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Let $f : \mathbb R^n → \mathbb R$ be a continuously differentiable function with non-zero gradient. Then, according to the implicit function theorem a level set defines a $n-1$ dimensional manifold $M$ and the gradient of $f$ is perpendicular to $M$.

Now, assume that we have a second function $g : \mathbb R^n → \mathbb R$ also continuously differentiable with non-zero gradient.

How to prove or refute precisely that if for all points in $M$ (defined by the level set of $f$) the direction of the gradient of $f$ and $g$ is identical (but not the length), then $M$ is also a manifold defined by a level set of $g$.

copper.hat
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Its_me
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  • Are the level sets of such a function necessarily connected? – copper.hat Nov 10 '15 at 19:27
  • No, therefore the restriction to connected components of $M$ in the answer below. But one can add additional constraints for $f$ and its domain to ensure that there's a single component. – Its_me Nov 10 '15 at 19:42
  • Do you know if there is a standard example showing a disconnected level set for such a function? – copper.hat Nov 10 '15 at 19:43
  • I'm not sure I understand, is $f$ a map from the reals to the reals? If so, we have $f'(0) = 0$. The question requires $f$ to have a non zero gradient. – copper.hat Nov 10 '15 at 19:51
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    You are right. I was not fast enough to cancle my incorrect answer before your reply. A correct one is $f(x,y) = 1 - \exp(-x^2) + \tanh(y)/100$. So this function has no minima/maxima but because of the second term everwhere non-zero gradient. – Its_me Nov 10 '15 at 20:02
  • Excellent, thanks! – copper.hat Nov 10 '15 at 20:41

1 Answers1

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Suppose that $M=f^{-1}(a), a\in R$, for every $x$ in $M$, we have to show that $g$ is constant on the connected component of $M$ which contains $x$. Let $c:I\rightarrow M$ be a differentiable path such that $c(0)=x$, ${d\over{dt}}g(c(t))=\nabla g_{c(t)}.c'(t)=u(t)\nabla f_{c(t)}c'(t)=0$ since the restriction of f to M is constant and $\nabla f$ is proportional to $\nabla g$. This implies that $g$ is constant on $c$ and henceforth on the connected component of $M$ which contains $x$.

Tsemo Aristide
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