# 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! • Consider any function with$f(1)=c$for some$|c|>1$. Then$\lim_{x\to0^+}f(x)$needn't exist. Commented Apr 28, 2018 at 12:23 ## 3 Answers 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. 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$, one just has to match the interval ends as smoothly as desired. • @AryamanJal Thanks, fixed Commented Apr 28, 2018 at 12:40 • Oh your answer is essentially the same as mine (although yours is definitely better :)), I didn't get a notification when you posted your answer :( – orlp Commented Apr 28, 2018 at 12:41$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)\$.

• If there is a formal proof I will be much more appreciate. Commented Apr 28, 2018 at 12:41
• @TonyMa See Hagen von Eitzen's answer for a more general and formal approach with the same idea.
– orlp
Commented Apr 28, 2018 at 12:43