Proposition 18.34.22.4F99D774A693 Put $\rho =\log (3/2)$. The map $f\mapsto \mu _{v+f/m}$ extends holomorphically to the ball $\{ f\in C(X,\mathbb C):\| f\| _\infty <\rho \} $, with values in complex measures of total variation at most three. For real $q\in C(X)$ and $k\geq 1$,
961FF4CC03F6
18.34.23
\begin{equation} \left\| \left.\frac{\mathrm{d}^ k}{\mathrm{d}t^ k}\right|_{t=0} \mu _{v+tq}\right\| _{\mathrm{TV}} \leq 3k!\left(\frac{m\operatorname {osc}_ Xq}{2\rho }\right)^ k. \end{equation}
The radius $\rho $ and the bound three do not depend on $m$, $N_ m$ or the real base weight $v$.
Proof.
On $V_ m$ use $T(f)=P_ v(\mathrm{e}^{-f}\, \cdot \, )$, the operator in the proof of proposition 18.34.5 for the perturbation $f/m$ of the weight $v$. That proof gives $\| T(f)^{-1}\| \leq 2$ on the ball. The extended measure has the density
\[ \frac{\mathrm{e}^{-mv(x)-f(x)}}{N_ m} \ell _ x\bigl (T(f)^{-1}k_{v,x}\bigr ) \]
with respect to $\nu $. By the Riesz identity at the weight $v$, its absolute value is at most $2\mathrm{e}^{\| f\| _\infty }B_ v(x)/N_ m\leq 3B_ v(x)/N_ m$, so its total variation is at most three. Holomorphy follows from the same finite-dimensional construction and the bounded linear passage from densities to measures. For a complex constant $c$ we have $T(f+c)=\mathrm{e}^{-c}T(f)$, and the resulting factor $\mathrm{e}^{c}$ cancels the factor $\mathrm{e}^{-c}$ in the density; thus adding constants to $f$ does not change the extension. For nonconstant real $q$, replace $q$ by $q-c$ with $c=(\sup _ Xq+\inf _ Xq)/2$, so that $\| q-c\| _\infty =\operatorname {osc}_ Xq/2$. Then $t\mapsto \mu _{v+tq}$ extends holomorphically to the disk $|t|<\rho /(m\| q-c\| _\infty )$, with total variation at most three. The Cauchy estimates for Banach-valued holomorphic functions, applied on circles of radius less than $\rho /(m\| q-c\| _\infty )$, followed by letting the radius increase to that value, prove (18.34.23). If $q$ is constant, the measure does not depend on $t$, and all derivatives vanish.
$\square$
Pome
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