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This question already has an answer here:

$a,b \in (0,1)$ and $f:[0,1] \to \mathbb R$ is continuous functions s.t. $ \int_0^x f(x)dx=\int_0^{ax}f(x)dx+ \int_0^{bx}f(x)dx$ . Knowing that $a+b=1$, we have to prove that $f$ is constant.

Using the derivative,we get: $f(x)=af(ax)+bf(bx)$

I managed to do it for the case $a=b=1/2$, but I don't know how to make it with $a,b$ arbitrary and $a,b \in (0,1)$ $a+b=1$

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marked as duplicate by John Omielan, John Hughes, Lord Shark the Unknown, Lee David Chung Lin, Cesareo Mar 14 at 9:30

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I think it should read:

$\int_0^1 f(t)dt=\int_0^{ax}f(t)dt+ \int_0^{bx}f(t)dt.$

Then $ f(x)=af(ax)+bf(bx)$ is not correct. Using derivatives, you get $0=af(ax)+bf(bx)$, since

$\frac{d}{dx}\int_0^1 f(t)dt=0.$

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  • $\begingroup$ Sorry. I corrected. $\endgroup$ – Gaboru Mar 13 at 11:24
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$\int_0^1 f(t)dt = lim_{a->0}\int_0^{ax}f(t)dt + lim_{a->0}\int_0^{bx}f(t)dt = \int_0^{x}f(t)dt$.

Taking derivatives of both sides, 0 = f(x).

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  • $\begingroup$ Sorry. I corrected. $\endgroup$ – Gaboru Mar 13 at 11:24

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