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Let $U$ be an open set and $f : U \subset\mathbb{R}^d \to \mathbb{R}^m$ be a differentiable function with $df=0, \forall x \in U.$

Show that $f$ is locally constant.

So my idea was, given some arbitrary $x_0 \in U$, first taking some ball $B_{\varepsilon}(x_0) \subset U$ which exists since $U$ is open, we can assume $f$ isn't constant there i.e there is $y_o$ s.t $f(x_0) \neq f(y_0)$.

Then defining $\gamma(t) = (1-t)x_0 +t y_0$, we have $f(\gamma(0)) \neq f(\gamma(1))$ meaning they are different in at least one coordiante, so without loss of generality $f^1(\gamma(0)) \neq f^1(\gamma(1))$. and then using lagrange for an $\mathbb{R} \to \mathbb{R}$ function we have $(f^1(\gamma(c)))' \neq 0$.

But on the other hand using chain rule $(f^1(\gamma(c)))' = d(f^1(\gamma(c))) = \underbrace{df^1(\gamma(c))}_\text{0 since $df=0$} \circ d \gamma(c) = 0$

I don't really like my solution since it relies on coordinates, if someone would like to share a different solution it would be great.

paxtibimarce
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1 Answers1

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Your solution seems correct to me. I think we could rely directly on mean value ineq., because of convexity of euclidean balls, no matter the norm : $$ \|f(b) - f(a)\| \leq \sup\| df\||b-a\| = 0$$

purple
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