# Method of solving the functional equation $f(2x)=f(x)$ using Lagrange's Mean Value Theorem

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?

Thanks!

• That seems wrong, yes, particularly since you don't know the function is differentiable yet, and the value $f'(c)=0$ only for one $c$. Apr 14, 2016 at 17:01
• 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 $f'(c)=0$. Apr 14, 2016 at 17:03
• 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? Apr 14, 2016 at 17:05
• @ThomasAndrews There'd even be non-constant, nowhere differentiable such functions Apr 14, 2016 at 17:07

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 $x$).
• 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! Apr 14, 2016 at 17:13
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.