If the derivative is $0$, then $f$ is constant in a banach space My question is simple. Take a differentiable function $f: U \subset \mathbb{E} \rightarrow \mathbb{F}$, where $\mathbb{E}, \mathbb{F}$ are banach spaces and $U$ is an open connected subset of $\mathbb{E}$. If $f'=0$, is $f$ constant?
My question comes from the analogue fact in $\mathbb{R}^n$. The way I see the proof of this in $\mathbb{R}^n$ is showing that is locally constant, and to do that I make a "coordinates path" from a point to other in the neighbourhood and use the partial derivatives being $0$ to conclude, but I cannot think a way to generalize this. 
 A: Here is an outline of a proof (which seems correct to me): First show it is locally constant, and then use the fact that $U$ is connected. Choose some point $y\in U$ and take some point $x$ in some ball centered at $y$ which is contained in $U\,.$ Also take some $x^*\in \mathbb{F}^*.$ Then $x^*[f((1-t)x+t(y-x))]$ can be considered as a function from $[0,1]$ to $\mathbb{R}$ so you can use the result from one dimensional analysis that $f'=0\implies f=$const., combined with the fact that the dual space of a Banach space always separates points to show that $f$ is locally constant.
A: In Banach spaces we still have the concept of directional derivatives (more fancifully called the Gâteaux derivatives). In much the same way as in finite dimensions, one can show that if the Fréchet derivative is zero at a point, then the Gâteaux derivatives at that point are zero in every direction. Since balls in Banach spaces are convex, applying the fundamental theorem of calculus for one-variable Banach-valued functions to the Gâteaux derivatives will show $f$ to be locally constant.
