Lemma 11.9.2.FD645EB1208B For a smooth complex norm on $F_0$ with square $G$, put
\[ a_ s(x)=G(e_ s),\qquad N(x)=\sum _ sa_ s(x),\qquad D_ jG(v)=\mathrm d_ vG(P_ jv), \]
where $e_ s$ is any $k_ s$-unit vector in the fiber of $F_ s$ at $x$; complex homogeneity makes $a_ s$ well defined. Then
\[ G(v)\leq N|v|_ k^2,\qquad |\mathrm d_ vG(w)|\leq 2N|v|_ k|w|_ k. \]
On the support of $\rho _ s$,
\[ |D_ jG|\leq 2\varepsilon N\quad (j\ne s),\qquad |D_ sG-2a_ s|\leq 10\varepsilon N. \]
Proof.
The triangle inequality and the Cauchy–Schwarz inequality give $p(v)\leq \sum _ s|v_ s|_ k\sqrt{a_ s}\leq \sqrt N|v|_ k$. Differentiating $|p(v+tw)-p(v)|\leq |t|p(w)$ in the real variable $t$ gives $|\mathrm d_ vp(w)|\leq p(w)$, and both global bounds follow.
For $v$ in the support of $\rho _ s$, set $w=(I-P_ s)v$ and $e_ s=P_ sv/|P_ sv|_ k$. Then $|w|_ k<\varepsilon $ and $|v-e_ s|_ k\leq |w|_ k+1-|P_ sv|_ k<2\varepsilon $. The segment from $e_ s$ to $v$ stays in the unit ball and avoids zero, so integrating the derivative bound along it gives $|G(v)-a_ s|\leq 4\varepsilon N$. For $j\ne s$, we have $|P_ jv|_ k<\varepsilon $, which gives the first axis bound. Euler's identity $\mathrm d_ vG(v)=2G(v)$ yields
\[ D_ sG-2a_ s=2(G(v)-a_ s)-\sum _{j\ne s}D_ jG. \]
The sum on the right is $\mathrm d_ vG(w)$, whose absolute value is at most $2\varepsilon N$. This proves the second axis bound.
$\square$
Pome
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