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I've been trying to prove to myself that if $\Omega$ is an open connected set in $\mathbb{R}^n$, then if $f\colon\Omega\to\mathbb{R}^m$ is a differentiable function such that $f'(x)=0$ for all $x\in\Omega$, then $f$ is constant.

I've reduced the problem to just showing $f$ is locally constant on $\Omega$. Given $x_0\in\Omega$, I know that $$ \lim_{x\to x_0}\frac{\|f(x)-f(x_0)-f'(x_0)(x-x_0)\|}{\|x-x_0\|}=\lim_{x\to x_0}\frac{\|f(x)-f(x_0)\|}{\|x-x_0\|}=0. $$

This implies $\lim_{x\to x_0}\|f(x)-f(x_0)\|=0$. So for any $\epsilon>0$, there exists some open ball $B(x_0,\delta)$ around $x_0$ in $\Omega$ such that $\|f(x)-f(x_0)\|<\epsilon$ for $x\in B(x_0,\delta)$. But since the choice of the open ball changes with $\epsilon$, I don't think I can conclude $\|f(x)-f(x_0)\|=0$ for $x\in B$. So this doesn't convince me that there actually exists a neighborhood of $x_0$ on which $f$ is constant.

How can you make the leap from zero derivative to locally constant?

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    $\begingroup$ Have you come across the mean value theorem? $\endgroup$ Jul 19, 2013 at 16:52
  • $\begingroup$ Restrict to an open ball contained in the domain so that segments $[x_0,x]$ be contained in the domain of $f$. Then consider the function $f((1-t)x_0+tx)$ of the real variable $t$. $\endgroup$
    – Julien
    Jul 19, 2013 at 17:04
  • $\begingroup$ @julien Thanks. Is the point that if $g(t)=f((1-t)x_0+tx)$, then there exists $c\in (0,1)$ such that $g(1)-g(0)=g'(c)$. But $g(1)=f(x)$, $g(0)=f(x_0)$, and $g'(c)=0$ since $f$ has zero derivative, which implies $f(x)=f(x_0)$. Since there is a segment contained in the ball from $x_0$ to any other $x$, this shows $f$ is constant on the ball? $\endgroup$ Jul 19, 2013 at 17:31
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    $\begingroup$ Simply, since $g(t)=f((1-t)x_0+tx)$ is differentiable on an open interval containing $[0,1]$ with zero (whence continuous) derivative by chain rule, FTC yields $f(x)-f(x_0)=g(1)-g(0)=\int_0^1g'(t)dt=0$. Of course, the MVT works as well, but attention: you need to consider $t\longmapsto (g(t),v)$ for a fixed vector $v$ as Peter did. $\endgroup$
    – Julien
    Jul 19, 2013 at 17:34
  • $\begingroup$ @julien Thank you for explaining. $\endgroup$ Jul 19, 2013 at 17:38

2 Answers 2

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Recall that an open connected set is $\Bbb R^n$ is polygonally connected. The generalized mean value theorem says that if $f:\Omega\subseteq \Bbb R^n\to\Bbb R^m$ is differentiable in the open set $\Omega$ then for each ${\bf a,b}\in\Omega$ such that $\mathscr L({\bf a},{\bf b})\subseteq \Omega$ and any $\bf w\in\Bbb R^m$ there exists a $\bf z$ in the line joining $\bf a$ and $\bf b$ such that $${\bf w}\cdot (f({\bf b})-f({\bf a}))={\bf w} \cdot {\rm D}f({\bf z})({{\bf b}}-{\bf a})$$

Since the derivative vanishes, this says that the dot product of $f({\bf b})-f({\bf a})$ with every vector in $\Bbb R^m$ is zero, so we must have $f({\bf b})=f({\bf a})$ for each ${\bf b},{\bf a}\in\Omega$.


I think it is worth adding a proof of both claims. First, let's show that

Every open connected set $\Omega$ in the Euclidean space is polygonally connected.

P Fix a point $x\in \Omega$, and let $S$ denote the set of points that can be joined by a polygonal path to $x$. Note $S$ is nonempty, for $x\in S$. Let $T=\Omega\smallsetminus S$. Then $S\cup T=\Omega, S\cap T=\varnothing$. We will show both $S$ and $T$ are open, which will force $T=\varnothing$; as desired.

Let $a\in S$, and join $a$ to $x$ by a polygonal path. Since $\Omega$ is open, there exists a ball $B(a)$ such that $B(a)\subseteq \Omega$. But if $a'\in B(a)$; we can join $a$ to $a'$ and subsequently $a'$ to $x$ by a polygonal path, since $B(a)$ is convex. Thus $B(a)\subseteq S$, and $S$ is open.

Now let $b\in T$. Since $\Omega$ is open, there is a ball $B(b)\subseteq \Omega$. If we could join a point $b'\in B(b)$ to $x$, then we would join $b$ to $x$, since $B(b)$ is convex. Since this cannot be possible, we have $B(b)\subseteq T$, and $T$ is open.

But then, since $\Omega$ is connected and $S\neq \varnothing$, we must have $T=\varnothing$. Since $x$ was arbitrary, this proves the claim. $\blacktriangle$


The proof of the generalized MVT is a consequence of the usual unidimensional mean value theorem. Pick ${\bf x},{\bf y}$ and let ${\bf u}={\bf y}-{\bf x}$. Pick $\bf w$ and define $$F(t)={\bf w}\cdot f({\bf x}+t{\bf u})$$

so that for $t\in(-\delta,1+\delta)$ with $\delta >0$ small enough we have ${\bf x}+t{\bf u}\in \Omega$. Apply the mean value theorem to $F$, whose derivative is $$F'(t)={\bf w}\cdot {\rm D}f({\bf x}+t{\bf u})({\bf u})$$

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A cute direct proof (styled after path connectedness ones) is to pick any point $x$ in the domain and consider the set $U$ on which $f$ takes the value $f(x)$ and the set $V$ on which it does not take this value. These sets are non empty if the hypothesis fails. Clearly $V$ is open, so we are done if $U$ is open.

It is enough to prove that around any $y\in U$ there is a ball also in $U$. But choosing any ball $U'$ containing $y$, we note that any other point $x'\in U'$ is connected to $y$ by a line, and that the function restricted to that line is differentiable with derivative zero.

Now it is a standard 1D problem. If we consider a function $g(t)=f(y+t(x'-y))-f(y)-\Delta t$ on that line with $\Delta$ chosen such that $g$ vanishes at either end, then it achieves its maximum/minimum at some point and its derivative vanishes there. This implies $\Delta$ is zero and so $f$ took the same value at either end. (This is a compressed proof/reminder of MVT and Rolle.)

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