Let $f:\mathbb R \to \mathbb R$ be differentiable such that $f(x/2)=f(x)/2$ for any $x\in \mathbb R$. How can I prove that $f(x)=f'(0)x$, for any $x\in \mathbb R$? It seems easy, but I don't know why, I couldn't prove it.

I need help, thanks.


Since $f$ is differentiable at $0$, the limit

$$ \lim_{n\to\infty}\frac{f(2^{-n}x)-f(0)}{2^{-n}x} $$

must exist for all $x\ne0$ and be independent of $x$. The first term doesn't depend on $n$ due to the given functional equation, so we must have $f(0)=0$ and $f(2^{-n}x)/(2^{-n}x)=f'(0)$. The result follows by substituting $n=0$.

  • $\begingroup$ Great solution! $\endgroup$
    – Adar Hefer
    Feb 11 '13 at 21:06
  • 1
    $\begingroup$ Great. Note that $f(0)=f(0)/2$ so $f(0)=0$ from the equation. $\endgroup$
    – Julien
    Feb 11 '13 at 21:11

Essentially the same answer as joriki, with a slightlly different argument.

Note that $f(0)=f(0)/2$ by the functional equation so $f(0)=0$.

Now for all $x\neq 0$, an easy induction shows: $$ \frac{f(x)}{x}=\frac{f(x/2^n)}{x/2^n}=\frac{f(x/2^n)-f(0)}{x/2^n-0}. $$ Taking the limit as $n$ tends to $+\infty$, $$ \frac{f(x)}{x}=f'(0). $$


Using the Taylor formula with Lagrange reminder, there is a $c\in \mathbb R$ such that $f(x)=f(\frac {x}{2}+\frac {x}{2})=f(\frac {x}{2})+f'(c)\cdot\frac{x}{2}\implies f(x)=\frac {f(x)}{2}+f'(c)\cdot\frac{x}{2}\implies f(x)=f'(c)\cdot x$

Now we derive the last expression, we have $f'(x)=f'(c)$, for every $x\in \mathbb R$. then in particular $f'(0)=f'(c)$.

Then $f(x)=f'(0)\cdot x$


Let $g(x)=f(x)/x$. Then $f$ differentiable implies that $g$ is continuous.

Then $f(x)=xg(x)$ and hence $g(x/2)=g(x)$ for all $x\in \mathbb R$.

It follows that $g(0)=\lim_n g(x/2^n)=g(x)$ for all $x \in \mathbb R$.

I think that this solution is one of the above ones disguised..


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.