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Let $f: \mathbb{R}^d \to \mathbb{R}^m $ be a function sucht that $Df(a) = 0 $ for all $a \in U \subseteq \mathbb{R}^d$. Also, suppose $U$ is open and path-connected.

Question: IS $f$ constant on $U$ ??

  • I think you don't need $U$ to be open – dani_s Mar 06 '14 at 10:13
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    @dani_s No, there are counterexamples if you drop the openness. E.g., there is a homeomorphism $f:\mathbb R^2\to\mathbb R^2$ that transforms an arc of snowflake into a line segment, and has zero derivative at every point of the snowflake. – user127096 Mar 08 '14 at 04:35

4 Answers4

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Here is little more precise answer: Let $a, b \in U$ be arbitrary. We are done if we can show $f(a) = f(b)$. Since $U$ is path-connected, we can find a $C^1$-path (obtained from a continuous path by a suitable regularization procedure) $\gamma :[0, 1] \to U$ from $a$ to $b$. Then

\begin{align*} f(b) -f(a) &= f(\gamma(1))- f(\gamma (0)) = \int_0^1 \frac{\operatorname{d}}{\operatorname{d}t} f(\gamma(t)) \operatorname{d}t = \int_0^1 Df_{\gamma(t)} (\dot{\gamma}(t)) \operatorname{d}t = 0, \end{align*} where we have used the chain rule and $Df =0$. This argument also works if the path is only piecewise $C^1$ by applying it to the segments.

Dragoslav
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Although for the function of several variables there is no Mean Value Theorem, there is another similar result.

Suppose $f$ maps a convex open set $E \subset \mathbb{R}^d$ into $\mathbb{R}^m$, $f$ is differentiable in $E$, and there is a real number $M$ s.t. $\left\| {f'\left( x \right)} \right\| \leqslant M$ for every $x \in E$. Then $\left| {f\left( a \right) - f\left( b\right)} \right|\leqslant M\left| {a - b} \right|$ for all $a,b \in E$.

In this case, we can put $M=0$ since $Df(a) = 0$.

Junefi
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  • But here the domain need not be convex so you can not apply mean value theorem. – XYZABC Dec 02 '17 at 03:32
  • @XYZABC It doesn't matter, since the argument works locally: it works in any open ball contained in the domain $U$. Since $U$ is connected, that $f$ is locally constant implies that it is constant. – Andrés E. Caicedo Nov 27 '18 at 12:38
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Yes: pick any two points and join them with a piecewise-linear path. On each segment, $f$ is constant, and hence the values at the initial and final points coincide.

Siminore
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This answer is similar to that of Dragoslav, except it uses the mean value theorem.

We want to show that every coordinate function $f_i : U \to \mathbb{R}$ is constant. Since $U$ is path-connected, it is enough to show that $$ f_i(\gamma(0)) = f_i(\gamma(1)) $$ holds for every path $\gamma : [0,1] \to U$. The map $f_i \circ \gamma$ is a function of a single real variable, so we can apply the ordinary mean value theorem to obtain $$ (f_i \circ \gamma)(1) - (f_i \circ \gamma)(0) = (f_i \circ \gamma)'(\xi)(1 - 0) $$ for some $\xi \in [0,1]$. Furthermore, the chain rule for total derivatives yields $$ (f_i \circ \gamma)'(\xi) = \underbrace{Df_i(\gamma(\xi))}_{= \, 0} \gamma'(\xi) = 0 $$ (note that the $f_i$ are totally differentiable iff $f$ is totally differentiable).

Silenus
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