Pome

Lemma 11.12.3BD7ECB0F942E

Lemma 11.12.3.BD7ECB0F942E Let $k$ be a smooth Hermitian metric on $E$, and let $G$ be a positive smooth complex homogeneous function of degree two on $E^\times $, extended by zero. Then its polar square $Q$, defined as in proposition 11.12.2, has a constant $c>0$ such that

\[ Q(x,\xi )-c|\xi |_{k^*}^2 \]

is real convex on every dual fiber. Neither strong pseudoconvexity nor real convexity of $G$ is required here.

Proof. The real fiber Hessian of $G$ is bounded above by $2L$ times $k$ on the compact unit sphere bundle of $k$, for some $L>0$, and homogeneity extends this bound to every nonzero vector. Also, $G$ is $C^1$ at zero with derivative zero, since its value is $O(|v|^2)$ and its first fiber derivative is $O(|v|)$, uniformly over the base. Integrating the Hessian estimate along a segment gives

7BB708C6E5FD 11.12.4
\begin{equation} G(v+w)\leq G(v)+\mathrm d_ vG(w)+L|w|_ k^2; \end{equation}

a segment passing through zero is handled by splitting it there and using the continuity of the first derivative.

Complex homogeneity and completing the square on each complex line give the identity

E21FC7F3990F 11.12.5
\begin{equation} Q(x,\xi )=\max _{v\in E_ x} \{ 2\operatorname {Re}\xi (v)-G(x,v)\} . \end{equation}

The maximum exists because $G(v)\geq a|v|_ k^2$ for a uniform constant $a>0$. Fix a maximizer $v_0$ for $\xi _0$. Differentiation, including the case $\xi _0=v_0=0$, gives $\mathrm dG(v_0)=2\operatorname {Re}\xi _0$. Evaluate the expression in (11.12.5) for $\xi _0+\zeta $ at $v_0+w$ and apply (11.12.4). The terms involving $\xi _0(w)$ cancel, and we obtain

\[ Q(\xi _0+\zeta )\geq Q(\xi _0)+2\operatorname {Re}\zeta (v_0) +2\operatorname {Re}\zeta (w)-L|w|_ k^2. \]

Taking the supremum over $w$ yields

\[ Q(\xi _0+\zeta )\geq Q(\xi _0)+2\operatorname {Re}\zeta (v_0) +L^{-1}|\zeta |_{k^*}^2. \]

Set $c=L^{-1}$ and $\Phi =Q-c|\cdot |_{k^*}^2$. After subtracting the square, this inequality gives a real affine function on the whole dual fiber that lies below $\Phi $ and equals $\Phi $ at $\xi _0$. Hence $\Phi $ is the supremum of such affine functions, and therefore convex. Neither uniqueness of the maximizer nor real convexity of $G$ has been used. $\square$

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