Let $f: \left[0,\infty\right)$ $\rightarrow\left[0,\infty\right)$ be a continuous function so that $f(f(x))=x^2, \forall x\in \left[0,\infty\right).$ Show that: $$\int_{0}^{1}f^2\left(x\right)dx\geq\frac{3}{13} $$ I noticed $f$ is injective, so it is strictly monotone. Also, $f^2(x)=f(x^2), \forall x \geq 0$, but by substituting $x^2$ I'm getting nowhere. Do you have any ideas?
Asked
Active
Viewed 96 times
1
-
@Teddy38, The issue is that $f(x) = x^{\sqrt{2}}$ is not the only solution. – Sangchul Lee Mar 01 '18 at 21:09
-
@ Sangchul Lee Ok, fair enough. – Teddy38 Mar 02 '18 at 21:02