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Let $f \colon \mathbb R \to \mathbb R$ be a continuous function. Let's define $ g(x) := \int_1^2 f(xt)dt. $

Prove that $g \equiv 0 \Rightarrow f \equiv 0$.

Well, I show you what I have done. First of all, I've noted that $g$ is differentiable for every $x \in \mathbb R \setminus\{0\}$. Indeed, $ g(x) = \frac{1}{x}\int_x^{2x} f(y)dy $ hence (by the fundamental theorem of calculus) $ \frac{dg}{dx} = -\frac{1}{x^2}\int_x^{2x} f(y)dy + \frac{1}{x}\left[2f(2x)-f(x)\right] = -\frac{1}{x}g(x) + \frac{1}{x}\left[2f(2x)-f(x)\right] $ So if $g(x)=0$ for every $x \in \mathbb R$ we must have $ \frac{dg}{dx} = 0 \Leftrightarrow 2f(2x)-f(x) = 0 $ i.e. $ f(x)=\frac{1}{2}f\left(\frac{1}{2}x\right) $ for every $x \in \mathbb R$. Is it correct what I've done so far? How would you conclude? I don't manage to prove $f \equiv 0$: I've just noted that $f(0)$ must be $0$, but no more...

Thanks in advance.

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    @ArturoMagidin Out of topic, but clicking the title with mouseball opens it in a new tab for me regardless of latex Title only or not. With Chrome.2012-07-23

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Yes, it's correct so far. Now show by induction that for each integer $n$, $f(x)=\frac 1{2^n}f\left(\frac x{2^n}\right).$ Since $f$ is continuous (it has already been used when you showed that $g$ was differentiable) on $[0,1]$, we have that $\frac x{2^n}\to 0$ when $n\to +\infty$ for a fixed $x$, hence $f(x)=\lim_{n\to +\infty}\frac 1{2^n}f\left(\frac x{2^n}\right)=\lim_{n\to +\infty}\frac 1{2^n}f\left(0\right)=0.$

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    Don't feel stupid! You have done the greatest part of the job, and an other eye saw how to conclude.2012-07-23