$\partial f = 0 \implies f$ is constant

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?

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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

1 Answer

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 $f$.

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