Let $f\in L^1([0.1])$, non-negative such that $f, f^2$ and $f^3$ have same $L^1([0,1])$ norm. Then, I want to show for a.e. $x$, either $f=0$ or $f=1$.
If $f\in[0,1]$, then just by comparing directly 1st and 2nd powers, we have $f=0$ or $1$ a.e. But if $f>1$, I have no idea on continuing and how to use 3rd power.
On
The existing answer, pointing out that $\int f(f-1)^2=0$, is very cool and very clever. But it depends strongly on the fact that the exponents are $1$, $2$, and $3$; in particular it doesn't say anything about the obvious generalization of the result in question. Here's a different argument that's maybe less of a "trick", and which in any case does generalize:
If we say $g=f^{1/2}$ and $h=f^{3/2}$ then we have $$\int gh=||g||_2||h||_2.$$ Hence $h$ must be a scalar multiple of $g$, which implies that $f=c\chi_A$, and now of course it follows that $c=1$.
There's a generalization that "must" be true if the original result is true; this argument does suffice for the generalization:
If $f\ge0$, $1<\alpha<\beta$ and $\int f=\int f^\alpha=\int f^\beta$ then $f=\chi_A$.
Sketch: Solving a few equations shows that there exist $a,b>0$ and $p\in(1,\infty)$ such that $a+b=\alpha$, $ap=1$ and $bp'=\beta$. Now if $g=f^a$ and $h=f^b$ we have $$\int gh=||g||_p||h||_{p'}.$$The condition for equality in Holder's inequality shows that...
You have $$ \int f^3 - 2f^2 + f = 0$$ which can be arranged to find $$\int f (f-1)^2 = 0.$$ Since $f(f-1)^2 \ge 0$ it follows that $f(f-1)^2 = 0$ almost everywhere and thus $f=0$ or $1$ almost everywhere.