# $|\nabla f (x)| =1$ implies $f$ linear?

Suppose $f: \mathbb{R}^n \to \mathbb{R}$ is smooth and $|\nabla f (x)| = 1$. Must $f$ be linear (up to an additive constant)? That is, must we have $f(x) = a\cdot x +b$ for constant $a,b\in\mathbb{R}^n$ with $|a| = 1$?

• If it is true, then it's not for any simple local reason. If we remove just one point from the domain, then it's not true anymore, because the distance to the removed point is then a solution. Jun 10, 2015 at 1:38
• It is true, but the argument I can come up with for it is based on the fact that the integral curves of a constant norm gradient are geodesics, which leads to geodesic crossing problems if you have a non-constant direction of the gradient
– jxnh
Jun 10, 2015 at 1:41
• Actually the sketch of a proof that I'm thinking of only works for $\mathbb{R}^2$, though i suspect the result is true generally.
– jxnh
Jun 10, 2015 at 2:13
• Does this answer your question? Function whose gradient is of constant norm Apr 13, 2021 at 14:18

1. By the mean value theorem, $f$ is $1$-Lipschitz.
2. Trajectories of steepest ascent/descent, i.e., solutions of the ODE $x'(t)=\nabla f(x(t))$ both forward and backward in time, are straight lines. Indeed, along such a curve we have $$|x(t)-x(s)| \ge |f(x(t))-f(x(s))|=|t-s|$$ which, the curve being unit speed, implies that it's a line.
3. Let $\Gamma$ be any level surface of $f$ (WLOG, $G=\{f=0\}$), and $L$ a normal line to this set. By item 2 above, $L$ a path of steepest ascent/descent. Choose coordinates so that $L$ is the $x_1$-axis. The restriction of $f$ to $L$ is either $x_1$ or $-x_1$. The $1$-Lipschitz condition implies $\operatorname{dist} (te_1,\Gamma)=|t|$ for all $t\in\mathbb{R}$. Hence, $\Gamma=\{x:x_1=0\}$.
4. Since all level surfaces are planes, and they are disjoint, they must be parallel. After rotation, $f$ is a function of $x_1$ only: more specifically, either $x_1$ or $-x_1$.