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

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.)

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 $$U$$...
• um, I think you mean "if $B$ is a ball centered at $x$ and contained in $U$ " (not $A$) May 3, 2020 at 22:15