Question: Let $h(t)$ be a smooth curve in a matrix Lie group *G*, with $h(0)=e$. Let $v \in T\_e G$. Then it holds that

$$\left.\frac{d}{dt}\left(Ad\_{h(t)}(v)\right)\right|\_{t=0} = ad\_{\left.\frac{d}{dt} h(t) \right|\_{t=0}}(v).$$

... I have been trying to figure out if this holds in a general Lie group. Any hints would be greatly appreciated.

Answer: 
Geometrically, for $|z| = 1$, $|2z - 1|$ is the distance between a point on the circle of radius $2$ centered at the origin (in red), and the point $(1,0)$ (in blue). [![](https://i.stack.imgur.com/IWU9cm.png)](https://i.stack.imgur.com/IWU9cm.png)

By looking at the attached image, we can see that the maximum such distance is $3$. This can more easily be seen by considering a plot of the circle of radius $3$ centered at $(1,0)$ (in green). The point $(-2,0)$ is a distance of $3$ away from $(1,0)$, and each other point on the circle of radius $2$ is closer to $(1,0)$.

Thus, $$\sup\_{|z| = 1} |2z - 1| = 3.$$
Note that $\:|2z-1|=2\Bigl|z-\frac12\Bigr|$, i.e. geometrically, twice the distance between the affix of $z$ and the point $\;A\bigl(\frac12,0\bigr)$ in the plane. It is attained at the intersection point of the diameter through $A$ (i.e. the $x$ axis) which has a negative abscissa: $(-1,0)$. Thus 
$$\sup\_{|z|=1}|2z-1|=2\cdot\frac 32=\color{red}3.$$