I'm interested in numerical methods on $SO(3)$ manifold, and working on a particular problem using the exponential coordinates. These can be computed using the Rodrigues' formula:
$$
R(u) := exp(u_\times)=Id +\frac{sin(a)}{a}u_\times +\frac{1-cos(a)}{a^2}u_\times^2
$$
with $a:=|u|, u\in \mathbb{R}^3$ and where $u_\times \in so(3)$ is the cross-product matrix of vector $u$.\
Its differential with respect to $u$ is:
$$
D_u[R(u)X] = -[R(u)X]_\times T(u)
$$

for any vector $X\in \mathbb{R}^3$, where
$$
T(u) := \int_0^1R(su)ds = Id +\frac{1-cos(a)}{a^2}u_\times+\frac{a-sin(a)}{a^3}u_\times^2
$$

We can restate this by saying that the directional derivative of $R(u)$ in the direction $Y$ is:
$$
[D_u R]Y = [T(u)Y]_\times R(u)
$$

Both $R$ and $T$ have these properties:
$$
||R(u)-R(v)|| \le|u-v| \\
||T(u)-T(v)|| \le \frac{1}{2}|u-v|
$$
for any $u$ and $v$, where I use the operator norm (subordinate norm) of the Euclidean norm.

To prove the convergence of Newton's iterations for the numerical method I'm using (conditions of Kantorovich) I need to estimate a bound on the second derivative (which is really hard to compute explicitly as far as I know), or a bound on:
$$
||D_u[R(u)X] - D_u[R(v)X]|| = sup_{|Y|=1}\left([R(u)X] \times [T(u)Y] - [R(v)X] \times [T(v)Y]\right)
$$
I found experimentally (using a program) that\
$$
||[R(u)X]_\times T(u) - [R(v)X]_\times T(v)|| \le \left|X\right| \left|u-v\right|
$$
or equivalently that:
$$
\left|[R(u)X] \times [T(u)Y] - [R(v)X] \times [T(v)Y]\right| \le \left|X\right| \left|Y\right| \left|u-v\right|
$$
for any vectors $X$ and $Y$. Any hints on how to prove this last inequality would be greatly appreciated. Thanks for your time!
(This is a re-post from stackexchange.math).