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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 ?

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

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