The determinant of f is not invertible when f is zero when the norm of the function is constant.

Question

Let $f:U\subset \mathbb{R^n}\rightarrow \mathbb{R}^n$ differentiable on the open $U$. If $|f(x)|$ is constant, then $Df(a)$ is not invertible for every $a\in U$.

How can I prove that?

Answer 1

If $\|f(x)\| = 0$, then clearly $Df(x)$ is zero for all $x$.

Suppose $\|f(x)\| \neq 0$. Let $\phi(x) = {1 \over 2} \|f(x)\|^2$. Since $\phi$ is constant, we have $D \phi(x) = f(x)^T {\partial f(x) \over \partial x} = 0$. If ${\partial f(x) \over \partial x} $ was invertible this would imply that $\|f(x)\|= 0$, a contradiction. Hence ${\partial f(x) \over \partial x} $ is singular everywhere.