Find $f'(0)$ if $f(x)+f(2x)=x\space\space\forall x$ Find $f'(0)$ if $f(x)+f(2x)=x\space\space\forall x$
If we assume $f'(0)\in\mathbb R$, then obviously, $f'(0)=\frac{1}{3}$.
But what if we don't assume the derivative exists?
I get this question when I am taking an exam, and I then asked the professor about whether there exist $f$ such that $f'(0)$ doesn't exist but satisfies the condition, but he says he think both exist or not is likely...
What I have tried:
If $f(x)+f(2x)=x\space\space\forall x\in\mathbb R$, $f(x)=\frac{x}{3}$ is the only nice function I have ever thought up$.
If we restrict $dom(f)\in\mathbb R$, I think that any polynomial besides $\frac{x}{3}$ doesn't satisfy the condition, and somethings like $\vert x\vert$ doesn't help neither.
On the other hand, I have tried to think about the equivalence statements with $f(x)+f(2x)=x\space\space\forall x$ in order to prove $f$ must be specific kind of problem s.t.$f'(0)$ exists.
$f(2x)+f(4x)=2x$, so $f(4x)-f(x)=x$.
In general, $$f(2^{2^n}x)-f(x)=x\prod_{k=1}^{n-1}(2^{2^k}+1)\forall n\in\mathbb N, x\in\mathbb R$$
But I don't think this helps.
Can we find $f(2x)-f(x)$ by given condition? I havn't get an idea. But while I am thinking of it, I observe that $g(x)=f(2x)-f(x)$ is on its own satisfying $g(2x)+g(x)=x$.
Will thinking about $f\circ f\circ f\circ f\circ f\circ\dots$ be useful? $f\biggl(f(x)+f(2x)\biggr)+f\biggl(2\bigl(f(x)+f(2x)\bigr)\biggr)=x\space\space\forall x$
Any help will be appreciate. Thank you!
 A: Consider the function
$$
f(x)=\left\{\begin{array}{}\sin(\pi\log_2(|x|))+\frac x3&\text{if }x\ne0\\ 0&\text{if }x=0\end{array}\right.
$$
That is, the conditions given do not imply that $f'(0)$ exists.
A: We find $f(0)=0$. On $\Bbb R\setminus \{0\}$ we can define the relation 
$$x\sim y\:\iff \exists k\in\Bbb Z\colon x=2^ky. $$
This is an equivalence relation and each equivalence class has exactly one representative in $A:=[1,2)\cup(-2,-1]$.
Clearly, we can find $f$ satisfying the conditions of the problem statement by defining it arbitrarily on $A$ and then extending it using the functional equation. Note that then the function $g(x):=f(x)-\frac x3$ obeys the functional equation $g(x)+g(2x)=0$ and that consequently $|g(x)|$ is constant on the equivalence classes of $\sim$. In particular, if for some $a\in A$ we have $g(a)\ne 0$, it follows that $g(4^{-n}a)=g(a)$ and $g(2\cdot 4^{-n}a)=-g(a)$ for all $n$ and hence $\lim_{x\to 0}g(x) $ does not exist; consequently, $g'(0)$ and $f'(0)$ do not exist either in that situation. In other words, if we assume that $f$ (and hence $g$) is continuous at $x=0$, it follows that $g\equiv 0$ and $f(x)=\frac x3$. All other starting definitions of $f$ on $A$ lead to a function not differentiable or even continuous at $x=0$. Note that it is well possible to achieve that $x=0$ is the only discontinuity of $f$, one just has to match the interval ends as smoothly as desired.
A: $f(x)$ is separated into independent subsets defined by equivalence relationship $a = b$ given by $\exists k : 2^k \cdot a = b$.
This is due to two reorderings of the original equation, which can be iterated:
$$f(2x) = x - f(x)$$
$$f(x/2) = x/2 - f(x)$$
Let $f(1) = 1$ by definition. Then we have iteration:
0.5 -0.5
0.25 0.75
0.125 -0.625
0.0625 0.6875
0.03125 -0.65625
0.015625 0.671875
0.0078125 -0.6640625
0.00390625 0.66796875
0.001953125 -0.666015625

It is clear that there is no derivative at $f(0)$.
