What is the Lipschitz constant of the differential of the matrix exponential $\mathfrak{so}(3)\to \mathrm{SO}(3)$ I'm interested in numerical methods on $\mathrm{SO}(3)$ manifold, and working on a particular problem using the exponential coordinates:
$$
R(u) := \exp(u_\times)
$$
with $u\in \mathbb{R}^3$ and where $u_\times \in \mathfrak{so}(3)$ is the cross-product matrix of vector $u$.
The directional derivative of $R(u)$ in the direction $Y$ is:
$$
[D_u R]Y = [T(u)Y]_\times R(u)
$$
for any vector $Y\in \mathbb{R}^3$, where
$$
T(u) := \int_0^1R(su)ds
$$
Both $R$ and $T$ are Lipschitz continuous with constants $1$ and $\tfrac{1}{2}$ respectively:
$$
\|R(u)-R(v)\| \le|u-v| \\
\|T(u)-T(v)\| \le \tfrac{1}{2}|u-v|
$$
for any $u$ and $v$, where I use the operator norm (subordinate norm) of the Euclidean norm.
To find the convergence bounds 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 the Lipschitz constant of the differential. Experimentally (using a program) I found that it is $1$. How to prove this?
 A: Found the proof! It's done using the integral definition of $T$:
$$
T(v) = \int_0^1 R(su) ds = \lim_{n\rightarrow \infty} \frac{1}{n}\sum_{i=1}^n R\left(\tfrac{i}{n}v\right)
$$
So for any vectors $X$ and $Y$:
\begin{align*}
&\biggl|\left[\mathrm{D}_v \left(R(v)X\right)\right]Y - \left[\mathrm{D}_u \left(R(u)X\right)\right]Y\biggr| =  \biggl|\left[R(u)X\right] \times \left[T(u)Y\right] - \left[R(v)X\right] \times \left[T(v)Y\right]\biggr| \\
&\le \lim_{n\rightarrow \infty} \frac{1}{n} \sum_{i=1}^n \biggl|\left[R(u)X\right] \times \left[R(\tfrac{iu}{n})Y\right] - \left[R(v)X\right] \times \left[R(\tfrac{iv}{n})Y\right]\biggr|
\end{align*}
and we only need to prove that each summand is less than $|u-v| |X| |Y|$:
\begin{align*}
    & \biggl|\left[R(u)X\right] \times \left[R(\tfrac{iu}{n})Y\right] - \left[R(v)X\right] \times \left[R(\tfrac{iv}{n})Y\right]\biggr| \\
    &= \biggl|R\left(\tfrac{i}{n}u\right)\left[\left(R\left(\tfrac{n-i}{n}u\right)X\right) \times Y\right] - R\left(\tfrac{i}{n}v\right) \left[\left(R\left(\tfrac{n-i}{n}v\right)X\right) \times Y\right]\biggr| \\
& \le \biggl|\left[ R\left(\tfrac{i}{n}u\right) - R\left(\tfrac{i}{n}v\right) \right]\left[\left(R\left(\tfrac{n-i}{n}u\right)X\right) \times Y\right]\biggr| \\
&+ \biggl| R\left(\tfrac{i}{n}v\right) \left[\left(\left(R\left(\tfrac{n-i}{n}v\right) - R\left(\tfrac{n-i}{n}u\right)\right)X\right) \times Y\right]\biggr| \\
    &\le \bigl|\tfrac{i}{n}u - \tfrac{i}{n}v\bigr| \bigl|X\bigr| \bigl|Y\bigr| + \bigl|\tfrac{n-i}{n}u - \tfrac{n-i}{n}v\bigr| \bigl|X\bigr| \bigl|Y\bigr| \\
    &= \bigl|u - v\bigr|\bigl|X\bigr| \bigl|Y\bigr| 
\end{align*}
where we used the invariance of vector products under rotations, the triangle inequality and that $R$ is $1$-Lipschitz (see this question).
Since $X$ and $Y$ are arbitrary,
$$
\left\|\mathrm{D}_v R(v) - \mathrm{D}_u R(u)\right\| \le |u-v|
$$
in the subordinate norm.
