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Let $f:U\to\mathbb{R}$ where $U$ is an open and connected subset of $\mathbb{R}^n$ such that $\nabla f=0$ in $U$. Does it imply the function $f$ is constant in $U$?

I know it is true for a convex $U$, but could not prove neither disprove for an arbitrary open set $U$.

Can some one please help with a counterexample if false.

Thanks.

Math
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  • Yes, Thanks a lot. – Math Jan 30 '23 at 08:55
  • Convexity allows you to use Lagrange's theorem, but is not a necessary condition. If you considerer any point $u \in U$, since $U$ is open, there will be a ball containing only elements of $U$, centered in $u$. Since the ball is convex, you conclude that $f$ will be constant on each of those balls. Being connected now plays an important role... – PierreCarre Jan 30 '23 at 08:57
  • Thank you very much. – Math Jan 30 '23 at 09:03

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