To prove Theorem 2 we need to establish appropriate a priori bounds on a solution to our equation. We proceed in five steps.

Step 1: Preliminaries

We want to appeal to some of the standard body of theory for compact Kahler manifolds, that is, where we consider a fixed reference metric $\omega_{0}$ on a compact manifold and another metric $\omega=\omega_{0}=i\overline{\partial}\psi$. Our problem differs a little from that usually considered in the literature. To fit into a general setting we could consider a fixed smooth function $G$ of $p$-variables, a compact Kahler manifold $X$ with $p$ fixed holomorphic vector fields $v_{\alpha}$ and a function $\psi$ which satisfies an equation

$$(\omega_{0}+i\overline{\partial}\psi)^{n}=\exp(\psi+G(\nabla_{1}\psi,\dots,\nabla_{p}\psi))$$

where $\nabla_{\alpha}\psi$ denotes the derivative of $\psi$ along the vector field $v_{\alpha}$. Then the modification by Tian and Zhu ([32], Section 5, especially Prop. 5.1) of the standard argument of Yau, shows that in this situation an $L^{\infty}$ bound on $\psi$ leads to bounds on all higher derivatives. (Apart from this the proof we give is self-contained.)

In our toric setting, we choose some fixed admissible Kahler potential $\phi_{0}$ on ${\bf R}^{n}$ with Legendre transform $u_{0}$. Then we consider some general Kahler potential $\phi$, with Legendre transform $u$ and set $\psi=\phi-\phi_{0}$. So an $L^{\infty}$ bound on $\psi$ on the compact toric manifold is identical to an $L^{\infty}$ bound on $\phi-\phi_{0}$ on ${\bf R}^{n}$. Now a general property of the Legendre transform is that it is an isometry with respect to the $L^{\infty}$ distance: that is to say

$$\sup_{\underline{t}\in{\bf R}^{n}}|\phi(\underline{t})-\phi_{0}(\underline{t})|=\sup_{x\in P}|u(x)-u_{0}(x)|.$$

This is an elementary exercise.

In our situation, $u_{0}$ is a fixed continuous function on $\overline{P}$ so an $L^{\infty}$ bound on the function $\psi$ on the compact Kahler manifold is equivalent to an $L^{\infty}$ bound on the “unknown” symplectic potential $u$.

In sum, we see that to prove our proposition it suffices to establish an a priori $L^{\infty}$ bound on symplectic potentials $u$ satisfying a differential inequality

$$|L-h|\leq C,$$

(22)

for fixed $C$. Of course for this to make sense we have to normalise the non-uniqueness under the addition of linear functions, but we can do this very simply by restricting to functions $u$ whose derivative vanishes at the origin. i.e are minimised at the origin. We write $m=-u(0)$ and $M=\max_{\overline{P}}(u-u(0))$ and our problem comes down to obtaining upper and lower bounds on $m$ and an upper bound on $M$.

Step 2

Here we get a lower bound on $m=-u(0)$. Let the polytope $P$ be contained in the $R_{1}$ ball about $0$ in ${\bf R}^{n}$ and fix $R_{0}>0$ to be (say) half the distance from $0$ to the boundary of $P$. We will work in “generalised” polar coordinates $(r,\theta)$ on $P\subset{\bf R}^{n}$, so

$$h=r\frac{\partial u}{\partial r}-u(0)=r\frac{\partial u}{\partial r}+m.$$

Now let $\Omega\subset P$ be the set where $|\nabla u|\leq 1$. Then for $x\in\Omega$ we have $|h(x)-m|\leq R_{1}$ and the basic assumption (22) gives $L\leq m+R_{1}+C$ so

$$\det(u_{ij})\leq\exp(m+R_{1}+C).$$

But the integral of $\det(u_{ij})$ over $\Omega$ gives the volume $\omega_{n}$ of the unit ball in ${\bf R}^{n}$ so

$$\exp(m+R_{1}+C){\rm Vol}(\Omega)\geq\omega_{n}.$$

Since the volume of $\Omega$ cannot exceed the volume of $P$ this gives a lower bound on $m$.

Step 3

Here we obtain a bound on local averages of $h$, away from the origin. The bound depends on $M$ but, crucially, is $O(\log M)$.

For $x\in P$ let $d(x)$ be the distance to the boundary. We consider points where $d(x)\leq R_{0}/2$ and let $B_{x}$ be the ball of radius $d(x)/2$ centred at $x$. So $B_{x}$ is contained in $P$ and if $y\in B_{x}$ the norm $|y|$ is greater than $R_{0}$. Thus on $B_{x}$ we have

$$\frac{\partial u}{\partial r}\leq\frac{1}{R_{0}}(h-m).$$

Now we have an obvious bound, at any point $y$,

$$|\nabla u|\leq\frac{M}{d(y)}.$$

For $y\in B_{x}$ the distance $d(y)$ is at least $d(x)/2$, so $|\nabla u|\leq 2M/d(x)$ on $B_{x}$. This means that the derivative of $u$ maps $B_{x}$ into a ball of radius $2M/d(x)$ hence

$$\int_{B_{x}}\det\nabla^{2}u\ d\underline{x}\leq\omega_{n}\left(\frac{2M}{d(x)}\right)^{n}.$$

Thus we have a bound on the average, in an obvious notation,

$${\rm Av}(\det\nabla^{2}u,B_{x})\leq\frac{(2M)^{n}}{d(x)^{2n}}.$$

Now the concavity of the logarithm means that

$${\rm Av}(\log\det\nabla^{2}u,B_{x})\leq\log({\rm Av}(\det\nabla^{2}u,B_{x})),$$

so

$${\rm Av}(L,B_{x})\leq\log\left(\frac{(2M)^{n}}{d(x)^{2n}}\right)=n\log(2M)-2n\log d(x).$$

Now ${\rm Av}(h,B_{x})\leq{\rm Av}(L,B_{x})+C$ and ${\rm Av}(\frac{\partial u}{\partial r},B_{x})\leq R_{0}^{-1}{\rm Av}(h,B_{x})-m/R_{0}$. Putting this together we get

$${\rm Av}(\frac{\partial u}{\partial r},B_{x})\leq c_{1}\log M+c_{2}-c_{3}\log d-\frac{m}{R_{0}},$$

(23)

for known $c_{i}$.

Step 4

Here we give an elementary geometric argument to relate the average value of the radial derivative $\partial_{r}u=\frac{\partial u}{\partial r}$ to the growth of the function $u$, using convexity. We will write $\kappa_{i}$ for positive constants depending on the Euclidean geometry of the polytope $P$.

For $\delta\geq 0$ consider the slightly smaller polytope $(1-\delta)P$. Fix $\delta_{0}$ so that if $\delta<\delta_{0}$ this polytope contains the ball of radius $R_{0}$ about the origin. Let $M(\delta)$ be the maximum value of $u-u(0)$ on $(1-\delta)\overline{P}$, so $M(\delta)$ increases to $M$ as $\delta$ decreases to $0$. For each vertex $p$ on $P$ let $f_{p}(\delta)=u((1-\delta)p)-u(0)$. Then clearly

$$M(\delta)=\max_{p}f_{p}(\delta).$$

Suppose that at a given small $\delta$ the maximum is attained by $f_{p}$, for a certain vertex $p$. We want to show that the derivative $f^{\prime}_{p}(\delta)$ satisfies a bound of the same form as our bound on the local averages of $\partial_{r}u$. To see this consider the point $p^{\prime}=(1-\frac{\delta}{2})p$. It is obvious that $p^{\prime}$ is contained in the interior of the convex hull of $p$ and $(1-\delta)\overline{P}$. It will be equally clear to the reader who draws a diagram that if $q$ is any point within distance $\kappa_{1}\delta$ of $p$ then $p^{\prime}$ is in the interior of the convex hull of $q$ and $(1-\delta)\overline{P}$. Thus a convex set containing $(1-\delta)\overline{P}$ and with $p^{\prime}$ on its boundary cannot contain any point within distance $\kappa_{1}\delta$ of $p$.

With this discussion in place we can quickly complete the proof. Let $Z$ be the value of the radial derivative $\partial_{r}u$ at the point $(1-\delta)p$. Then

$$u(p^{\prime})\geq u((1-\delta)p)+\kappa_{2}Z\delta$$

Let $K$ be the closed convex set of points $x\in\overline{P}$ where $u(x)\leq u(p^{\prime})$. By the principle above, $K$ cannot meet the $\kappa_{1}\delta$ ball about $p$. Let $\sigma$ be any ray from the origin through a point $q$ which is within $\kappa_{1}\delta$ of $p$. Then there are $t<t^{\prime}<1$ such that $tq$ is in the boundary of $(1-\delta)P$ and $t^{\prime}q$ is in the boundary of $K$. Since $u(tq)\leq u((1-\delta p))$ the increase in $u$ along the segment from $tq$ to $t^{\prime}q$ is at least $\kappa_{2}Z\delta$. But the length of this segment is at most $O(\delta)$ and the radial derivative is increasing, so we see that the radial derivative $\partial_{r}u$ is at least $\kappa_{3}Z$ at the point $t^{\prime}q$, and hence a fortiori at $q$. Now by comparing with the average of the radial derivative over a suitable ball of radius $\kappa_{4}\delta$ we deduce that, after adjusting the constants $c_{i}$ appropriately, we have

$$M^{\prime}(\delta)\geq-(c_{1}\log M+c_{2}-c_{3}\log\delta-\frac{\kappa{5}m}{R_{0}}.$$

(24)

Step 5

Since the logarithm function is integrable around $0$ we deduce from (24), by integrating over $\delta$, that

$$M\leq M(\delta_{0})+C^{\prime}(\log M+1)-\epsilon m,$$

for known $\epsilon,C^{\prime}>0$. The convexity of $u$ gives

$$M(\delta_{0})\leq(1-\delta_{0})M.$$

So

$$\delta_{0}M\leq(C(\log M+1)-\epsilon m).$$

Since $\log M$ is $o(M)$ for large $M$ this has no solutions if $m$ is large, so we get an upper bound on $m$. On the other hand, the lower bound on $m$ obtained in Step 1 gives an upper bound on $M$ and we are finished.