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The fact that a differentiable function is perpendicular to its level sets has been asked before. See: Gradient is perpendicular to level set and implicit function theorem and Why is the gradient always perpendicular to level curves?. Underlying every proof that I have seen of this fact is the assumption that the implicit function theorem can be applied. However, the implicit function theorem requires continuous differentiability. Is it the case then that this fact holds only under the assumption that the function be continuously differentiable, rather than just differentiable?

Spencer Gibson
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  • I'm pretty sure if the level curve even exists and has a tangent, then that tangent is necessarily perpendicular to the gradient, because otherwise the directional derivative in the direction of that tangent would be nonzero, which it can't be. So it seems to me that it is a matter of whether the question even makes sense. – Ian Jan 19 '23 at 19:47
  • @Ian To conclude that the level set is a level curve, the implicit function theorem is usually invoked, right? If the function isn't continuously differentiable then this cannot always be done. – Spencer Gibson Jan 19 '23 at 19:52
  • Technically a level set of a two variable function does not have to be a single curve at all, it could have more than one connected component for example. But the IFT ensures that the level set to a $C^1$ function, away from a point where the gradient vanishes, is locally the graph of a function. (At some points you may be forced to write it as one of $x(y)$ or $y(x)$ if one of the partial derivatives vanishes) The question is whether the tangent to the level set at a particular point necessarily makes sense without a $C^1$ hypothesis, and I am not entirely sure whether that is the case. – Ian Jan 19 '23 at 19:56
  • @Ian Yes you're right, it can have more than one component. I'm interested to see if anyone else comments on whether we have to use the IFT to ensure that locally the level set is a graph. Personally, I have never seen a proof that doesn't assume this. – Spencer Gibson Jan 19 '23 at 19:58
  • It doesn't actually have to be the graph of a function at all, the question is whether it has a tangent at just this one point. – Ian Jan 19 '23 at 19:58

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