TITLE: Solve these functional equations: $\int_0^1\!{f(x)^2\, \mathrm{dx}}= \int_0^1\!{f(x)^3\, \mathrm{dx}}= \int_0^1\!{f(x)^4\, \mathrm{dx}}$
QUESTION [19 upvotes]: Let $f : [0,1] \to \mathbb{R}$ be a continuous function such that
  $$\int_0^1\!{f(x)^2\, \mathrm{dx}}= \int_0^1\!{f(x)^3\, \mathrm{dx}}= \int_0^1\!{f(x)^4\, \mathrm{dx}}$$
  Determine all such functions $f$.

So far, I've managed to show that $\int_0^1\!{f(x)^t\, \mathrm{dx}}$ is constant for $t \geq 2$. Help would be appreciated.

REPLY [4 votes]: By Holder's Inequality,
$$\left(\int_{[0,1]}ff^2dx\right)^2\leq \int_{[0,1]}f^2dx \int_{[0,1]}f^4 dx=\left(\int_{[0,1]}f^3dx\right)^2$$
Equality occurs in this inequality iff $f=af^2$ for some $a\in\mathbb{R}$ i.e., Linearly dependence of $f,f^2$..
$f=af^2$ implies $f^2=af^3$. Considering the integrals and their equality, we see that $f=f^2$.
Which in turn implies that $f(x)=0$ or $1$ for $x\in [0,1]$