\rho \log (\rho \mathrm{/}{p}_{\beta })$ integrable. Expanding $\varphi$ and using $\int \rho =\int p_{\beta }=1$ gives $D(\rho \mathrm{\parallel }{p}_{\beta })=\int p_{\beta }\varphi (r)d{\nu }_N=\int \rho \log \rho d{\nu }_N+\beta ⟨H_{N,a}{⟩}_{\rho }+\log Z_N\ge 0.$

Rearranging step 2.1 and using the entropy convention [F2] gives the stated inequality. Equality forces $\varphi (r)=0$ almost everywhere for the probability measure $p_{\beta }d{\nu }_N$, so $r=1$ almost everywhere by [F3], and $\rho =p_{\beta }$. Conversely, if $\rho =p_{\beta }$ almost everywhere then $D(\rho \mathrm{\parallel }{p}_{\beta })=0$ in step 2.1 and the inequality is an equality. At equal mean energy this is the unique entropy-maximizing density.

Put $\psi (x)=(1+x)\log (1+x)-x$. For $\mathrm{\mid }x\mathrm{\mid }\le 1\mathrm{/}2$, $\psi (0)={\psi }^{\mathrm{\prime }}(0)=0$ and ${\psi }^{\mathrm{\prime }\mathrm{\prime }}(x)=1\mathrm{/}(1+x)$, hence $\mathrm{\mid }{\psi }^{\mathrm{\prime }\mathrm{\prime }}(x)-1\mathrm{\mid }\le 2\mathrm{\mid }x\mathrm{\mid }$. Integrating this second-derivative bound twice from zero gives $\mathrm{\mid }\psi (x)-x^2\mathrm{/}2\mathrm{\mid }\le \mathrm{\mid }x{\mathrm{\mid }}^3\mathrm{/}3\le 2\mathrm{\mid }x{\mathrm{\mid }}^3\mathrm{/}3$. The two centering conditions make ${\rho }_t$ normalized and energy preserving, while $\mathrm{\mid }tu\mathrm{\mid }\le 1\mathrm{/}2$ makes it nonnegative. Moreover [F1] gives $\int \mathrm{\mid }{H}_{N,a}\mathrm{\mid }{p}_{\beta }<\mathrm{\infty }$ and $\int p_{\beta }\mathrm{\mid }\log p_{\beta }\mathrm{\mid }<\mathrm{\infty }$ by $\log p_{\beta }=-\beta H_{N,a}-\log Z_N$; since $1+tu$ and $\log (1+tu)$ are bounded, these imply $\int \mathrm{\mid }{H}_{N,a}\mathrm{\mid }{\rho }_t<\mathrm{\infty }$ and $\int \mathrm{\mid }{\rho }_t\log {\rho }_t\mathrm{\mid }<\mathrm{\infty }$. Thus ${\rho }_t$ belongs to the stated density class. Therefore $D({\rho }_t\mathrm{\parallel }{p}_{\beta })=\int p_{\beta }\psi (tu)d{\nu }_N=\frac{t^2}{2}\int p_{\beta }{u}^{2}d{\nu }_N+R_t$ with the stated bound. Since $u$ is nonzero in $L^2(p_{\beta }d{\nu }_N)$, the quadratic term dominates the cubic bound for sufficiently small nonzero $t$. The equal-energy identity in step 2.1 then gives $S({\rho }_t)<S(p_{\beta })$.