Let $f:\mathbb{R}^n\to \mathbb{R}$ be a $C^1$ function, if $|\nabla f|=1,$ is it true that $f$ is surjective?

Question

Suppose $f:\mathbb{R}^n\to\mathbb{R}$ is a $C^1$ function, I wonder whether or not $f$ is surjective provided that $|\nabla f(x)|=1$ for any $x\in\mathbb{R}^n$ ?

When $n=1,$ the condition $|f^\prime|=1$ implies that $f$ is a non-degenerate linear function, so clearly $f(\mathbb{R})=\mathbb{R}.$ In general dimension, I wish to show that $f(\mathbb{R}^n)$ is both open and closed in $\mathbb{R}.$ Using rank theorem, we obtain that $f(\mathbb{R}^n)$ is an open set in $\mathbb{R}$, but now I have no idea to show that $f(\mathbb{R}^n)$ is closed. Can someone help me?

Here is an answer for the case were $f$ is $C^\infty$. $|\nabla f (x)| =1$ implies $f$ linear? — Sassatelli Giulio, May 19 '24 at 09:27
I don't see in the answer there the hypothesis $C^\infty$ being used, but the argument is mostly a sketch. — Sassatelli Giulio, May 19 '24 at 09:37
Not sure why Hyperbolic PDE Friend deleted his answer, but the function $f(x_1,x_2)=\sin(x_1)+\cos(x_2)$ is a counterexample. — K.defaoite, May 19 '24 at 17:55
@K.defaoite the gradient of your $f$ vanishes e.g., in $(x_1, x_2) = (\pi/2,0)$. Note that $\sin^2(x) + \cos^2(x) = 1$ relies on the fact that the argument of the two trigonometric functions is the same. — noam.szyfer, May 21 '24 at 05:13

score 4 · Accepted Answer · answered May 19 '24 at 10:07

If $f$ is only $C^1$ then its gradient field $X$ will be only continuous, and therefore one cannot apply the usual existence-and-uniqueness theorem for the integral curve of the field. However, there is a theorem asserting the existence only of a solution, called Peano's existence theorem, and even a global version thereof under hypotheses that would be satisfied if the field is of unit length at each point. Therefore taking an integral curve of the field, we obtain a curve along which the function increases with unit speed and therefore must take all real values.

Let $f:\mathbb{R}^n\to \mathbb{R}$ be a $C^1$ function, if $|\nabla f|=1,$ is it true that $f$ is surjective?

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