# integration of iterates vs. integration of powers

Suppose $$f:[0,1]\to [0,1]$$ is continuous and has the property $$\forall n\in\Bbb N\;\;\int_0^1f^n(x)\;dx=\int_0^1(f(x))^n\;dx$$ where $$f^n=f\circ\cdots\circ f=f\circ f^{n-1}$$ is the $$n^{\text{th}}$$ composition of $$f$$ on itself ; i.e. $$f^1(x)=f(x), f^2(x)=f(f(x)),\dots$$

Must $$f$$ be the zero function (or the constant function given by $$f(x)=1$$) ? Or might $$f$$ be a non-constant function ?

• The question is not well-defined unless you mean $f\colon [0,1]\to[0,1]$, since otherwise you cannot compose $f$ with itself. Dec 7 '19 at 5:14
• to avoid confusion, I suggest the notation $f^{\circ n}=f\circ f\circ \cdots \circ f=f\circ f^{\circ (n-1)}$, and $f^n=f\cdot f\cdots f=f\cdot f^{n-1}$ Dec 7 '19 at 5:29
• That's a very bad notation, $f^n$ denotes always what you wrote as $f^{\circ n}$. Dec 7 '19 at 22:46
• standard notation is best Dec 8 '19 at 1:30