A continuous function integral inequality

Question

Let $m$ be a positive integer. $f\colon[0,\infty)\to[0,\infty)$ is a continuous function such that $f(f(x))=x^m,\forall x\in[0,\infty)$. Show that

$$\int_0^1f^2(x)\,dx\ge\frac{2m-1}{m^2+6m-3}$$

Answer 1

lemma1:

$$f(f(x))=x^a,a\in N^{+}$$ then $f(x)$ is increasing

poof: otherwise we Assume there exsit $u\neq v$,such $f(u)=f(v)$,then $$u^a=f(f(u))=f(f(v))=v^a\Longrightarrow u=v$$ contradiction.

lemma 2:

$$f(f(x))=x^a,a\in N^{+}$$ then $$f(0)=0$$

poof: let $f(0)=b$,then $f(f(0))=0\Longrightarrow f(b)=0$

and since

$$f(f(b))=b^m\Longrightarrow b=b^a\Longrightarrow b=0 or b=1$$ if $f(0)=1$,and $f(f(x))=x^m\Longrightarrow f(1)=1$,and $f(x)$ is increasing

contradiction. since $$f(f(x))=x^a\Longrightarrow f(x)=f^{-1}(x^a)$$ so $$\int_{0}^{1}x^{a-1}f(x)dx=\int_{0}^{1}x^{a-1}f^{-1}{x^a}dx=\dfrac{1}{a}\int_{0}^{1} t^{\frac{a-1}{a}}f^{-1}(t)t^{\frac{1}{a}-1}dt=\dfrac{1}{a}\int_{0}^{1}f^{-1}(x)dx$$ so $$\int_{0}^{1}f^{-1}(x)dx=\int_{0}^{1}ax^{a-1}f(x)dx$$ since $f(x)$ is increasing and $f(0)=0$ ,so by use Young inequality

http://en.wikipedia.org/wiki/Young's_inequality

we have $$\int_{0}^{1}f(x)dx+\int_{0}^{1}f^{-1}(x)dx\ge 1\Longrightarrow \int_{0}^{1}f(x)dx+\int_{0}^{1}ax^{a-1}f(x)dx\ge 1$$ and use Cauchy-Schwarz inequality,we have \begin{align*} 1&\le\int_{0}^{1}f(x)(1+ax^{a-1})dx\le\left(\int_{0}^{1}f^2(x)dx\int_{0}^{1}(1+ax^{a-1})^2dx\right)^{\frac{1}{2}}\\ &=\left(\int_{0}^{1}f^2(x)dx\right)^{\frac{1}{2}}\left(\int_{0}^{1}(1+2ax^{a-1}+a^2x^{2a-2})dx\right)^{\frac{1}{2}}\\ &=\left(\int_{0}^{1}f^2(x)dx\right)^{\frac{1}{2}}\left(1+2+\dfrac{a^2}{2a-1}\right)^{\frac{1}{2}}= =\left(\int_{0}^{1}f^2(x)dx\right)^{\frac{1}{2}}\left(\dfrac{a^2+6a-3}{2a-1}\right)^{\frac{1}{2}} \end{align*} so $$\int_{0}^{1}f^2(x)dx\ge\dfrac{2a-1}{a^2+6a-3}$$