The condition that a toric manifold $X$ be Fano, with $L=K_{X}^{-1}$ is easily stated in terms of the polytope $P$. There is a preferred “centre” $\nu_{0}\in P$ such that for each face $\lambda_{r}(p_{0})-c_{r}=1$. This follows because the wedge product of the vector fields generating the action is a meromorphic $n$-form on $X$ with a simple pole along each of the divisors corresponding to the faces. Then the inverse is a section of $K_{X}^{-1}$ and is a multiple of the standard basis element $s_{\nu_{0}}$. This centre is also the centre of mass of $(\partial P,d\sigma)$.

In this Section we discuss a Theorem of Wang and Zhu [34].

We will begin by giving a proof which is somewhat different to that of Wang and Zhu (although it borrows ideas from that paper and from [32]), working largely with the symplectic description. We can assume that the centre $\nu_{0}$ is the origin. Given a symplectic potential $u$ we write

$$h=x^{i}u_{i}-u,$$

and

$$L=\log\det\nabla^{2}u.$$

These are smooth functions on $P$ but both tend to infinity at the boundary. Note that $h$ depends on a choice of origin in ${\bf R}^{n}$. Of course $h$ is just the composite of the Kahler potential $\phi$ with the derivative of $u$, mapping $P$ to ${\bf R}^{n}$. The assumption that the toric manifold $X$ be Fano is equivalent to the fact that, for any admissible $u$, the difference $L-h$ is a smooth function on $\overline{P}$. The condition that $u$ describe a Kahler-Ricci soliton is that

$$L-h=\sum c_{i}x^{i},$$

(11)

for constants $c_{i}$ (which of course specify the relevant holomorphic vector field on the Kahler manifold). Just as in our discussion of extremal metrics, it is natural in this context to consider more generally an equation $L-h=A$ for some prescribed smooth function $A$ on $\overline{P}$. Again, much as for the extremal case, there are elementary constraints that we need to impose on $A$. For any symplectic potential $u$ we consider the integrals

$$\int_{P}x^{i}e^{L-h}d\underline{x},$$

for $i=1,\dots,n$. Transforming the integral to the dual space, it becomes

$$\int_{{\bf R}^{n}}\frac{\partial\phi}{\partial t_{i}}e^{-\phi}d\underline{t}=-\int_{{\bf R}^{[}n}\frac{\partial e^{-\phi}}{\partial t_{i}}=0.$$

So a necessary condition that the equation $L-h=A$ has a solution is that, for each $i$,

$$\int_{P}x^{i}e^{A}d\underline{x}=0.$$

(12)

This fixes the constants $c_{i}$ in (11). To see this, consider the function of $\underline{c}\in{\bf R}^{n}$:

$$F(\underline{c})=\int_{P}e^{\sum c_{i}x^{i}}d\underline{x}$$

This is convex and proper (since the origin lies in $P$) and so has a unique critical point. But the derivative of $F$ with respect to $c_{i}$ is

$$\int_{P}x^{i}e^{\sum c_{i}x^{i}}\ d\underline{x}.$$

So the unique critical point of $F$ gives exactly the constants $c_{i}$ required to satisfy the constraint.

In sum, the theorem of Wang and Zhu follows from

An equivalent statement is

For any smooth function $A$ on $\overline{P}$ there are constants $\gamma_{i}$ and an admissible potential $u$ such that $L-h=A+\sum\gamma_{i}x^{i}$. The $\gamma_{i}$ are unique and $u$ is unique up to the addition of a linear function.

The equivalence of the statements follows from the same argument as above.

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.

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.

We will now discuss briefly the original approach of Wang and Zhu. For simplicity we will just consider the case when the Futaki invariant vanishes, so we seek a Kahler-Einstein metric. Recall from the above that the vanishing Futaki invariant is equivalent to fact that the centre of mass of the polytope $P$ is the preferred centre, which we are taking as $0\in{\bf R}^{n}$.

Wang and Zhu use the continuity method with respect to the family of equations

$$\det(\nabla^{2}\phi)=\exp(-(s\phi+(1-s)f)),$$

(25)

where $f$ is a fixed admissible Kahler potential and $0\leq s<1$. We discuss first the case when $s=0$. Then the equation in question is just the toric case of the “prescribed volume form” equation, solved, for general Kahler manifolds, by Yau. But let us see how to give a simple proof in this special situation. As we have seen it suffices to bound the $L^{\infty}$ norm of the symplectic potential $u$ corresponding to $\phi$. We can apply the Sobolev inequality so for each $p>n$ there is a $c_{p}$ such that

$${\rm Osc}_{P}(u)\leq c_{p}\|\nabla u\|_{L^{p}}.$$

So we conclude that in our problem it suffices to find $C$ such that there is some point $x\in P$ with $-C\leq u(x)\leq C$ and $\|\nabla u\|_{L^{p}}\leq C.$

The equation (25) with $s=0$ is degenerate, in that we can obviously change $\phi$ by the addition of a constant, so we may normalise $u$ to be zero at some point. Thus all we need to do is bound the $L^{p}$ norm of $\nabla u$. But for this we simply write

$$\int_{P}|\nabla u|^{p}\ d\underline{x}=\int_{{\bf R}^{n}}|\underline{t}|^{p}\det\nabla^{2}\phi\ d\underline{t}=\int_{{\bf R}^{n}}|\underline{t}|^{p}e^{-f}\ d\underline{t}<\infty.$$

This concludes the proof of the $L^{\infty}$ estimate for the case $s=0$. (Here we have not used the fact that $f$ is convex, so by deforming $f$ one can prove the toric case of Yau’s Theorem: the existence of a solution for any $f$.)

Now we go on to the main case, when $s>0$. It suffices to obtain estimates for $s\geq s_{0}$ for some fixed $s_{0}>0$.

Set $w=s\phi+(1-s)f$. Then $w$ is another admissible function and

$$\det(\nabla^{2}w)\geq s_{0}^{n}\det(\nabla^{2}\phi)=s_{0}^{n}e^{-w}.$$

Let the minimal value of $w$ be $m$, attained at a point $\zeta\in{\bf R}^{n}$. The first main step in the proof is

The foundation of the approach of Wang and Zhu is the following fact.

Wang and Zhu prove this using a comparison argument. It can also be shown using the elementary geometry of the derivative of $v$ (see [17] Prop. 3.2.3), but both approaches depend on the fact that after a unimodular affine transformation we can suppose that there are concentric balls

$$B(R_{1})\subset K\subset B(R_{2}),$$

with the ratio $R_{2}/R_{1}$ of the radii bounded by a fixed constant depending on the dimension. Notice that a reverse inequality holds. If in the same situation $\det(\nabla^{2}v)\leq\Lambda$ then ${\rm Vol}(K)\geq C\Lambda^{-1/2}$ ([17], Cor. 3.2.4).

With this background in place we can proceed to explain the proof of Wang and Zhu. Let $m$ be the minimal value of the function $w$ and set $v=w-m$. Then $\det(\nabla^{2}v)\geq\lambda=t_{0}^{n}e^{m+1}$ on the set $K$ where $v\leq 1$. So we deduce that

$${\rm Vol}(K)\leq C\lambda^{-1/2}=C^{\prime}e^{m/2},$$

(26)

say. For each positive $h$ let $K_{h}$ be the set $\{v\leq h\}$ and $V(h)={\rm Vol}(K_{h})$. Then convexity implies that $K_{h}$ is contained in the dilate of $K$ by factor $h$ about the minimum point of $v$. Thus

$$V(h)={\rm Vol}(K_{h})\leq h^{n}{\rm Vol}(K)\leq h^{n}C^{\prime}e^{m/2}.$$

By the co-area formula

$$\int_{{\bf R}^{n}}e^{-w}\ d\underline{t}=\int_{0}^{\infty}e^{-h}V(h)\ dh.$$

Now the volume form $\det(\nabla^{2}\phi)$ is at most $e^{-m}e^{-v}$ and its integral is the volume of our manifold $X$. So

$${\rm Vol}(X)\leq e^{-m}\int_{0}^{\infty}C^{\prime}e^{m/2}e^{-h}h^{n}\ dh=C^{\prime\prime}e^{-m/2},$$

say. We see that

$$m\leq m_{0}=2\log(I_{0}/C^{\prime\prime}),$$

(27)

and then deduce from (26) that

$${\rm Vol}(K)\leq C^{\prime}e^{m_{0}/2}.$$

(28)

Now we use the fact that $|\nabla w|\leq b$ say. This means that the distance from the boundary of $K$ to the minimum point $\zeta$. is at least $b^{-1}$, so $K$ contains a ball of this fixed radius about $\zeta$. If $K$ contains a point $\zeta^{\prime}$ with $|\zeta-\zeta^{\prime}|=R$ for large $R$, then the volume of $K$ would be large, contradicting the bound (28). So we conclude that $K$ is contained in the ball $\{\zeta^{\prime}:|\zeta^{\prime}-\zeta|\leq R_{0}\}$ for some fixed $R_{0}$. But then convexity implies that

$$|\xi-\zeta|\leq R_{0}^{-1}v(\xi).$$

This completes the proof of Proposition 1.

The second main step is to show that $|\zeta|$ is not large. This is where the hypothesis that the the Futaki invariant vanishes is used. Consider the derivative $Df$ of the fixed admissible function $f$. This is a vector-valued function on ${\bf R}^{n}$, which gives a proper map to the open polytope $P$. The crucial thing is an identity

$$\int_{{\bf R}^{n}}Dfe^{-w}\ d\underline{t}=0.$$

(29)

To see this, consider one component $\frac{\partial f}{\partial t_{a}}=f^{a}$ of $Df$, and observe first that

$$\int_{{\bf R}^{n}}\left((1-s)\frac{\partial f}{\partial t_{a}}+s\frac{\partial\phi}{\partial t_{a}}\right)e^{-w}\ d\underline{t}=\int_{{\bf R}^{n}}\frac{\partial w}{\partial t_{a}}e^{-w}\ d\underline{t}=0.$$

So it is the same to show that

$$\int_{{\bf R}^{n}}\frac{\partial\phi}{\partial t_{a}}e^{-w}\ d\underline{t}=0.$$

But this integral is

$$\int_{\bf R}^{n}\frac{\partial\phi}{\partial t_{a}}\det(\phi_{ab})\ d\underline{t}$$

which is the same as

$$\int_{P}x^{a}\ d\underline{x}$$

and this vanishes by our hypothesis.

Consider a codimension-$1$ face of $P$ defined by an equation $\lambda_{r}(x)=c_{r}$. Let $g_{r}$ be the function

$$g_{r}(\underline{t})=\log(\lambda_{r}(Df(\underline{t}))-c_{r}).$$

It is easy to check that the derivative of $g_{r}$ is bounded on ${\bf R}^{n}$. Suppose $|\zeta|$ is large. This means that $Df(\zeta)$ is close to the boundary of $P$, so there is some $r$ for which $g_{r}(\zeta)$ is very negative $g_{r}(\zeta)\leq-M$ say, for $M$ large. Then the bound on the derivative of $g_{r}$ means that we can find a constant $\sigma$ such that on the ball $B$ of radius $\sigma M$ about $\zeta$ we have $g_{r}\leq-M/2$. Thus $\lambda_{r}(Df)\geq c_{r}/2$ say, on $B$, if $M$ is large enough. Equally, it follows from Proposition 1 that when $M$ is large the integral of $e^{-w}$ over ${\bf R}^{n}\setminus B$ is small. This shows that

$$\int_{{\bf R}^{n}}\lambda_{r}(DF)e^{-w}>0,$$

if $M$ is large, which is a contradiction to the identity (29) above.

It is now easy to complete the proof. Since $\zeta$ is bounded we have

$$w(\underline{t})\geq\epsilon|\underline{t}|-c$$

and the bound on the $L^{p}$ norm of $\nabla u$ follows just as before. Then it is straightforward to get upper and lower bounds on $u$ at some point, for example the point corresponding to $\zeta$.