Htykuut

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 $U$ ??

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.

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 $left| {fleft( a right) - fleft( bright)} right|leqslant Mleft| {a - b} right|$ for all $a,b in E$.

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

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.