A continuously differentiable function that can't be solution of an ODE Is it possible to find a continuously differentiable real function $f$ defined on  $\mathbb R$ such that $f$ can't be the solution of an ODE $f^\prime(x) = G(f(x))$ in any neighborhood of $0$?
 A: You don't mean an arbitrary ODE, you mean an autonomous first order ODE. And for that, the main point is that if the solution to the ODE IVP is unique, then there cannot be any nontrivial critical points, i.e. if $f'=0$ at any point then $f$ is constant. If $G$ is not nice enough to ensure uniqueness (e.g. $G(x)=|x|^{1/2}$) then even that property goes out the window.
A: The notation $G[a,b] \to \mathbb R$ isn't clear to me. If the domain of G is the set of function from [a,b] to R, then one can simply declare G(f) to be f', at which point f is a solution of f' = G(f).
EDIT: With you change, it seems like any function with different derivatives at the same y value would qualify. For instance, if f(x) = sin(x), then at x = 0, f(x) = 0 and f('(x) = 1. At x = pi, f(x) = 0 and f'(x) = -1. Since G(0) can't be both 1 and -1, there's no G that gives f as a solution over an interval that includes both 0 and pi.
A: First let us summerize what has been mentioned in the comments/other answers:


*

*as Ian mentioned, if we demand $G$ to be such that uniqueness of solution holds (e.g. locally Lipschitz) then it is not possible to find such a $G$.

*as MathematicsStudent1122 mentioned in a comment, $C^1$ functions are locally invertible if the derivative is different from zero. 


Thus, for given $f \in C^1$ a right-hand side of an autonomous ODE which is solved by $f$ is given by
$$
G(x) = \begin{cases} 
0 , & \text{if} \,\, f'(x) = 0 \\
f' \circ \left(f{\big|_{B_r(x)}}\right)^{-1}(x), & \text{otherwise}.
\end{cases}
$$
where $B_r(x)$ is the ball of radius $r$ around $x$ such that $f$ is invertible on $B_r(x)$.
