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I have a function $f:\mathbb{R}^N \to \mathbb{R}$ and I know that on its level sets $f^{-1}(z)$ the norm of its gradient is constant. What can I say about this function? $$ ||\nabla_x f(x)|| = \text{const} \qquad \qquad \forall x \in f^{-1}(z) := \left\{x \in \mathbb{R}^N \, :\, f(x) = z\right\} \qquad \forall \in \mathbb{R} $$

Related questions are this and this. However, they consider the norm of the gradient to be constant for every $x$ in the domain. I know that this is true only on each level set.

Euler_Salter
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    I'm not sure whether it may say much. For example, $f(x) = |x|$ satisfies this as $||\nabla_x f(x)|| = 1$. Another function $f(x) = x^2$ also satisfies this - for every level set (ie ${a,-a}$) $||\nabla_x f(x)|| = 2|x|$. In fact these two satisfy the property for every level set, whereas you know if only for some level set(s). – Rahul Madhavan Apr 13 '21 at 11:25
  • @RahulMadhavan what if this property were satisfied for every level set? – Euler_Salter Apr 13 '21 at 11:26
  • @RahulMadhavan I have found the following paper which seems like it might be related but It's hard to decipher https://projecteuclid.org/journals/kodai-mathematical-journal/volume-19/issue-1/On-Riemannian-manifolds-admitting-a-function-whose-gradient-is-of/10.2996/kmj/1138043545.full – Euler_Salter Apr 13 '21 at 11:27
  • I changed the question so that the function must have gradient of constant norm on all its level sets – Euler_Salter Apr 13 '21 at 11:58
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    Is $f$ continuously-differentiable? Do you know there are no critical points? – Andrew D. Hwang Apr 13 '21 at 14:21
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    I am happy to assume any regularity condition such as continuously-differentiable and no critical points – Euler_Salter Apr 13 '21 at 14:41
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    One result you could say is that the level sets are spaced "evenly". That is, on one level set $A = f^{-1}(a)$, you're the same distance from another level set $B = f^{-1}(b)$ no matter where you are along $A$. However, since you're only assuming constant norm along each set, at first I can only see that this is true in some "infinitesimal" non-exact sense, i.e. as $b \to a$. – Chris Apr 13 '21 at 14:43
  • In case it matters, it makes a substantial difference in the outcome whether or not there are critical points, just as it matters whether or not the domain is all of $\mathbf{R}^{n}$ or (say) $\mathbf{R}^{n}$ with one point removed. – Andrew D. Hwang Apr 13 '21 at 14:45
  • @AndrewD.Hwang Domain will definitely be all of $\mathbb{R}^n$! – Euler_Salter Apr 13 '21 at 14:47
  • @AndrewD.Hwang How did you manage to obtain the fact that level sets are evenly spaced? – Euler_Salter Apr 13 '21 at 14:48
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    That was "Chris". :) But the idea is that the gradient flow sends each regular level to another regular level, precisely because the norm of the gradient is constant on the levels. – Andrew D. Hwang Apr 13 '21 at 14:58
  • My bad, will correct now! – Euler_Salter Apr 13 '21 at 15:00

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If $f$ is $C^{1}$, then because the norm of the gradient is constant on levels of $f$, there is a continuous, real-valued function $\lambda$ of one variable satisfying $$ \|\nabla f(x)\| = \lambda\bigl(f(x)\bigr)\qquad\text{for all $x$.} $$ If $f$ has no critical points, then $\lambda > 0$. Let $\Lambda$ be an antiderivative for $1/\lambda$ and let $g = \Lambda \circ f$. By the chain rule, $$ \|(\nabla g)(x)\| = \bigl|\Lambda'\bigl(f(x)\bigr)\bigr| \cdot \|\nabla f(x)\| = 1. $$ By the solution of the linked question, $g$ is affine. Consequently, $f$ is constant on hyperplanes, and is therefore effectively a function of one variable.

If $f$ has critical points, more can happen. For example, $f$ could be a function of distance-squared from an affine subspace (a point up through a hyperplane).

[Musings: Offhand I don't have a proof that's all, but along the lines of Chris' comment this is what one expects; I'd be inclined to check whether the regular levels of $f$ have constant principal curvatures.]

Andrew D. Hwang
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  • By critical points do you mean like a maximum/minimum? – Euler_Salter Apr 13 '21 at 15:02
  • Yes; specifically, points where the gradient vanishes. – Andrew D. Hwang Apr 13 '21 at 15:04
  • Ooh I am sorry I thought of something else. I apologise for that, there can be some critical values. There will always be a finite number of critical values and most likely only one. Do you think you could say more about it then? – Euler_Salter Apr 13 '21 at 15:10
  • Every spherically-symmetric function (i.e., a function of distance from some point) has your property, and by the musings I'd guess that's all, but offhand I don't have a proof. If this answer isn't relevant I'm happy to delete it. In any case, this information is crucial to add to your question. :) – Andrew D. Hwang Apr 13 '21 at 15:13
  • What do you mean exactly by function of distance-squared or spherically-symmetric function? – Euler_Salter Apr 13 '21 at 15:34
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    Up to a translation (to move the critical point to the origin), a function of $x_{1}^{2} + \cdots + x_{n}^{2}$ (distance-squared to the origin). A function of a sum of only some of these terms also has your property; the levels are then cylinders coaxial with some affine subspace of $\mathbf{R}^{n}$. – Andrew D. Hwang Apr 13 '21 at 16:42