analysis - $f:[0,1]rightarrow [0,1]$ and $fcirc f$ are not identically zero but $f(f(f(x)))=0$ for all $xin [0,1]$. Does $f$ exist?

Let $f:[0,1]\rightarrow [0,1]$ be a non-identically zero function such that $f\circ f$ is not identically zero but $f(f(f(x)))=0$ for all $x\in [0,1]$. Does there exist such a function?
I am thinking that such function does not exist but don't know how to start.

If $f$ is continuous, then we know it has a fixed point, say $x_0$. If $x_0\neq 0$, this contradicts the assumption that $f(f(f(x)))=0$. Otherwise it does not give us more information about $f$ (even if $f$ assumed to be continuous).

Please give some hints to solve.

Thanks in advance.

Answer

Hint: Split the domain into three intervals of equal length and construct a piecewise function (there does exist such a function that is continuous).