Theorem 11.17.2854B49CB03B9

Proof. Step 1: basic bounds. Put $A_ x=\overline\Omega \cap E_ x$. By compactness of $X$ there are $0<a_0<b_0$ such that, in every fiber, the closed $k$-ball of radius $a_0$ is contained in $\Omega $ and $\overline\Omega $ lies in the open ball of radius $b_0$. The maximum defining $p$ is attained, and

As a supremum of absolute values of linear functionals, $p$ is complex homogeneous and satisfies the triangle inequality.

Step 2: support points. Fix $\xi _0\ne 0$ and a maximizing point $v_0\in A_{x_0}$. It lies on $\partial \Omega $, since at an interior point a small vertical displacement would increase $|\xi _0(v)|$. Multiply $\xi _0$ by a complex number of modulus one to obtain $\eta $ with $\eta (v_0)=p(x_0,\xi _0)=a>0$. Then $\ell =\operatorname {Re}\eta $ satisfies $\ell \leq a$ on $A_{x_0}$. Choose a smooth defining function $f$ near the boundary, with $\Omega =\{ f<0\} $ and $\mathrm df\ne 0$ there.

Let $V=T^{1,0}_{v_0}E_{x_0}$ be the vertical tangent space and $T=\ker \partial f(v_0)$. Choose an $n$-dimensional complex subspace $W\subset T$ on which $\mathcal L_ f$ is negative definite. We first show that the vertical differential $\mathrm d_ vf(v_0)$ is nonzero. Otherwise $V\subset T$, and counting dimensions gives $W\cap V\ne \{ 0\} $. A vector in this intersection has negative vertical Levi form, and the two corresponding real directions show that the real vertical Hessian of $f$ has a negative direction. Its negative cone is open, hence contains a direction $w$ with $\ell (w)>0$. Since the vertical first derivative vanishes, Taylor expansion gives $f(v_0+tw)<0$ for small $t>0$, contradicting $\ell \leq a$ on $A_{x_0}$.

The boundary of $A_{x_0}$ is therefore smooth at $v_0$. Since the supporting functional $\ell $ attains its maximum on $A_{x_0}$ there,

the sign follows because an inward direction along which $\ell $ increases would contradict the maximality. On the affine hyperplane $\{ \ell =a\} $ of the fiber, the restriction of $f$ has a local minimum zero at $v_0$: a point there with $f<0$ would allow an additional displacement increasing $\ell $ within $\Omega $. The real vertical Hessian of $f$ is thus nonnegative on $\ker \mathrm d\ell $. Consequently $\mathcal L_ f$ is nonnegative on the complex space $V\cap T$, of dimension $r-1$. Hence $W\cap V=\{ 0\} $, and the projection $W\to T^{1,0}_{x_0}X$ is an isomorphism.

Step 3: plurisubharmonicity. In local coordinates $z$ centered at $x_0$, choose a holomorphic section $\sigma _0$ through $v_0$ with tangent space $W$. The function $f\circ \sigma _0$ has zero first derivative and negative definite mixed Hessian. Its quadratic expansion is

where $P$ is a holomorphic quadratic polynomial. Choose a vertical vector $b$ with $\partial _ vf(b)=1$, and replace $\sigma _0$ by $\sigma =\sigma _0-bP$. This preserves the first jet and cancels the pure quadratic part. Thus, after shrinking the chart,

for some $\epsilon >0$. In particular $\sigma (z)\in A_ z$ throughout the chart.

Compactness of $\overline\Omega $ makes $p$ upper semicontinuous: for convergent covectors, the corresponding maximizing vectors have a subsequence whose limit lies in the limiting fiber of $\overline\Omega $. The supporting section gives the opposite inequality $p(z,\xi )\geq |\xi (\sigma (z))|$, with equality at $(x_0,\xi _0)$. Therefore $p$ is continuous off the zero section, and its upper bound gives continuity at zero. The holomorphic pairing $\xi (\sigma (z))$ is nonzero near the contact point, and its logarithmic absolute value is a pluriharmonic function lying below $\log p$ and equal to it at $(x_0,\xi _0)$. The mean-value property on small holomorphic disks therefore gives the submean inequality for $\log p$ at every point off the zero section. Thus $\log p$, and then $Q=\exp (2\log p)$ by Jensen's inequality, are plurisubharmonic there. At the zero section the continuous nonnegative function $Q$ vanishes, so its submean inequality on disks also holds there.

Step 4: uniform convexity in the fibers. Consider all maximizing triples $(x,\xi ,v)$ with $|\xi |_{k^*}=1$. They form a compact set, since $p$ is continuous, and by Step 2 the vertical differential $\mathrm d_ vf$ does not vanish at these support points. Thus there are uniform constants $m>0$, $\rho >0$ and $C\geq 1$ such that, for vertical vectors $w$,

Choose $0<\delta <\min \{ a_0/2,\rho /2,m/(2C)\} $. At a support point, write $N$ for the outward real vertical unit normal and set $c=v-\delta N$. If $|w+\delta N|_ k\leq \delta $, then

All these points lie in the neighborhood where $f$ is defined, so the closed ball $\overline B_ k(c,\delta )$ is contained in $A_ x$.

The support function of this ball is $s_ c(\xi )=|\xi (c)|+\delta |\xi |_{k^*}\leq p(x,\xi )$. For a maximizing covector of unit norm, with its phase normalized as in Step 2, the real part $\operatorname {Re}\eta $ has real dual norm one and its metric dual vector is $N$. Hence $\operatorname {Re}\eta (N)=1$, and since $|\eta (N)|\leq 1$, also $\eta (N)=1$. We obtain $\eta (c)=a-\delta >0$, and therefore $s_ c(\xi )=a$. Homogeneity now gives

where $c$ ranges over the centers of the supporting balls in the fiber. Every such center satisfies $|c|_ k\leq b_0+\delta $, so by lemma 11.17.1 all the functions $s_ c^2-\gamma |\xi |_{k^*}^2$ are convex for the same $\gamma >0$. Their supremum is finite, and it proves the asserted strong convexity of $Q$.

Step 5: smoothing. Finally, if $E$ is ample, proposition 11.11.2 gives the smooth strictly plurisubharmonic homogeneous function on $E^*$ required in proposition 11.10.5, and applying that proposition to $Q$ gives the smooth norm. $\square$