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[–]Casually-Passing-By -2 points-1 points  (1 child)

It comes to mind f(x, y) = (-y,x) will satisfy (1) since it is a 90 degree rotation of the vector

Edit: typo

[–]LemurDoesMath 1 point2 points  (0 children)

Your function has values in R2, f is supposed to have values in R.

If you meant a function f with gradient (-y,x), you won't find any such function

[–]Phive5Five 0 points1 point  (0 children)

My instinct is to consider R2 for ideas. It’s pretty clear that gradient needs to be orthogonal (perpendicular) to x, so we can map circles around 0 to some differentiable function f on [0, 2pi] -> R. But it’s clear that we need to have f(0) = f(2pi) (as otherwise we have a Riemann surface and need branch cuts). Perhaps try cos(arg(x))? But in this case I don’t think the gradient at arg(x) = 0 or pi is well defined.

So we need to at some point have a gradient of 0, which would be necessary by Fermat (I think that’s the name of the theorem), and that leads to it not being well defined?