Proposition 11.12.2672C05CD55F8

Proof. The ratio is invariant under complex rescaling of $v$, so the maximum can be taken over the compact unit sphere in a local frame, which proves continuity. Positive upper and lower bounds for $G$ on that sphere compare $Q$ with a Hermitian squared dual norm; this gives positivity off zero and continuity at zero, and complex homogeneity is clear. Moreover,

Each term satisfies the triangle inequality in $\xi $, and so does their supremum.

On a coordinate ball, set $G_\varepsilon =\mathrm e^{-\varepsilon |z|^2}G$. By lemma 11.12.1, its Kobayashi curvature is strictly positive, and its polar square is $Q_\varepsilon =\mathrm e^{\varepsilon |z|^2}Q$. At a point $(x_0,\xi _0)$ with $\xi _0\ne 0$, choose a maximizing line and a vector $v_0$ on it. The lemma, applied to $G_\varepsilon $, gives a holomorphic section $\sigma $ such that

with equality at $(x_0,\xi _0)$. The pairing $\xi (\sigma (z))$ is nonzero near that point, so the right-hand side is pluriharmonic there. Restriction to sufficiently small holomorphic disks centered at the point therefore gives the submean inequality for $\log Q_\varepsilon $. This argument applies at every point off the zero section, and together with continuity it shows that $\log Q_\varepsilon $ is plurisubharmonic there.

As $\varepsilon \downarrow 0$, the functions $\log Q_\varepsilon =\log Q+\varepsilon |z|^2$ converge locally uniformly off the zero section. Passing to the limit in the submean inequalities on disks proves that $\log Q$ is plurisubharmonic off the zero section. Since the exponential is increasing and convex, Jensen's inequality shows that $Q=\exp (\log Q)$ is plurisubharmonic there. At a point of the zero section, $Q=0$ and $Q\geq 0$, so every centered disk satisfies the submean inequality there. Together with continuity and the submean inequalities at nonzero points, this proves plurisubharmonicity on the whole total space of $E^*$. $\square$