Let $f:[0, \infty) \to [0,\infty)$ be a differentiable function with $f'$ continuous. If $f(f(x))=x^2$, prove that $$\int_0^1 (f'(x))^2dx \geq \frac{30}{31}$$ without explicitly finding $f.$

Since we are not allowed to determine $f$, I tried to apply the Cauchy Schwarz inequality for integrals in some ways, but it was not strong enough. I also noticed that $f(x)=x^\sqrt{2}$ verifies the given identity and I tried to get a lower bound for the integral from $$0 \leq \int_0^1 (f'(x)-\sqrt{2}x^{\sqrt{2}-1})^2dx$$ but this wasn't good enough either.

I also managed to prove that $f$ is strictly increasing and $f(0)=0, f(1)=1.$

Since $f(f(x))=x^2$ we get that $f$ is injective. Since it is also continuous it must be strictly increasing or strictly decreasing. If the latter, let $f(0)=a \Rightarrow f(a)=0 \Rightarrow f(x)<f(a)=0$ for $x>a$. and since $f \geq 0$ we would get a contradiction.

Hence $f$ is strictly increasing and $f(0)=0$ since $f$ is obviously surjective. If $f(1)<1 \Rightarrow 1 = f(f(1))<f(1)$ and if $f(1)>1 \Rightarrow 1=f(f(1))>f(1)$, both of which are absurd. Thus $f(1)=1.$

  • $\begingroup$ Andrew, how did you manage to prove $f(0) = 0$, $f(1)=1$, and that $f$ is strictly increasing? $\endgroup$ – amsmath Apr 19 at 17:04
  • $\begingroup$ @amsmath I added the explanation. $\endgroup$ – AndrewC Apr 19 at 17:24
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    $\begingroup$ Nice reasoning, Andrew. $\endgroup$ – amsmath Apr 19 at 17:30
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    $\begingroup$ Have you tried applying Cauchy Scwartz to $f'(x)$ and $g(x)=x^{\sqrt{2}}$? $\endgroup$ – N. S. Apr 19 at 17:47
  • $\begingroup$ @N.S. Yes, but the inequality I obtained wasn't strong enough. Would you mind providing your proof? $\endgroup$ – AndrewC Apr 19 at 18:25

You have proved that $f(0) = 0$ and $f(1) = 1$. Hence, $1 = \int_0^1f'(x)\,dx\,\le\,\left(\int_0^11\,dx\right)^{1/2}\left(\int_0^1(f'(x))^2\,dx\right)^{1/2}$ and thus $\int_0^1(f'(x))^2\,dx\ge 1 > \frac{30}{31}$.

  • $\begingroup$ Thank you! Sometimes we don't see what's right in front of us. $\endgroup$ – AndrewC Apr 19 at 18:24
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    $\begingroup$ @AndrewC Yeah, that's what I thought, too. ;o) Maybe your teacher had another reasoning in mind, which gives the slightly worse estimate 30/31. $\endgroup$ – amsmath Apr 19 at 18:32

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