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Let $f$ be a continuous function whose domain includes $[0,1]$, such that $0 \le f(x) \le 1$ for all $x \in [0,1]$, and such that $f(f(x)) = 1$ for all $x \in [0,1]$. Prove that $\int_0^1 f(x)\,dx > \frac34$.

Here's all that I have, from the Mean Value Theorem, we have some $c\in[0,1]$, and $a$, such that $$a=f(c)=\int_0^1 f(x)dx.$$ By the Extreme Value Theorem, there exist some $m$, $n\in[0,1]$ such that $$f(m)\ge f(x)\ge f(n).$$ I'm stuck here. Is this the right approach? Where do I go from here?

I also got to know what the very fact that $f(f(x))=1$ shows that there is some $x$ such that $f(x)=1$ because the range of $f(x)$ is the domain of $f(x)$ (which I'm still trying to understand; I know what it means, I'm just trying to take it in).

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    $\begingroup$ To help you play with the ideas, here's an example of a family of functions nestled at the "minimum" of that inequality: $$\begin{cases} 1 & \frac{1}{2} \leq x \leq 1 \\ \frac{1+(2x)^n}{2} & 0 \leq x < \frac{1}{2} \\ \end{cases} $$ In the limit $n\to\infty$ one retrieves $\frac{3}{4}$ but the limiting function is not continuous. $\endgroup$ – Ninad Munshi Jul 9 '20 at 3:20
  • $\begingroup$ Can you not say that since $f(f(x))=1$ then $f(x)=f^{-1}(1)$ and so: $$\int_0^1f(x)dx=\int_0^1f^{-1}(1)dx=f^{-1}(1)$$ $\endgroup$ – Henry Lee Jul 9 '20 at 14:49
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$f(1)=f(f(f(1)))=(f\circ f) (f(1))=1$

$f([0,1])=[a,1]$ for some $a >0$ since the image is connected hence an interval ending at $1$ and compact hence the interval is closed while obviously $f([a,1])=1$ so $a >0$

But now on $[0,a], f(x) \ge a$ so $\int_0^1f(x)dx=\int_0^af(x)dx+\int_a^1f(x)dx \ge a^2+1-a \ge 3/4$ and we cannot have equality since then $a=1/2$ and because $f(1/2)=1, f(x) \to 1, x \to 1/2, x<1/2$ so $f$ cannot be identically $1/2$ on $[0,1/2)$ and it is bigger on at least a small interval near $1/2$

Note that by choosing that interval very small and making $f$ linear there (and $1/2$ before, $1$ after) we can get the integral $3/4+\epsilon$ so the result is sharp.

Done!

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COMMENT.- This is just to expose a way more geometric than analytical.

The only fixed point of $f$ is $1$ because if not then $f(x)=x\Rightarrow f(x)=1$ contradiction. Then $f(0)=a\gt0$ and the graph of $f$ is above the diagonal $y = x$.

There is an automatic way of plotting two (correlated) points of the function: to each point $(x, y)$, with $x\gt y$, corresponds a point $(y, 1)$.

I hope the attached figure it is enough to show the reasoning in this comment (the shadow region corresponds to the obvious hyperbola).

enter image description here

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