Let $\phi$ a differentiable function on $\Omega \in \mathbb{R}^n$, open connected set.

then if $\frac{\partial\phi}{\partial x_i}=0 \; for \; i=1,..,n$.

we have $\phi = constant$, because $\Omega$ is connected!

I didn't get why $\Omega $ should be connected.

  • 5
    $\begingroup$ Because if the derivative is $0$, the function is only con,stant on each cinnected component of the domain, but the constant may be different from one connected component to another. $\endgroup$
    – Bernard
    Mar 24, 2019 at 22:48
  • 2
    $\begingroup$ An easy example: $f(x)$ defined on $(0,1)\cup (2,3)$ by $f(x)=1$ for $x\in (0,1)$ and $f(x)=2$ for $x\in (2,3)$. It is defined on an open set, the derivative is zero everywhere but the function is not constant. Connectedness of the domain solves this problem because on a connected set a locally constant function is always globally constant. $\endgroup$
    – Mark
    Mar 24, 2019 at 22:53
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    $\begingroup$ See also: math.stackexchange.com/q/447524/9464 $\endgroup$
    – user9464
    Mar 25, 2019 at 0:04

2 Answers 2


If $U$ and $V$ are disjoint open sets, $\phi(x)=1$ for all $x \in U$ and $\phi(x)=0$ for all $x \in V$ then all the partial derivatives are $0$ but $\phi$ is not a constant function on $U \cup V$.


Since $\Omega$ is open and connected, any two points $x,y \in \Omega$ can be joined by a path with a finite number of segments that are parallel to the axes. Then the mean value theorem shows that $\phi(x)=\phi(y)$ and hence $\phi$ is constant on $\Omega$.

  • $\begingroup$ I think OP's question is why connectedness is necessary. $\endgroup$ Mar 24, 2019 at 23:13
  • $\begingroup$ @KaviRamaMurthy: The title was asking why the (partial) derivative(s) needed to be zero... $\endgroup$
    – copper.hat
    Mar 24, 2019 at 23:20
  • $\begingroup$ Sorry. I didn't notice that what OP says in the question is very different from what he says in the title. $\endgroup$ Mar 24, 2019 at 23:21
  • $\begingroup$ @KaviRamaMurthy: Similarly, I didn't read the content carefully :-). The wording of the question could be improved. $\endgroup$
    – copper.hat
    Mar 24, 2019 at 23:25

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