If $Df(c)=0$ $\forall c\in V$ then $f$ is constant on $V$

Question

Theorem:

Suppose that $V$ is open and connected in $\Bbb R^n$

$f:V\to \Bbb R^m$ is differentiable on $V$

I want to prove this theorem with mean value theorem for real valued functions.

How to prove this?

No. My instructor want to use MVT for real valued functions. Please can you show me the proof? @Potato — 1190, Jul 20 '13 at 05:57
@Sharkos I saw thissolution now. But he solved it by using MVT FOR VECTOR VALUED FUNCTIONS. but I want to use MVT for REAL VALUED FUNCTİONS. :( — 1190, Jul 20 '13 at 07:25
@B11b Well I didn't... Restrict to a line between points, done. — not all wrong, Jul 20 '13 at 07:37
@B11b Hand on, real valued functions? That's a completely trivial distinction, work componentwise and it's the same. — not all wrong, Jul 20 '13 at 07:39

score 2 · Answer 1 · answered Jul 20 '13 at 06:00

If $a\not=b$, the line passing through $a$ and $b$ must contain a point $c$ where the derivative of the restriction $g$ of $f$ to that line parametrised by length along that line is $g'(c)=\frac{g(a)-g(c)}{a-b}$. If $D(f)=0$ everywhere, then $\displaystyle g'(c)$ is always $0$.

For any point $c$ in an open set, points $a$ and $b$ can be chosen in an arbitrarily small neighbourhood of $c$ with the line between then passing through $c$, and since differentiable functions must be continuous, $g(a)=g(b)$ and hence $f(a)=f(b)$ everywhere in $V$.

If $Df(c)=0$ $\forall c\in V$ then $f$ is constant on $V$

1 Answers1