$Df = 0$ on open and connected set $\implies f$ is a constant function

Question

Suppose $(E, \parallel \parallel),(F, \parallel \parallel)$ are Banach spaces, $U \subset E$ is an open, connected set, and $f : U \to F$ such that $Df = 0$. Prove that $f$ is a constant function.

Using the MVT, then for $y \in [x,y] := \{ (1-t)x+ty \mid 0 \leq t \leq 1\}$, $$ \| f(x)-f(y) \| \leq \sup_{c \in [x,y]} \|Df(c)(y-x) \| = 0. $$ So $f(x) = f(y), \ \forall y \in [x,y]$.

How do I show that $f(x)=f(y)$ for all $y \in U$? (I know it has to do with the connectedness of $U$, but I don't know how.)

Answer 1

Fix some $x_0 \in U$ and consider the set $A = \{x : f(x) = f(x_0)\}$. Your argument shows that if $x \in A$ and $B$ is a ball centered at $x$ and contained in $U$, then $f(y) = f(x)$ for all $y \in B$ (since balls are convex), and thus $B \subset A$. So you have shown that $A$ is an open subset of $U$. By continuity of $f$, it is also a closed subset of $U$...