# Proving that a function is constant from functional equation [duplicate]

$$a,b \in (0,1)$$ 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 $$a,b \in (0,1)$$ $$a+b=1$$

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

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

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

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

• Sorry. I corrected. – Gaboru Mar 13 at 11:24

$$\int_0^1 f(t)dt = lim_{a->0}\int_0^{ax}f(t)dt + lim_{a->0}\int_0^{bx}f(t)dt = \int_0^{x}f(t)dt$$.

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

• Sorry. I corrected. – Gaboru Mar 13 at 11:24