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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!

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  • $\begingroup$ Consider any function with $f(1)=c$ for some $|c|>1$. Then $\lim_{x\to0^+}f(x)$ needn't exist. $\endgroup$ Commented Apr 28, 2018 at 12:23

3 Answers 3

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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.

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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.

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  • $\begingroup$ @AryamanJal Thanks, fixed $\endgroup$ Commented Apr 28, 2018 at 12:40
  • $\begingroup$ 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 :( $\endgroup$
    – orlp
    Commented Apr 28, 2018 at 12:41
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$f(x)$ 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 $f(0)$.

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

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