# Range of continuous function $f:[0,1] \rightarrow [0,1]$ satisfying $f(f(x))=1$.

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

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