A function with zero gradient is locally constant

Question

A mapping $f:X\to Y$ is defined to be locally constant if $\forall x\in X$, there exists a neighbourhood $V(x)$ containing $x$ such that $a\in V(x)\implies f(a)=x_0$ for some constant $x_0$. In other words, every point in that neighbourhood maps to the same image.

My book says that if $\text{grad }f=0$ for all $x \in X$, then $f$ is locally constant.

Could someone give a proof of this fact? It has been confusing me for some time now!

They meant $grad f=0$ in a neighborhood of $x$ (or, maybe even everywhere in the domain). — Moishe Kohan, Sep 12 '14 at 15:38
With the edit, your book is then true of course ;-) I remove my preceding comments. — Jean-Claude Arbaut, Sep 13 '14 at 05:59

score 7 · Answer 1 · answered Sep 12 '14 at 19:10

7

This is a special case of the mean value inequality for functions of several variables: if $|\nabla f|\le M$ on the line segment from $a$ to $b$, then $$ |f(a)-f(b)|\le M|a-b| $$ The proof amounts to applying the usual mean value theorem to the one-variable function $$g(t)=\langle f(a+t(b-a)),b-a\rangle$$ which has $g(1)-g(0)=|f(b)-f(a)|^2$ and $|g'(t)| =| \langle f'(t), b-a\rangle|\le M|b-a|$.

In your case $M=0$.

answered Sep 12 '14 at 19:10

Is $\langle x,y\rangle$ the inner product of $x$ and $y$? – Sep 13 '14 at 06:56
Yes, it must be. – Henno Brandsma Sep 13 '14 at 07:15
@AyushKhaitan Yes, it is. – Sep 13 '14 at 15:36

A function with zero gradient is locally constant

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