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 $\zeta$.