$\| X'(t) \| = (\| X(t) \|)'$

Question

Let $X : \mathbb{R} \to \mathbb{R}^n$ be a $C^1$ function. Let $\| .\|$ be the norm : $\| v \| = \max_{1 \leq i \leq N} \mid v_i \mid$. Then is it true that : $$\| X'(t) \| = (\| X(t) \|)'$$ ?

I am wondering if in general if I have any function $f : \mathbb{R}^n \to \mathbb{R}^p$ and a norm $N$ on a : $\mathbb{R}^p$ then is it always possible to invert the norm and the differential operator or the norm and in the integral?

Thank you.

Answer 1

For $n=1$ the identity function $X(x)=x$ is a $C^{1}$ function. In this case $|X(x)|$ is not even differentiable at $0$.

Answer 2

To show that this is not always the case, let's give a clear example in two dimension. To that end, we interpret $X'(t)$ as the velocity of a trajectory in $\mathbb R^n$, so that $\vert\vert X'(t)\vert\vert$ represents the speed.

Consider now trajectories moving along the surface of a sphere, i.e. $\vert\vert X(t)\vert\vert = 1$ at all times. They may attain varous speeds, i.e. different and even growing $\vert\vert X'(t)\vert\vert$'s.

Example:

$X(t)=(\cos(\omega(t)), \sin(\omega(t)))$

$\implies\vert \vert X(t)\vert \vert = \cos(\omega(t))^2+\sin(\omega(t))^2=1$

$\implies\vert \vert X(t)\vert \vert' = 0$

while

$X'(t)=\omega'(t)\cdot(-\sin(\omega(t)), \cos(\omega(t)))$

$\implies\vert \vert X'(t)\vert \vert = \omega'(t)$

You can also play this the other way around, e.g. going with $\omega(t)=c\cdot t$ and move the circle away from the origin: In this case, $\vert \vert X(t)\vert \vert' $ will vary and $\vert \vert X'(t)\vert \vert$ actually wont.