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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.

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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 $n=0$.

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

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

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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..

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