Let $\nabla f(x) = 0$. Show that $f$ is a constant function.

Question

Let $f:\mathbb{R^2} \to \mathbb{R}$ and $\partial_1f(x,y)=0, \partial_2f(x,y)=0$ for all $x,y \in \mathbb{R^2}.$ Show that $f$ is a constant function.

From the problem statement I got that $\nabla f(x) = (\partial_1f(x,y), \partial_2f(x,y)) = (0,0).$

And now $$f(y)-f(x) = \nabla f(x)\cdot(y-x)+||y-x||\varepsilon(y-x)$$

which will result in $$f(y)-f(x)=0\cdot(y-x)+||y-x||\varepsilon(y-x) =||y-x||\varepsilon(y-x).$$

What should I do with the term $||y-x||\varepsilon(y-x)$ that I'm left with? By the definition $\varepsilon(y-x)$ goes to $0$ when $y\to x$, but how do I show this?

Seems that all the answers use the mean-value theorem. Isn't there a way to show this from the approach I was marching on? — , Sep 22 '20 at 11:26
@Daniel: Even for functions $h: \Bbb R \to \Bbb R$ is the mean-value theorem the tool to show that $h' = 0$ implies that $h$ is constant. — Martin R, Sep 22 '20 at 11:27

score 2 · Answer 1 · answered Sep 22 '20 at 11:58

2

For any $x,y\in \mathbb{R}^2$ you have

$$ f(y)-f(x)=\int_0^1 \frac{d}{dt}(f(x+t(y-x)) dt=\int_0^1 \nabla f(x+t(y-x))\cdot (y-x) dt=0 $$

because, as you assume, $\nabla f=0$ in $\mathbb{R}^2$.

answered Sep 22 '20 at 11:58

Kosh

1,490

Let $\nabla f(x) = 0$. Show that $f$ is a constant function.

1 Answers1