Derivative of a norm vs norm of a derivative Consider a vector-valued function of the time, say $$v: \tau\in\mathbb{R}\to\mathbb{R}_N.$$
Suppose that for $\tau=t$, the function is equal to the zero vector, i.e. $$v(t)=0_N.$$ Denote as $\alpha\in\mathbb{R}_N$ the value of the time derivative of $v$ in $\tau=t$, i.e. $$\dot{v}(t)=\frac{d}{d\tau}v(\tau)|_{\tau=t}=\alpha.$$
Of course we can write that $$||\dot{v}(t)||=||\frac{d}{d\tau}v(\tau)|_{\tau=t}||=||\alpha||.$$
But I wonder if I can say that $$(\frac{d}{d\tau}||v(\tau)||)_{\tau=t} \le ||\alpha||.$$
And if so, why?
 A: Assuming $v\neq 0$, note that
$$
2\|v\|\frac{d}{dt} \|v\| = \frac{d}{dt} \|v\|^2=\frac{d}{dt} v\cdot v=2v\cdot \dot\alpha.
$$
It follows from the Cauchy-Schwarz inequality that
$$
\frac{d}{dt}\|v\|\leq \|\dot \alpha\|.
$$
If you do not assume $v\neq 0$, it may be that $\| v\|$ is not classically differentiable. Assuming that $v$ is once continuously differentiable, observe however that the set $\{t|v(t)\neq 0\}$ is open with at most countably many connected components (because $\mathbb R$ is second countable), say $\{t|v(t)\neq 0\}=\cup _n I_n$ with $I_n$ disjoint open intervals. If $I_n=(a,b)$, we see from maximality of a component combined with continuity of $v$ that $v(a)=v(b)=0$. It follows that, for any smooth, compactly supported $\phi\in C_c^\infty(\mathbb R)$,
\begin{align*}
\int \|v(t)\| \phi'(t) dt&=\int_{\{v\neq 0\}} \|v(t)\| \phi'(t) dt=\sum_n \int_{I_n} \|v(t)\| \phi'(t) dt \\
&= -\sum_n \int_{I_n} \frac{v(t)\cdot v'(t)}{\|v(t)\|} \phi(t) dt= -\int_{\{v\neq 0\}} \frac{v(t)\cdot v'(t)}{\|v(t)\|} \phi(t) dt.
\end{align*}
What we have shown is that $t\mapsto \|v(t)\|$  is weakly differentiable with
$$
\frac{d}{dt}\|v(t)\|=\begin{cases}\frac{v(t)\cdot v'(t)}{\|v(t)\|} &\mbox{ if } v(t)\neq 0,\\
0 & \mbox{ if }v(t)=0.
\end{cases}
$$
Of course, this implies the inequality you propose!
