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.