Theorem 18.34.6F82E8E685FEA

Proof. The derivative formula. By proposition 18.34.5, $k_{v_ t,x}$ depends smoothly on $t$, for any fixed norm on $V_ m$. Fix a real parameter $t$ and set

Then $H(0)=I$, $H^{(j)}(0)=-\mathcal T_{t,j}$ by Faà di Bruno's formula, and $k_{v_{t+a},x}=H(a)^{-1}k_{v_ t,x}$ as in the proof of proposition 18.34.5. Since $H(a)-I=O(|a|)$, truncating the Neumann series of $H(a)^{-1}$ after $k$ powers leaves a remainder $O(|a|^{k+1})$. Expanding the truncated series to order $k$ in $a$ gives (18.34.7), with its ordered products and coefficients $k!/\alpha !$. Only a finite Taylor jet of the smooth curve is used.

Gevrey bounds. For nonnegative integers $n_\ell $ with $\sum \ell n_\ell =j$,

The sum $\sum \prod _\ell (n_\ell !)^{-1}$ over these tuples is the coefficient of $z^ j$ in $\exp (z/(1-z))$, and evaluation at $z=1/2$ shows that it is at most $\mathrm{e}\, 2^ j$. Together with the assumed bounds this gives

for some $L_0>0$. Since $P_{v_ t}$ is a contraction, each ordered product in (18.34.7) has operator norm at most $L_0^ k(\alpha !)^ s$. There are $2^{k-1}$ compositions of $k$, and $\alpha !\leq k!$. Since $v_ t$ is uniformly bounded for $t\in J$, the norms $\| \cdot \| _{m,v_ t,\nu }$ are uniformly equivalent, and the evaluations and Riesz vectors are bounded uniformly on $J\times X$. Hence the $t$-derivatives of the unweighted diagonal $x\mapsto \ell _ x(k_{v_ t,x})$ are bounded in $C(X)$ by $C(2L_0)^ k(k!)^ s$. The factor $\mathrm{e}^{-mv_ t}$ satisfies bounds of the same Gevrey order, by the estimate for $\mathcal B_ j(-mv_ t)$. Leibniz's formula and $j!(k-j)!\leq k!$ then give (18.34.8) for $B_{v_ t}$, and the bounded linear map $B\mapsto (r!/m^ r)B\nu $ gives the bound in total variation. $\square$