Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $E$ and $F$ be Banach spaces and let $X \subset E$ be a connected open subset. I am trying to show that if that derivative function $\partial f: X \rightarrow \mathcal{L}(E, F)$ is identically $0$ on $X$ then $f$ must be constant.

Since $X$ is open, for any given $x_0 \in X$, we can find an open ball about $x_0$, $\mathbb{B}_{\epsilon}(x_0)$, such that $\mathbb{B}_{\epsilon}(x_0) \subset X$. Since the open ball is connected, we can apply the integral mean value theorem to conclude that

$$ \|f(x) - f(x_0) \| = 0 \implies f(x) = f(x_0) \; \forall \; x \in \mathbb{B}_{\epsilon}(x_0) $$

It follows that $f$ is locally constant. Now, from this result we know that $f$ is constant on $X$. I am trying to follow an argument however that does not make use of this result. Let $y_0 = f(x_0)$. Then, since $\partial f$ exists, $f$ is continuous, and from this it follows that the (non-empty) fiber $f^{-1}(y_0)$ is closed. If we can show that $f^{-1}(y_0)$ is also open we can use the hypothesis that $X$ is connected to conclude that $f^{-1}(y_0) = X$ and so $f$ must be constant. I'm stuck on this point. Therefore, my question is:

Within the context of the preceding paragraph, how can we conclude that $f^{-1}(y_0)$ must be open?

share|cite|improve this question
BTW, once you start using the continuity method (open+closed+non-empty = whole space), your proof cannot really be different from the result you linked to, as you can simply rearrange the arguments given in that link into an argument by continuity. – Willie Wong Jul 27 '11 at 12:29
@Willie The argument I'm following, which purports to be a "proof" does not make use of any of these arguments but flatly concludes, as in "clearly" that $f^{-1}(y_0)$ is open which leads me to believe there is some basic fact that I'm overlooking. I can throw away that argument and simply use the result I linked to but would still like to know why it is "immediate" or "clear" that $f^{-1}(y_0)$ must be open. – ItsNotObvious Jul 27 '11 at 13:18
The definition of local constancy immediately implies that $f^{-1}(y_0)$ cannot contain a boundary point. Of course, "clearly" is a weasel word: what is clear to a 50 year old professor is not the same as what is clear to a 20 year old undergrad. – Willie Wong Jul 27 '11 at 14:22
@Willie Maybe I see this now: Because we can always find an open set $U$ such that $f(U) = z$ for some fixed $z$, $f^{-1}(f(U)) \subset f^{-1}(z)$ which implies $f^{-1}(z)$ is open. Correct? – ItsNotObvious Jul 27 '11 at 14:45
Aiye. That's the answer. – Willie Wong Jul 27 '11 at 16:23
up vote 4 down vote accepted

You use that $f$ is locally constant. =)

To show that a set is open, you need to show that every point in it is contained in an open ball. Analytically this is a local condition, since you can take the ball to be arbitrarily small. (Algebraically, however, one may choose a topology where open balls are no longer intuitively local.) So basically, you cannot get away from using some local property of $f$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.