Example 11.10.1.196112D1B85A On the trivial bundle $D\times \mathbb C^2$, where $D=\{ z\in \mathbb C:|z|<1/2\} $, put
\[ Q(z,v)=\max \left\{ |v_1|^2+|v_2|^2, \frac{|v_1|^2}{1-|z|^2}+|z|^2|v_2|^2 \right\} . \]
This is a continuous plurisubharmonic function, positive off the zero section and complex homogeneous of degree two. Its square root is a complex norm on every fiber, and for each fixed $z$, the real function $Q(z,\cdot )$ is strongly convex. Nevertheless, $Q$ is not locally strongly convex in the fibers near $(z,v)=(0,(1,0))$ in the sense defined above.
Proof.
Write $t=|z|^2$. Both branches are continuous, positive semidefinite Hermitian quadratic forms in $v$, and the first is positive definite. Their square roots satisfy the triangle inequality and complex homogeneity; their maximum therefore is a complex norm. In particular $Q$ is positive off zero and continuous, also on the zero section.
The first branch is plurisubharmonic on the total space. For the second, the locally uniformly convergent expansion
\[ \frac{|v_1|^2}{1-|z|^2}+|z|^2|v_2|^2 =\sum _{k=0}^{\infty }|z^ kv_1|^2+|zv_2|^2 \]
expresses it as a limit of sums of squared moduli of holomorphic functions. Thus it too is plurisubharmonic, and so is their maximum.
For $0<t<1/4$, subtracting $t|v|^2$ gives
\[ Q(z,v)-t|v|^2 =\max \left\{ (1-t)|v|^2, \left(\frac1{1-t}-t\right)|v_1|^2\right\} , \]
which is convex. At $z=0$, one has $Q(0,v)=|v|^2$. Hence each fixed fiber has a positive strong convexity margin.
To test a margin on a neighborhood of $(0,(1,0))$, fix any $c>0$ and choose a nonzero $z$ arbitrarily close to zero with $t<c$. At $v=(1,0)$ the second branch is strictly larger than the first, so it remains the maximum on a small fiber ball. Along the real line $v=(1,s)$ in this ball,
\[ Q(z,(1,s))-c|(1,s)|^2 =\frac1{1-t}-c+(t-c)s^2. \]
Its second derivative is negative. Thus no positive $c$ makes $Q-c|v|^2$ convex on the fiber balls in any such total-space neighborhood. A fiber-linear change of smooth frame and its inverse are uniformly bounded on a smaller base neighborhood, so a positive quadratic margin in another frame would give one in this frame. This proves the failure of the local margin.
$\square$
Pome
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