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Problem Statement: Suppose $U \subset \mathbb{R}^n$ is open, and $f:U\rightarrow \mathbb{R}$ is a differentiable scalar field such that $\frac{d}{dx}f(x)=0$ for all $x\in U$, show that if $U$ is connected then $f$ is constant. Hint: fix $a \in U$, show that $\left\{x\in U : f(x) = f(a)\right\}$ is clopen.

We had a previous part that I feel might be helpful here. We had to show that if we have a function from an open ball to $\mathbb{R}$, and all directional derivatives are $0$ for all $x$ in the ball, that $f$ is constant. I am confident in that proof, I give it below:

The hypothesis of the mean value theorem is satisfied. We have that $A = B(a) \subseteq \mathbb{R}^n$ is open, we have that $f$ is differentiable on A because all of the directional derivatives exist and equal $0$. We also have that, since the open ball is convex, that $A$ contains a line segment with endpoints $\vec{a}$ and $\vec{a}+\vec{h}$ therefore there is a point $\vec{c}=\vec{a}+t_o \vec{h}$ with $0 < t_0 < 1$ such that $$f(\vec{a}+\vec{h}) - f(\vec{a}) = Df(\vec{c})\cdot \vec{h}$$ however from the problem statement we know that $Df(\vec{c})\cdot \vec{h}$ is zero, therefore $$f(\vec{a}+\vec{h}) - f(\vec{a}) = 0 \implies f(\vec{a} + \vec{h}) = f(\vec{a})$$ This implies that is a constant since the choice of $\vec{a}$ and $\vec{h}$ is arbitrary.

This shows immediately, to me, that the function should be constant because if the derivative is $0$ then all the directional derivatives should be zero and because the set is open, it is the union of open balls. Therefore what I have proven before holds here, but I feel like it shouldn't be that easy. Also I don't see how the hint helps me. If $a$ is the only point in U so that $f(x)=f(a)$ then the set is definitely not clopen, and even if the set is clopen then I have no clue how that helps me. Any help would be appreciated.

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  • $\begingroup$ The set described in the hint is clearly closed. To show that it is open, you may use mean value theorem to prove the function is locally constant. $\endgroup$ Sep 29, 2016 at 17:43
  • $\begingroup$ I'm having a bit of trouble showing that. If a is the only point that has f (x)=f (a) then it is not open. I can't really even use the mean value theorem on this because it's only one point. I can show that a neighborhood around a has the function as constant on it, that seems trivial, but that isn't the set the hint gave me so I don't know $\endgroup$
    – Ben Ray
    Sep 29, 2016 at 17:49
  • $\begingroup$ You don't need to make such assumption because as it turns out, that will never happen. Such a neighborhood of any point in the hint's set will be contained in the latter. $\endgroup$ Sep 29, 2016 at 17:57
  • $\begingroup$ Oh, I see because we have that the derivative is 0 so that f(a+h)-f(a) is 0, and so there is a neighborhood around a that has f(x)=f(a) for all x in the neighborhood. $\endgroup$
    – Ben Ray
    Sep 29, 2016 at 17:59
  • $\begingroup$ Yes. Just use that reasoning for every other point of the set. $\endgroup$ Sep 29, 2016 at 18:05

2 Answers 2

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Let $S := \{ x \in U :f(x)=f(a)\}$. First, we have $S=f^{-1}(\{f(a)\})$. Since $f$ is differentiable, $f$ is continuous. As $S$ is the inverse image of an closed set ($\{f(a)\}$) by a continuous application, we know that $S$ is closed. Let us show that $S$ is open as well.

Let $x_0 \in S$. Because $x_0 \subset U$ where $U$ is open, there exists $\xi$ such that $D_{\xi} := D(x_0, \xi) \subset U$. Therefore, if $y \in D_{\xi}$ we can write:

$|f(y) - f(x_0)| \leq \max_{x\in[y,x_0]}|Df(x)| |y - x_0| = 0$ which yields $f(y) = f(x_0)$ so that $D_{\xi} \subset S$, proving that $S$ is open as well.

Hence $S$ not being empty and being clopen implies that $S = U$, thus proving that $f$ is constant.

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  • $\begingroup$ Okay I see now that the connected assumption is necessary in saying that S=U since in a connected space, the only clopen sets are empty and the universe set. We have shown that S is nonempty and clopen so it has to be the U set. Is there a good counterexample for this where U is not connected, but everything else holds and we have that f is not constant? $\endgroup$
    – Ben Ray
    Sep 29, 2016 at 18:43
  • $\begingroup$ I was thinking that if $U = \left\{(x,y) : xy > 0\right\}$ and we define $f(x,y) = c$ if $x,y > 0$ and $f(x,y) =-c$ if $(x,y) < 0$ which isn't (big question mark here) a constant function, but everything else holds. The function is only locally constant in this case, not globally. $\endgroup$
    – Ben Ray
    Sep 29, 2016 at 18:47
  • $\begingroup$ Let $f(x) = -1$ if $x \in \mathbb{R}^*_-$ and $f(x) = 1$ if $x \in \mathbb{R}^*_+$ is a good counterexample ;). We have $f \in C^{\infty}(\mathbb{R}^*)$ and $Df(x) = 0$ for every $x \neq 0$, but $f$ is not constant. $\endgroup$
    – Hermès
    Sep 29, 2016 at 19:01
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Hint: Let $a\in U,$ consider an open ball $B(a,r)\subset U$, for $y\in B(a,r)$, define $f_y:[0,1]\rightarrow R$ by $f_y(t)=f(a+ty)$. Show that $f$ is differentiable and its differential is zero. So you can say that for every $t\in [0,1], f_y(t)=f_y(0)=f(a)=f_y(1)=f(y)$ conclude that $\{y: f(y)=f(a)\}$ is open.

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