Continuous function $f:[0,1] \rightarrow [0,1]$ satisfies $${f(f(x))=1},{\forall x \in [0,1]}.$$ Find range of $$\int_{0}^{1} f(x) dx.$$

I found that $f(x) \in [0,1]$ so $$f(f(f(x)))=f(1)$$ and $$f(1)=1$$. And also maximum is $1$, when $f(x)=1$ . But I couldn't imagine another case and minimum case.....

My opinion:

Thm.Continuous function $f$ that $f:[0,1] \rightarrow [0,1]$ satisfied $${f(f(x))=1},{\forall x \in [0,1]}$$ Than $$f(x)=1$$.

Proof) $f:[0,1] \rightarrow [0,1]$ is continuous, So $\delta_h (\epsilon)>0$ is exists such that $$\mid {x-h} \mid < \delta_h (\epsilon) \Rightarrow \mid {f(x)-f(h)} \mid <\epsilon$$ $f(f(h))\in [0,1]$ and there are $\delta_{f(h)}(\epsilon)>0$ such that $$\mid x-f(h)\mid <\delta_{f(h)}(\epsilon) \Rightarrow\mid f(x)-f(f(h))\mid <\epsilon$$ So $$\mid f(x)-f(f(h))\mid = \mid f(x) -1 \mid <\epsilon$$

I found out it that makes my opinion wrong, thanks for help

  • $\begingroup$ @mfl $f(t) \in [0,1]$ so f(x)=t than f(f(f(x)))=f(f(t))=f(1) $\endgroup$
    – user366725
    Feb 12 '19 at 17:23
  • $\begingroup$ Suppose $f$ is a polynomial. $f(f(x)) = 1$, does that mean $f$ is just a constant (zero degree)? $\endgroup$ Feb 12 '19 at 17:27
  • $\begingroup$ @Expikx I agree about that, but I can't find other case for nonpolynomial continuous fiuction $\endgroup$
    – user366725
    Feb 12 '19 at 17:32

Let $m$ be the minimum of $f$. Then $f([0,1])=[m,1]$, so $f=1$ on $[m,1]$; and $f: [0,m] \rightarrow [m,1]$ is continuous surjective.

Thus, the integral of $f$ is more than $(1-m)*1+m*m=1-m+m^2$ and less than $(1-m)*1+m*1=1$.

Any value between can be reached by taking $f(x)=m+c_tx^t$ between $0$ and $m$ where $c_t=(1-m)/m^t$, for $t > 0$, and the integral of $f$ is $1-m+m^2+((1-m)m)/(1+t)=1-\frac{t}{1+t}m(1-m)$.

So the possible integrals of $f$ range from $3/4$ to $1$, $1$ included, $3/4$ excluded.


Well, as $f$ is a continuous function, it follows that Im$(f[0,1]) = [a,b]$ for some $a,b$, and as $f(f(x)) = 1$ it follows that there exists an $x$ s.t. $f(x)$ is 1, so $b$ must be 1. So Im$(f) = [a,1]$ and furthermore, for each $x \in [a,1]$, it follws that $f(x) = 1$ and for each $x \le a$, that $f(x) \ge a$.

So for some $a \in [0,1]$, it follows that $f$ must satisfy

$$\int_0^1 f(x) dx \geq (a \times a) \ + \ (1-a) \times 1 \ = \ 1-a+a^2 \doteq h(a) $$

However, $h(a) \geq \frac{3}{4}$ and $h(1/2)=3/4$, so $\int_0^1 f(x)dx$ must be at least $\frac{3}{4}$.

And this bound is "almost" tight: for each such $a$ and $\epsilon >0$ there is indeed an $g$ that satisfies $\in_0^1 g(x) dx \le 1-a+a^2$ and $g(g(x))=1$; set $g(x) = a$ for all $x<a-\epsilon$; $g(x) = a+\frac{(x-a+\epsilon)}{\epsilon}$ for all $x \in [a-\epsilon, a]$ [i.e., $f(x)$ ramps up rapidly from $a$ to 1 in the interval $[a-\epsilon, a]$ ], and $g(x)=1$ for all $x \in [a,1]$.

So $\int f dx \in (\frac{3}{4}, 1]$.


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