Total derivative vanishes implies function is constant

Question

I came across this problem:

Let $f: \mathbb{R}^n \to \mathbb{R}^m$ be a totally differentiable function, whose derivative $Df$ vanishes for all $x\in \mathbb{R}^n$. Show that $f$ is constant. Hint: Consider the line segment $L$ connecting two arbitrary points $x_{0},y_{0}\in\mathbb{R}^n$ and show that for suitable points $x_{i},y_{i}\in L$, $|f(x_{i})-f(y_{i})|$ is sufficiently small. Use compactness to proof the statement.

We aren't supposed to use partial derivatives like this: Functions where the total derivative is zero. Also, we don't have a mean value theorem for multi variable functions and we didn't introduce connectedness yet (as used in the second answer above).

For the first part of the hint, one could use continuity of $f$ to force $|f(x_{i})-f(y_{i})|$ small enough. But this would require $\delta$ to be (very) small aswell. Regarding compactness, my first guess was to define a sequence of points $\in L$ and use sequential compactness in some sense. None of my thoughts worked out so far.

Regards cerocius

Answer 1

Suppose you have some number $\epsilon_1$. For each point $p$, you can take the set of points such that $|\frac{f(x)-f(p)}{x-p}|<\epsilon_1$. This is then a family of open sets. If you can prove that this family covers the space, then by compactness, there is a finite subcover. Thus, there is a finite number of sets that cover the path going from $x_0$ to $y_0$. You can then apply the Triangle Inequality a finite number of times to find the upper bound of how much f varies over that path.

But this would require δ to be (very) small aswell.

Well, in reality, since f is constant, $\delta$ is infinite. But until we prove that f is constant, we have to pretend that $\delta$ might be going to zero. However, that's okay. $\delta$ going to zero means that the number of sets we need to cover the path goes to infinity. But because of the $x-p$ in the denominator, as the size of each set shrinks, the amount that f can vary within that set shrinks as well. Setting an upper limit on the total amount that f can vary over the whole path doesn't depend on the number of sets, but instead is limited by $\epsilon_1$, so if we let that go to zero, then f being constant follows.

Answer 2

Every partial derivative is equal to $0$ and $\Bbb R^n$ is path-connected. Moreover, for any two points in $\Bbb R^n$ we can find a path between them that consists of finitely many subpaths, each of which is parallel to an axis. Take one such subpath, suppose (w.l.o.g.) it is parallel to the $x$-axis. Say it connects $(x_0, \dots)$ to $(x_1, \dots)$. The $\dots$ stands for the other $n-1$ coordinates which we don't care about here. Then by the Fundamental Theorem of Calculus

$$f(x_1, \dots) - f(x_0, \dots) = \int_{x_0}^{x_1} \frac{\partial f}{\partial x}(t, \dots) dt = 0$$

as $\frac{\partial f}{\partial x} = 0$ everywhere. So $f(x_1, \dots) = f(x_0, \dots)$. Do this for every subpath to show that $f$ is constant on all of $\Bbb R^n$.

Answer 3

The fact that $Df$ vanishes everywhere implies that the derivative at $t=0$ of the vector-valued function of a single real variable $t \mapsto f(x + t(y_0 - x_0))$ vanishes for any $x$ on the line between $x_0$ and $y_0$, and hence every component of this derivative vector must be zero. If $f(x_0) \neq f(y_0)$, one of their components must differ, and this contradicts the MVT for real-valued functions of a single variable.