For any symplectic potential $u$ on our Fano polytope, centred at the origin, we write $\rho=L-h$. Now we define the following weighted norms, for functions $f,g$ on $P$:

$$\langle f,g\rangle_{u}=\int_{P}fge^{\rho}\ d\underline{x};$$

$$\langle\nabla f,\nabla g\rangle_{u}=\int_{P}f_{i}g_{a}u^{ia}e^{\rho}\ d\underline{x};$$

$$\langle\nabla^{2}f,\nabla^{2}g\rangle_{u}=\int_{P}f_{ij}g_{ab}u^{ia}u^{jb}e^{\rho}\ d\underline{x}.$$

The first variation of $\rho$ with respect to an infinitesimal variation $f$ in $u$ is $\delta\rho=\Box f$, where $\Box$ is the differential operator

$$\Box f=u^{ij}f_{ij}-x^{i}f_{i}+f.$$

(13)

Since

$$\rho_{j}=-u^{ia}_{a}u^{aj}-x^{a}u_{ja},$$

this can also be written as

$$\Box f=\left(u^{ij}f_{i}\right)_{j}-u{ij}\rho_{j}f_{i}+f,$$

(14)

from which it follows that

$$\langle\Box f,g\rangle_{u}=-\langle\nabla f,\nabla g\rangle_{u}+\langle f,g\rangle_{u}.$$

(15)

In particular, $\Box$ is self-adjoint with respect to the weighted norm.

Now define a functional by

$${\cal F}(u)=\int_{P}e^{\rho}\ d\underline{x}.$$

(16)

Then the first variation is

$$\delta{\cal F}=\int_{P}\Box fe^{\rho}d\underline{x}=\langle\Box f,1\rangle_{u}.$$

(17)

By the self-adjoint property we can also write this as

$$\delta{\cal F}=\langle f,\Box 1\rangle_{u}=\langle f,1\rangle_{u}=\int_{P}fe^{\rho}d\underline{x}.$$

(18)

This leads to two different expressions for the second variation of ${\cal F}$. If we put $u_{t}=u+tf,\rho_{t}=\rho(u_{t})$ and write $\Box_{t}$ for the operator defined by $u_{t}$ then

$$\frac{d}{dt}\Box_{t}f=-u^{ia}u^{jb}f_{ij}f_{ab}.$$

So,

$$\frac{d^{2}}{dt^{2}}{\cal F}(u_{t})=\frac{d}{dt}\int_{P}\Box_{t}fe^{\rho_{t}}d\underline{x}=\int_{P}\left(\Box_{t}f\Box_{t}f-u^{ia}u^{jb}f_{ij}f_{ab}\right)e^{\rho_{t}}d\underline{x},$$

which is equal to

$$\langle\Box_{t}f,\Box_{t}f\rangle_{u_{t}}-\langle\nabla^{2}f,\nabla^{2}f\rangle_{u_{t}}.$$

On the other hand

$$\frac{d^{2}}{dt^{2}}{\cal F}(u_{t})=\frac{d}{dt}\int_{P}fe^{\rho_{t}}d\underline{x}=\int_{P}f\Box_{t}fe^{\rho_{t}}d\underline{x}.$$

So, evaluating at $t=0$ and dropping $t$ from the notation, we have the identity

$$\langle\Box f,\Box f\rangle_{u}-\langle\nabla^{2}f,\nabla^{2}f\rangle_{u}=\langle f,\Box f\rangle_{u}.$$

(19)

Applying (15), with $g=\Box f$, this gives,

$$\langle\nabla f,\nabla\Box f\rangle_{u}=-\langle\nabla^{2}f,\nabla^{2}f\rangle_{u}.$$

(20)

It is obvious from the definition that $\Box$ vanishes on the linear functions and $\Box 1=1$. If $f$ is any eigenfunction of $\Box$, with eigenvalue $\lambda$, which is orthogonal to the linear functions and then constants, then $\nabla^{2}f$ is non-zero and the identity gives

$$\lambda\langle\nabla f,\nabla f\rangle_{u}=-\langle\nabla^{2}f,\nabla^{2}f\rangle_{u},$$

so $\lambda<0$. (This is a variant of the standard lower bound on the eigenvalues of the Laplacian on a manifold with positive Ricci curvature, the identity can of course be verified more directly, but the argument above avoids some laborious manipulation.) In sum, we have derived an inequality

$$\langle 1,f\rangle_{u}\langle 1,1\rangle_{u}-\langle f,\Box f\rangle_{u}\geq 0,$$

(21)

with equality if and only if $f$ is a linear function.

Now to apply this to our problem. First, we can use the continuity method for the equation $L-h=A+\sum\gamma_{i}x^{i}$, with respect to variations in $A$. The linearised equation is $\Box_{u}f=\delta A+\sum\delta\gamma_{i}x^{i}$. Since the cokernel of $\Box_{u}$ is identified with the linear functions this linearised equation has a solution and we can apply the implicit function theorem in the usual way.

Second, we obtain the uniqueness of solutions. Consider the functional $-\log{\cal F}$. Along a line $u_{t}=u+tf$ we have

$$\frac{d^{2}}{dt^{2}}\left(-\log{\cal F}\right)=\frac{1}{{\cal F}^{2}}({\cal F}{\cal F}^{\prime}-{\cal F}^{\prime\prime})$$

where ${\cal F}^{\prime},{\cal F}^{\prime\prime}$ denote the derivatives of ${\cal F}$. Evaluating at $t=0$ we have

$${\cal F}=\langle 1,1\rangle_{u},{\cal F}^{\prime}=\langle 1,f\rangle_{u},{\cal F}^{\prime\prime}=\langle f,\Box_{u}f\rangle_{u},$$

so our inequality (21) asserts that the second derivative of $-\log{\cal F}$ is positive, and strictly positive unless $f$ is affine-linear. Thus $-\log{\cal F}$ is a convex function. Now if $\rho=A+\sum\gamma_{i}x^{i}$ the $\gamma_{i}$ are determined by $A$, using the same argument as in the previous subsection. So we may as well suppose that $\gamma_{i}=0$. Then ${\cal F}(u)=C$ where $C$ is the integral of $e^{A}$. The equation $\rho=A$ is the Euler-Lagrange equation for critical points of the linear function

$$u\mapsto\int_{P}ue^{A}\ d\underline{x}$$

subject to the constraint $-\log{\cal F}=-\log C$. The convexity gives uniqueness, modulo linear functions.