# Given $f(x)$ is continuous on $[0,1]$ and $f(f(x))=1$ for $x\in[0,1]$. Prove that $\int_0^1 f(x)\,dx > \frac34$.

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).

• 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. – Ninad Munshi Jul 9 '20 at 3:20
• Can you not say that since $f(f(x))=1$ then $f(x)=f^{-1}(1)$ and so: $$\int_0^1f(x)dx=\int_0^1f^{-1}(1)dx=f^{-1}(1)$$ – Henry Lee Jul 9 '20 at 14:49

$$f(1)=f(f(f(1)))=(f\circ f) (f(1))=1$$

$$f([0,1])=[a,1]$$ 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!

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).