A problem i have goes as follows:

Let $f:\mathbb R\to\mathbb R$ be a continuous function satisfying $f(2x)=f(x),\;\forall\;x\in\mathbb R$. If $f(1)=3$, then the value of $\displaystyle \int_{-1}^1 f(f(f(x)))\,\mathrm dx$ is?

Now, their solution is as follows.

Consider $f(x)$ on the interval $[x_0,2x_0]$. Then: $$\dfrac{f(2x_0)-f(x_0)}{x_0}=f'(c),\text{ where }c\in(x_0,2x_0)$$ $$\therefore f'(c)=0$$ $$\therefore \color{red}{f(x)=\text{a constant}}$$ $$\vdots$$

How did they get to the step highlighted in red from the previous step? The function derivative is zero at some $x=c$, how does it imply that the function derivative is zero on the entire function?

Also, is there another method to solve the functional equation?


  • $\begingroup$ That seems wrong, yes, particularly since you don't know the function is differentiable yet, and the value $f'(c)=0$ only for one $c$. $\endgroup$ – Thomas Andrews Apr 14 '16 at 17:01
  • $\begingroup$ No, they did not prove that $f'(x_0)=0$ for all $x_0$, @YotamD They only proved that there is a $c\in (x_0,2x_0)$ where $f'(c)=0$. $\endgroup$ – Thomas Andrews Apr 14 '16 at 17:03
  • $\begingroup$ For example, if you changed the condition of this problem to restrict the function to positive real numbers, there is clearly a non-constant function $f(x)=\sin2\pi\log_2(x)$. So where does the proof fail in the case of functions on the positive reals? $\endgroup$ – Thomas Andrews Apr 14 '16 at 17:05
  • $\begingroup$ @ThomasAndrews There'd even be non-constant, nowhere differentiable such functions $\endgroup$ – Hagen von Eitzen Apr 14 '16 at 17:07

The red claim is true, but a non sequitur as stated. For example, there seems to be an application of Rolle's theorem, but the we do not even knwo if the premises hold. Also, one could (immediately) only conclude that some points have derivative zero.

However: The continuous function $f$ restricted to the compact interval $[1,2]$ attains its maximum $M$ and its minimum $m$. Then $M$ and $m$ are also attained in every interval $[2^{-k},2^{1-k}]$. By continuity, $f(0)=M=m$. We conclude that $f$ is constant (as the same argument works for negative $x$).

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  • $\begingroup$ Could you please explain to me how you concluded that the same value of $m$ and $M$ will be found for all intervals $[2^{-k},2^{1-k}]$? Also, did you conclude $f(0)=m=M$ by taking the limit $k\to 0$? You see I'm not well versed with this topic! $\endgroup$ – FreezingFire Apr 14 '16 at 17:13

In fact, it is enough to assume it is continuous at 0.

Since $f$ is continuous at 0, then for any $\epsilon$ there must exist $\delta$ so that when $|x|<\delta$ then $|f(x)-f(0)| < \epsilon$. Suppose the farthest value from $f(0)$ this function attains is $f(a)=f(0)+c$. Then take $\epsilon=c$ to see that $|f(x)-f(0)|<c$ when $|x|<\delta$. But, $c=|f(a)-f(0)|=|f(a/2)-f(0)|=\cdots=|f(a/2^b)-f(0)|$, where $b$ is large enough to get $a/2^b<\delta$. Contradiction.

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