Some definitions first. Let $A \subseteq \mathbb R^n$. Let $x,y \in A$. A path between $x$ and $y$ is a continuous function $f: [0,1] \rightarrow \mathbb{R}^n$ with $f(0) = x$ and $f(1) = y$. The set $A$ is path-connected when for every $x, y \in A$, there exists a $C^1$ path between $x$ and $y$.

Let $f: A \rightarrow \mathbb{R}^m$ be a function, with $A \subseteq \mathbb{R}^n$. Suppose that $f'(a) = 0$ for all $a \in A$. Now if $A$ is path-connected, then $f$ is constant.

In a proof I saw of this theorem the property that every path between two points is $C^1$ is used. My question is: is this necessary? If so, I'd like to see a counterexample. In other words, I'm looking for a function $f: A \rightarrow \mathbb{R}^m$ with zero derivative everywhere, $A$ such that there is a path between any two points (but the path is not necessarily $C^1$) and $f$ is NOT constant.

  • 4
    $\begingroup$ How do you define $f'$ if $A$ isn't open? If $A$ is open and connected then you can connect any two points by a smooth path. $\endgroup$
    – t.b.
    Oct 12 '11 at 11:55
  • $\begingroup$ That's true? Well, I guess that answers my question. $\endgroup$ Oct 12 '11 at 12:10
  • 2
    $\begingroup$ @m.k. Hint: if $A$ is an open connected subset of $\mathbb{R}^n$ and if $f:A\to \mathbb{R}^m$ is a differentiable function such that $f'(x)$ is a zero matrix for all $x\in A$, then $f$ is locally constant on $A$. Therefore, $f$ is constant on $A$ since $A$ is connected. $\endgroup$ Oct 12 '11 at 12:29

As pointed out in the comments, such a function cannot exist. In order to define “$f$ is differentiable”, you need your set to be open. Every pair of points in an open and connected set can be connected by a smooth path, see below. Your argument then shows that $f$ must be constant. (Alternatively, you can argue as Amitesh suggests)

Here's a proof of the fact that any two points of an open and connected set $U \subset \mathbb{R}^n$ can be connected by a piecewise smooth path:

Define an equivalence relation on $U$ by $x \sim y$ if and only if there is a piecewise smooth path $\gamma: [a,b] \to U$ such that $\gamma(a) = x$ and $\gamma(b) = y$.

Let $x \in U$ be arbitrary.

Notice that the equivalence class of $[x]$ of $x$ is open. If $y \in [x]$ there is some open ball $B_r(y) \subset U$. We may connect $y$ to any point $z \in B_{r}(y)$ using the straight line segment $(1-t)y + tz$, $t \in [0,1]$.

Since the complement of $[x]$ is a union of (open) equivalence classes, $[x]$ is closed. Thus, $[x]$ is open, closed and non-empty, hence all of $U$ by connectedness.

If you want to get “smooth paths” instead of only “piecewise smooth paths”, refine the argument slightly by allowing only paths $\gamma: [a,b] \to U$ such that $\gamma|_{[a,a+\varepsilon)} \equiv a$ and $\gamma|_{(b-\varepsilon,b]} \equiv b$ for some $\varepsilon \gt 0$. Then the concatenation of two such paths is still a smooth path. Instead of taking the straight line segment connecting $y$ and $z$ take a smooth function $f: [0,1] \to [0,1]$ such that $f|_{[0,\varepsilon)} \equiv 0$ and $f((1-\varepsilon, 1] \equiv 1$ and take the path $(1-f(t))y + f(t)z$.


This answer is late, but anyway: You may want to look at Hassler Whitney's paper "A function not constant on a connected set of critical points", Duke Math. J. 1 (1935), no. 4, 514--517.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.