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Let $U \subseteq \Bbb R^n$ be an open set and let $f:U \to \Bbb R^n$ be a differentiable function. if $U$ is path connected and $D_f=0$ for every point in $U$, show that $f$ is constant ($D_f$ is the differential of $f$).

I thought about proving that $f$ is locally constant and then showing that if $f$ is locally constant and $U$ is path connected then $f$ is global constant, but it seems like a lot of work. Is there something simpler / shorter?

user401516
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2 Answers2

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Hint: let $\gamma : [0,1] \to U$ be a piecewise linear path between two points $x,y \in U$. Now use the mean value theorem to study the difference $$f(\gamma(1)) - f(\gamma(0)).$$

Zestylemonzi
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  • Can you elaborate please? – user401516 Nov 04 '17 at 16:26
  • I meant to write piecewise linear instead of smooth - I've corrected it. Do you know the multivariable mean value theorem? https://en.m.wikipedia.org/wiki/Mean_value_theorem – Zestylemonzi Nov 04 '17 at 17:12
  • Thanks, it makes more sense now. but how do I know every 2 points have a piecewise linear path between them? I only know it is path connected – user401516 Nov 04 '17 at 17:19
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    You know that there is a path $\gamma$ between your two points in $U$. Then use that the image of your path is compact and so has a finite cover of open sets contained in $U$. You should be able to see how to construct your piecewise linear path now. – Zestylemonzi Nov 04 '17 at 21:43
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I just found ashortcut by @Sharkos answer in here : Why does zero derivative imply a function is locally constant? maybe thats what you looking for. However the standard proof achieved by using mean value theorem.

Let $p,q \in U$. Let $\gamma : [0,1] \rightarrow U$ be a path connecting $p=\gamma(0)$ and $q=\gamma(1)$. We can cover the image $\gamma[0,1]$ by finite number of open balls because $\gamma[0,1]$ compact. I'll leave it to you to show that $f$ is constant on each ball. Because the intersection of these balls is not empty we conclude that $f$ must be the same constant as we move from each ball to the next. After finite steps we conclude that $f(p)=f(q)$. So $f$ is constant on all $U$ $\square$.

$\textbf{EDIT :}$

However, to show that $f$ is constant on each ball use mean value theorem, that is if $\gamma(t) = a + t(b-a)$ is a line segment from $a$ to $b$ on a convex subset $U\subset \mathbb{R}^n$, then there exist $c \in \gamma(t)$ s.t
$$ f(b) -f(a) = Df \cdot (b-a) $$ By this you must able to show that (by Cauchy-Schwarz ineq.) $$ \text{if} \quad ||Df|| \leq M \quad \forall x\in U, \text{then} \quad |f(x)-f(y)| \leq M ||x-y|| \quad \forall x,y \in U $$ and therefore this result you can show that $f$ is constant on the open ball.

Kelvin Lois
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  • I missunderstood the part about $f$ being constant on each ball. Don't we need a function from $\Bbb R \to \Bbb R$ that is differentiable in order to use the mean value theorem? – user401516 Nov 04 '17 at 16:28
  • Yes. Dont we assume that f is differentiable ? We use generalize mean value theorem here. – Kelvin Lois Nov 04 '17 at 16:33
  • I didn't learn this theorem, I thought about using some trick to use the regular theorem, but I'm pretty stuck. – user401516 Nov 04 '17 at 16:37
  • Thats the standard proof i think. Maybe you can modified the question and show where you are and where you stuck. – Kelvin Lois Nov 04 '17 at 16:41
  • Ok, thank you. Another question: why is the intersection of the balls not empty? – user401516 Nov 05 '17 at 14:55
  • This is a result from topology if you want to be rigorous. from Willard's book theorem 26.15) which is says that if because $\gamma[0,1]$ is connected, the open cover that is the open balls have non empty intersection. – Kelvin Lois Nov 05 '17 at 15:08
  • A simpler argument using contradiction, by consider a particular open ball $B$ s.t it us has no intersection with any other balls.if this is the case, the points on the edges, that is $\bar{B} - B$, not covered by the balls. If its covered then there must be intersection between them because the points are limit points. – Kelvin Lois Nov 05 '17 at 15:12
  • Could I ask a follow up question? Why do we need to (using compactness) construct a finite number of balls? Can we not conclude $f(p)=f(q)$ if the cover is infinite? – cpahanson Aug 19 '22 at 07:59