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Suppose $f:U\to\mathbb R^n$ a differentiable function where $U\subseteq\Bbb R^n$ is an open set.

I want to prove that:

$|f(x)|$ is constant for every $x\in U$ $\implies \det \mathfrak J\equiv 0$.

where $\mathfrak J(x)$ is the Jacobian matrix at the point $x$.

My attempt

Let's define $g(x)=|f(x)|$.

So $g'(x)\cdot v=\frac{\langle f'(x)\cdot v,f(x)\rangle}{|f(x)|}=0$ for every $v\in U$.

Therefore

$$\langle f'(x)\cdot v,f(x)\rangle=0,\ \forall v\in U.$$

Can we use this fact to prove the question?

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    It's not $\forall v\in U$, $v$ should be a vector. And your last equation shows that image of $f'(x)$ lies in a $n-1$ dimensional subspace and thus cannot be full rank, so it determinant is zero. – 2017-01-07
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    What is the assumption about the "differentiability class" for $f$ ? If $f$ is only $C^1$ and then continuous, you can partition $U$ in $f^{-1}(]-\infty,0[)$, $f^{-1}(]0,+\infty[)$ and the remainder and look what happens. – 2017-01-07
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    @DuchampGérardH.E. I edited the question, thank you for the remark – 2017-01-07
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    @JohnMa so for every $v\in \mathbb R^n$? – 2017-01-07
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    Why does my last equation show that image of $f′(x)$ lies in a $n−1$ dimensional subspace? could explain a little bit more? thank you – 2017-01-07
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    Since all $f'(x) v$ are orthogonal to the vector $f(x)$. – 2017-01-07

2 Answers 2

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We know that $|f(x)|=C$ for some $C$. If $C=0$ then $f(x) =0$ for all $x \in U$, and so $Df(x) =0$, which means that $\det Df(x) =0$ for all $x$. We can then assume that $C\neq 0$.

We can square to get $|f(x)|^2 = C^2$ for all $x$. We then differentiate to get that $$ \partial_i |f(x)|^2 = 0 \text{ for all } i =1,\dotsc,n. $$ We then compute $$ \partial_i |f(x)|^2 = 2\sum_j \partial_i f_j(x) f_j(x) = 2\sum_j (Df(x))_{ji} f_j(x), $$ and hence $$ (Df(x))^T f(x) =0 \text{ for all }x. $$ Since $C \neq 0$ we know that $f(x) \neq 0$ for all $x \in U$, and thus the above tells us that $Df(x)^T$ has a nontrivial kernel for each $x$, which means that $Df(x)^T$ is singular, i.e. $\det Df(x)^T =0$. Since $\det Df(x) = \det Df(x)^T$ we then deduce that $\det Df(x) =0$ for all $x \in U$.

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    What do you mean by $\partial_i$? Thank you for your answer – 2017-01-07
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    The partial derivative with respect to $x_i$. In other words $x = (x_1,\dotsc,x_n) \in \mathbb{R}^n$ and $\partial_i = \partial/ \partial x_i$. – 2017-01-07
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If you had a non zero Jacobian at some point, you could apply there the inverse function theorem and conclude that the image of your function contains an open set. Yet it doesn't.