Let $\bar{D}_{2\rho}$ be a closed disk in $\mathbb{R}^2$ of radius $2\rho$, where $\rho>1$.
Let $\psi_0$ be a smooth map：
$$\psi_{0}:[0,1]\times\bar{D}_{2\rho}\rightarrow\hat{S},$$
where $\hat{S}$ is the set of $2\times2$ symmetric trace free matrix.
Define a map $\psi_{u_0}:[0,\delta]\times\bar{D}_{2\rho}\rightarrow\hat{S}$ by setting
$$\psi_{u_0}(\underline{u},v)=\frac{\delta^{\frac{1}{2}}}{|u_0|}\psi_0\left(\frac{\underline{u}}{\delta},v\right),$$
where $0<\delta<1.$
Now, denote $m(\underline{u},v):=\exp(\psi(\underline{u},v))$, where $\exp$ is matrix exponential, i.e.
$$\exp(A):=I+\sum_{k=1}^{\infty}\frac{1}{k!}A^k,$$
where $I$ is the identity matrix.
Denote $|A|$ to be the magnitude of the matrix $A$, i.e.,
$$|A|:=\sqrt{\sum_{i,j=1}^2|A_{ij}|^2}.$$
Let $\|m-I\|_{C_{\delta}^k([0,\delta]\times\bar{D}_{2\rho})}$ be the weighted $C^k$ norm defined by
$$\|m-I\|_{C_{\delta}^k([0,\delta]\times\bar{D}_{2\rho})}:=\max_{m+n\leq k}\sup_{(\underline{u},v)\in[0,\delta]\times\bar{D}_{2\rho}}\delta^n\left|\frac{\partial^m}{\partial v^m}\frac{\partial^n}{\partial\underline{u}^n}(m-I)(\underline{u},v)\right|$$
and define
$$\|\psi_0\|_{C^k([0,\delta]\times\bar{D}_{2\rho})}:=\max_{m+n\leq k}\sup_{(\underline{u},v)\in[0,\delta]\times\bar{D}_{2\rho}}\left|\frac{\partial^m}{\partial v^m}\frac{\partial^n}{\partial\underline{u}^n}\psi_0(\underline{u},v)\right|.$$
I would like to show that
$$\|m-I\|_{C_{\delta}^k([0,\delta]\times\bar{D}_{2\rho})}\leq\delta^{\frac{1}{2}}|u_0|^{-1}F_k\left(\|\psi_0\|_{C^k([0,\delta]\times\bar{D}_{2\rho})}\right),$$
where $F_k$ is a non-negative non-decreasing continuous function.
This question is about the inequality (2.51) on Page 74 of Christodoulou’s book, which can be found at https://arxiv.org/pdf/0805.3880.pdf
I tried to compute it directly but it seemed impossible to perform chain rule, but I cannot find an alternative way to show the inequality, could anyone please suggest how to obtain the inequality? Every comment and suggestion is really appreciated, thanks a lot!