11.17 Concave neighborhoods and dual norms4276F3B6339D

Proof. For $0<\tau \leq 1$, replace $s_ c$ by the norm

which is smooth off zero. Its Lipschitz constant is at most $L=\sqrt{B^2+1}+\delta $, uniformly in $c$ and $\tau $. All derivatives below are real derivatives on the dual space. At $\xi \ne 0$, set $u=\xi /|\xi |$ and decompose a vector as $w=\alpha u+b$ with $b$ orthogonal to $u$ for the real inner product. Write $s=s_{c,\tau }$ and $A=\mathrm ds_\xi (w)$. Since the first summand of $s$ is convex and $s\geq \delta |\xi |$,

Homogeneity gives $\mathrm ds_\xi (u)=s(\xi )/|\xi |\geq \delta $. Since $|\mathrm ds_\xi (b)|\leq L|b|$, we have

Take $\gamma =\delta ^4/(\delta ^2+2L^2)$. Then $\gamma \leq \delta ^2/2$ and $\gamma (1+2L^2/\delta ^2)=\delta ^2$, so $D^2(s^2)(w,w)\geq 2\gamma |w|^2$.

The homogeneous function $s^2-\gamma |\xi |^2$ of degree two extends as a $C^1$ function at zero with differential zero. Its restriction to every real line is convex, including at zero, because its second derivative is nonnegative away from zero and its first derivatives match there. Thus it is convex on the whole space. Finally,

and the locally uniform limit as $\tau \downarrow 0$ proves the assertion with the same $\gamma $. $\square$

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. $\square$

Proof. Suppose that $Y$ is $p$-convex for an integer $1\leq p<\kappa $. Then a smooth exhaustion function $\psi $ has at least $N-p+1$ positive Levi eigenvalues outside a compact set $K\subset Y$. Its restriction to the closed submanifold $Z$ is an exhaustion function, since its sublevel sets are closed subsets of the compact sublevel sets of $\psi $, and $K\cap Z$ is compact.

At a point of $Z\setminus K$, choose a complex subspace $P\subset T^{1,0}Y$ of dimension $N-p+1$ on which the Levi form of $\psi $ is positive definite. Then

The Levi form of $\psi |_ Z$ is the restriction of the Levi form of $\psi $, since the inclusion is holomorphic, so it has at least $D-p+1$ positive eigenvalues. Since $p<\kappa \leq D$, this shows that $Z$ is $p$-convex, which contradicts the definition of $\kappa $. $\square$

Proof. The chart of $Z$ where the $\mathcal O_ X$-coordinate is nonzero is identified with the total space of $E$, and $Y$ is its zero section. Suppose that $Z\setminus Y$ is $r$-convex. Since $\dim Z=n+r$, a smooth exhaustion function $\psi $ then has at least $n+1$ positive Levi eigenvalues outside a compact set $K$. Choose a relatively compact disk-bundle neighborhood $D\Subset E$ of $Y$. The set $Z\setminus D$ is compact and disjoint from $Y$. Choose a regular value $c$ of $\psi $ greater than its maximum on $Z\setminus D$ and on $K$.

Since $\{ \psi \leq c\} $ is compact in $Z\setminus Y$, the set $\Omega =\{ \psi >c\} \cup Y$ is an open neighborhood of $Y$. Its closure is a compact subset of $D$, and its boundary is the regular level set $\{ \psi =c\} $. The defining function $f=c-\psi $ has at least $n+1$ negative Levi eigenvalues there. Restriction to the complex tangent hyperplane removes at most one negative direction, leaving at least $n$. Since $E$ is ample, theorem 11.17.2 produces a smooth complex norm on $E^*$ with plurisubharmonic square, a contradiction. $\square$

Proof. Use theorem 11.15.5 to choose $E$ whose dual Dolbeault operator lies in the $C^4$ neighborhood of theorem 11.9.4; the split dual bundle satisfies the hypotheses of that theorem by proposition 11.13.4. Thus $E^*$ admits no smooth complex norm with plurisubharmonic square, and corollary 11.17.4 with $r=2$ gives the assertion. $\square$