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.


Note that $$\int_0^1 f(x)^2 (1 - f(x))^2 \; dx = \int_0^1 \left(f(x)^2 - 2 f(x)^3 + f(x)^4\right) \; dx = 0$$ so $f(x)(1-f(x))$ must always be $0$. Since $f$ is continuous, the only solutions are $f(x)=0$ and $f(x)=1$.


Calculate the difference of first and second integral then second and third.

$$\int_0^1f^2(x)-f^3(x)dx=0$$ $$\int_0^1f^3(x)-f^4(x)dx=0$$

Subtract both equations: $$\int_0^1f^2(x)-2f^3(x)+f^4(x)dx=0$$ $$\int_0^1f^2(x)\left[1-2f(x)+f^2(x)\right]dx=0$$ $$\int_0^1f^2(x)(1-f(x))^2dx=0$$

Look at the integrand it consists of the product of two squares. It can only become $0$ if $f(x)=0$ or $f(x)=1$.

  • 4
    $\begingroup$ Or some indicator function that jumps back and forth, but continuity precludes that. $\endgroup$
    – Teepeemm
    Apr 27 '16 at 1:13

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 $x\in [0,1]$

  • $\begingroup$ @BarryCipra : Yes.. That is now edited $\endgroup$
    – user311526
    May 7 '16 at 12:57

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