Zero derivative on a connected open set implies constant

Builds on Mean value inequality Differentiable function of several variables

Statement

Let $U \subseteq \mathbb{R}^n$ be a connected open set and let $f \colon U \to \mathbb{R}^m$ be differentiable on $U$ . Suppose that $Df(x) = 0$ for each $x \in U$ . Then $f$ is constant on $U$ .

Remarks

The proof fixes $a \in U$ and shows that the set $S = \{x \in U : f(x) = f(a)\}$ is open and closed in $U$ . It is open by the mean value inequality with $M = 0$ on small balls, and closed because $f$ is continuous. As $U$ is connected and $a \in S$ , $S = U$ . Connectedness is needed: on $U = (0, 1) \cup (2, 3)$ , the function that is $0$ on the first interval and $1$ on the second has derivative $0$ but is not constant. For real-valued $f$ , the condition $Df = 0$ means that all partial derivatives of $f$ vanish on $U$ .

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Euclidean space object
Function concept
Function of several variables concept
Limit of a function of several variables concept
Differentiable function of several variables definition
Mean value inequality statement
Zero derivative on a connected open set implies constant statement

Tag	`0BYZTFM0DJ`
URI	`m2://0BYZTFM0DJ`
Kind	statement, theorem
Writing	W0 stub
MSC	26B05
Status	This copy of the site has no build data. Run `m2 status 0BYZTFM0DJ`.
Uses	Mean value inequality, Differentiable function of several variables

Zero derivative on a connected open set implies constant

Statement

Remarks

Reading path

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