I'm interested in numerical methods on $\mathrm{SO}(3)$ manifold, and working on a particular problem using the exponential coordinates. These can be computed using the Rodrigues' formula: $$ R(u) := \exp(u_\times)=\mathrm{Id} +\frac{\sin(a)}{a}u_\times +\frac{1-\cos(a)}{a^2}u_\times^2 $$ with $a:=|u|$, $u\in \mathbb{R}^3$ and where $u_\times \in \mathfrak{so}(3)$ 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 $$ T(u) := \int_0^1R(su)ds = \mathrm{Id} +\frac{1-\cos(a)}{a^2}u_\times+\frac{a-\sin(a)}{a^3}u_\times^2 $$

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 $T$ are Lipschitz continuous: $$ \|R(u)-R(v)\| \le|u-v| \\ \|T(u)-T(v)\| \le \frac{1}{2}|u-v| $$ 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: $$ \|D_u[R(u)X] - D_v[R(v)X]\| = \sup_{|Y|=1}\left([R(u)X] \times [T(u)Y] - [R(v)X] \times [T(v)Y]\right) $$ that is I need to show that the derivative $D_u[R(u)X]$ is Lipschitz continuous in $u$, and find the best Lipschitz constant. Experimentally (using a program) I found that $$ \|[R(u)X]_\times T(u) - [R(v)X]_\times T(v)\| \le \left|X\right| \left|u-v\right| $$ or equivalently that: $$ \big|[R(u)X] \times [T(u)Y] - [R(v)X] \times [T(v)Y]\big| \le \left|X\right| \left|Y\right| \left|u-v\right| $$ for any vectors $X$ and $Y$. Any hints on how to prove this last inequality would be greatly appreciated. Thanks for your time!

Edit: It's possible to show a weaker statement using the triangle inequality: $$ \big|[R(u)X] \times [T(u)Y] - [R(v)X] \times [T(v)Y]\big| \le \frac{3}{2} \left|X\right| \left|Y\right| \left|u-v\right| $$ but experimentally this bound is never reached.

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?

I'm interested in numerical methods on $\mathrm{SO}(3)$ manifold, and working on a particular problem using the exponential coordinates. These can be computed using the Rodrigues' formula: $$ R(u) := \exp(u_\times)=\mathrm{Id} +\frac{\sin(a)}{a}u_\times +\frac{1-\cos(a)}{a^2}u_\times^2 $$ with $a:=|u|$, $u\in \mathbb{R}^3$ and where $u_\times \in \mathfrak{so}(3)$ 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 $$ T(u) := \int_0^1R(su)ds = \mathrm{Id} +\frac{1-\cos(a)}{a^2}u_\times+\frac{a-\sin(a)}{a^3}u_\times^2 $$

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 $T$ are Lipschitz continuous: $$ \|R(u)-R(v)\| \le|u-v| \\ \|T(u)-T(v)\| \le \frac{1}{2}|u-v| $$ 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: $$ \|D_u[R(u)X] - D_v[R(v)X]\| = \sup_{|Y|=1}\left([R(u)X] \times [T(u)Y] - [R(v)X] \times [T(v)Y]\right) $$ that is I need to show that the derivative $D_u[R(u)X]$ is Lipschitz continuous in $u$. 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: $$ \big|[R(u)X] \times [T(u)Y] - [R(v)X] \times [T(v)Y]\big| \le \left|X\right| \left|Y\right| \left|u-v\right| $$ for any vectors $X$ and $Y$. Any hints on how to prove this last inequality would be greatly appreciated. Thanks for your time!

Edit: It's possible to show a weaker statement using the triangle inequality: $$ \big|[R(u)X] \times [T(u)Y] - [R(v)X] \times [T(v)Y]\big| \le \frac{3}{2} \left|X\right| \left|Y\right| \left|u-v\right| $$ but experimentally this bound is never reached.

I'm interested in numerical methods on $\mathrm{SO}(3)$ manifold, and working on a particular problem using the exponential coordinates. These can be computed using the Rodrigues' formula: $$ R(u) := \exp(u_\times)=\mathrm{Id} +\frac{\sin(a)}{a}u_\times +\frac{1-\cos(a)}{a^2}u_\times^2 $$ with $a:=|u|$, $u\in \mathbb{R}^3$ and where $u_\times \in \mathfrak{so}(3)$ 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 $$ T(u) := \int_0^1R(su)ds = \mathrm{Id} +\frac{1-\cos(a)}{a^2}u_\times+\frac{a-\sin(a)}{a^3}u_\times^2 $$

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 $T$ are Lipschitz continuous: $$ \|R(u)-R(v)\| \le|u-v| \\ \|T(u)-T(v)\| \le \frac{1}{2}|u-v| $$ 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: $$ \|D_u[R(u)X] - D_v[R(v)X]\| = \sup_{|Y|=1}\left([R(u)X] \times [T(u)Y] - [R(v)X] \times [T(v)Y]\right) $$ that is I need to show that the derivative $D_u[R(u)X]$ is Lipschitz continuous in $u$, and find the best Lipschitz constant. Experimentally (using a program) I found that $$ \|[R(u)X]_\times T(u) - [R(v)X]_\times T(v)\| \le \left|X\right| \left|u-v\right| $$ or equivalently that: $$ \big|[R(u)X] \times [T(u)Y] - [R(v)X] \times [T(v)Y]\big| \le \left|X\right| \left|Y\right| \left|u-v\right| $$ for any vectors $X$ and $Y$. Any hints on how to prove this last inequality would be greatly appreciated. Thanks for your time!

Edit: It's possible to show a weaker statement using the triangle inequality: $$ \big|[R(u)X] \times [T(u)Y] - [R(v)X] \times [T(v)Y]\big| \le \frac{3}{2} \left|X\right| \left|Y\right| \left|u-v\right| $$ but experimentally this bound is never reached.