A continuous function whose left derivative at a point is awlays twice its right derivative is constant Let $f: \mathbb{R}\to\mathbb{R}$ be a continuous function such that for all $x \in \mathbb{R}$, the right derivative of $f$ at $x$ is twice the left derivative of $f$ at $x$.
Does it follow that $f$ is constant?
 A: It's true but the proof is not easy. Don't have time to type it all. Will just give a rough sketch. Since the left and right derivatives exist everywhere by fact 2 below they must coincide except on a countable set. But then $f'(x)=0$ except on a countable set. Then use fact 4 below applied to $f$ and $-f$ to conclude that $f$ is both increasing and decreasing and so constant.
Facts:


*

*The set of points of strict local minimum of a function $g$ is countable, where $x$ is a a point of strict local minimum if there is $\varepsilon>0$ such that $g(y)>g(x)$ for all $y\in(x-\varepsilon,x+\varepsilon)$, $y\neq x$.

*The left and right derivatives coincides except for countably many points. Idea: For $r\in\mathbb{Q}$ you have to prove that the set $F_{r}=\{x:~f_{-}^{\prime}(x)<r<f_{+}^{\prime}(x)\}$ is countable. Consider the function
$g(x)=f(x)-rx$. If $x\in F_{r}$ then $x$ is a strict minimum of $g$. Conclude
that $\{x:~f_{-}^{\prime}(x)<f_{+}^{\prime}(x)\}$ is countable.
Similarly, the set $\{x:~f_{+}^{\prime}(x)\neq f_{-}^{\prime}(x)\}$ is countable.

*Let $f:\mathbb{R}\rightarrow\mathbb{R}$
be continuous in $\mathbb{R}$ and differentiable for all $x\in \mathbb{R}$ except at most a
countable set $E$. Prove that if $f^{\prime}(x)>0$ for all $x\in \mathbb{R}\setminus E$, then $f$ is strictly increasing. Idea: If $f(b)<f(a)$, let $f(b)<t<f(a)$ with $t\notin f(E)$, 
consider the set $\{x\in [a,b]:\,f(x)=t\}$. This set has a maximum at $c$ with $f'(c)\le 0$ which is a contradiction. 

*Let $f:\mathbb{R}\rightarrow\mathbb{R}$
be continuous in $\mathbb{R}$ and differentiable for all $x\in \mathbb{R}$ except at most a
countable set $E$. Prove that if $f^{\prime}(x)\geq0$ for all $x\in \mathbb{R}\setminus E$, then $f$ is increasing. Idea: apply (3) to $f(x)+\varepsilon x$.
There might be a simpler proof...
