The gradient of a function has constant Euclidean length $1$

Question

Consider a function $f : \mathbb R^{2} \to \mathbb R$ that is defined on every point and is differentiable. Then it has a gradient $\nabla f$. Now, suppose that $|\nabla f(x,y)| = 1$ for all $x,y \in \mathbb R^{2}$. Then must the function be a linear function (and hence $\nabla f$ is constant)?

That was the question. So I was thinking that the answer should be no, but coming up with an example is hard. We can draw the field lines for the gradient, and it should be orthogonal to the level curves. And since the gradient is of length 1, it seem intuitive that for any level curves $f(x,y)=c$ and $f(x,y)=d$ if you draw any field lines between them the length along the field line connecting these curve should be $|d-c|$. So an easy idea here is to take concentric circle to be level curve, but this one can't be made into a function differentiable at the center of these circles.

Or perhaps I should start with field lines instead? Start with a family of field lines, cut a single level curve through it, declare that to be $0$, then going along each field lines and find the value of the function depending on the length. I could go with a family of parabola because that's the only thing that perhaps the length can be computed.

But it's going to be a mess, so I don't know if I should keep going with this. So is there a simpler way? Or am I heading in a completely wrong direction here?

EDIT: found an older question that answered this: $|\nabla f (x)| =1$ implies $f$ linear? . Thanks to the user below for providing me with the technical phrase to search for.

@RodrigodeAzevedo: it means the length, the length of gradient vector at all points are $1$. Unless you are referring to below, in which case it's absolute value. — calcstudent, Dec 21 '19 at 01:30
@JackyChong: sorry, I never heard of that before, looking it up now on Wikipedia. — calcstudent, Dec 21 '19 at 01:31
@RodrigodeAzevedo: what do you mean? It's the usual length of vector. In formula it is $\sqrt{x^{2}+y^{2}}$ — calcstudent, Dec 21 '19 at 01:58
@JackyChong: sorry I don't understand the wikipedia article, and the part that seem to answer the question lack citation. Can you explain it, or do you have a good source? — calcstudent, Dec 21 '19 at 02:35
Does this answer your question? $|\nabla f (x)| =1$ implies $f$ linear? — Chain Markov, Dec 23 '19 at 15:11
@YaniorWeg: yeah it's the same one I found afterward, it's in the EDIT. Took me a while to find them though, because there are no easy search terms. Though if you can rephrase it in a more intuitive and easier-to-understand manner you can post it as an answer. — calcstudent, Dec 23 '19 at 15:17
Does this answer your question? Function whose gradient is of constant norm — Andrew D. Hwang, Apr 13 '21 at 14:18

score 0 · Answer 1 · answered Dec 21 '19 at 02:13

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|grad(f)| = 1 => f_x^2+f_y^2 = 1 => f_x=cos(t) and f_y=sin(t)

We can pick a fixed t, which makes the derivatives constant forcing f to be linear, or try to let t vary.

answered Dec 21 '19 at 02:13

Abhishek Vangipuram

376

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You need a curl free condition in order to know that it's the gradient of a function, you can't just pick the gradient at will. – calcstudent Dec 21 '19 at 02:34

The gradient of a function has constant Euclidean length $1$

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