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 $|a| = 1$?
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8If 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. – hmakholm left over Monica Jun 10 '15 at 01:38
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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 '15 at 01:41
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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 '15 at 02:13
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2Does this answer your question? Function whose gradient is of constant norm – Andrew D. Hwang Apr 13 '21 at 14:18
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Yes, this is true: smooth global solutions of the eikonal equation are affine. The proof is similar to Solution of eikonal equation is locally the distance from a hypersurface, up to a constant. It goes like this:
- By the mean value theorem, $f$ is $1$-Lipschitz.
- 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.
- 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, $\Gamma=\{x:x_1=0\}$.
- 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 $-x_1$.
