Can derivative of a smooth norm be zero?

Question

Let us call a norm on $\mathbb{R}^n$ smooth if its restriction $\| \cdot \|:\mathbb{R}^n\setminus \{ 0 \} \to \mathbb{R}$ is a smooth map.

Is it true that its differential at (every non-zero point) is nonzero?

Remarks:

By homogeneity, it's enough to prove this for every point on the unit sphere.

Answer 1

Take $x \in \mathbb R^n \setminus \{0\}$. If $d \Vert \cdot \Vert_x = 0$, we would be able to find $\delta > 0$ such that $$\vert \Vert x+h \Vert - \Vert x \Vert -d \Vert \cdot \Vert_x(h)\vert=\vert \Vert x+h \Vert - \Vert x \Vert \vert \le \frac{\Vert h \Vert}{2}$$ for $\Vert h \Vert \le \delta$.

Taking $h = \frac{\delta}{\Vert x \Vert}x$ (for which $\Vert h \Vert = \delta$), we get the contradiction $$\vert \Vert x+h \Vert - \Vert x \Vert \vert = \vert \Vert x+\frac{\delta}{\Vert x \Vert}x \Vert - \Vert x \Vert \vert = \delta \le \frac{\delta}{2}$$ Hence the derivative of the norm cannot vanish at a nonzero point.

Answer 2

An easy alternative way to see the differential is always non-zero, is to take directional derivative in the "direction of our given point":

Denote $f=\| \cdot \|$. Then $df_x(x)=\frac{{d}}{dt}|_{t=0}f(x+tx)=\frac{{d}}{dt}|_{t=0}\|(t+1)x\|=\frac{{d}}{dt}|_{t=0}|t+1|\|x\|$

So $df_x(x)=\|x\|\cdot\frac{{d}}{dt}|_{t=0}(t+1)=\|x\|\neq0$ as required.