# If $g(x) := \int_1^2 f(xt)dt \equiv 0$ then $f \equiv 0$

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...

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Please don't make the titles consist of nothing but LaTeX code. –  Arturo Magidin Jul 23 '12 at 20:09
@ArturoMagidin: I apologize, I didn't know. Sorry. –  Romeo Jul 23 '12 at 20:11
No problem; it's just that it can cause some side-effects. E.g., if a title consists of nothing but LaTeX code, I cannot right-click and ask that the post be opened on a separate tab/window. –  Arturo Magidin Jul 23 '12 at 20:12
@ArturoMagidin It is also a bit ugly when I browse the question on my smartphone app in the train :) –  Bernhard Jul 23 '12 at 20:19
@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. –  Jean-Sébastien Jul 23 '12 at 20:21

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 $$f(x)=\lim_{n\to +\infty}\frac 1{2^n}f\left(\frac x{2^n}\right)=\lim_{n\to +\infty}\frac 1{2^n}f\left(0\right)=0.$$
Thanks for your answer. I showed what you suggest to me, but I still don't see how to finish. I mean: we are getting closer and closer to $0$ and $f$ is continuous but I don't manage how to write down a correct proof... Thanks for your kindness. –  Romeo Jul 23 '12 at 20:10