Prove that $\int_0^1 (f'(x))^2dx \geq \frac{30}{31}$ when $f(f(x))=x^2$

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 $$f(0)=a \Rightarrow f(a)=0 \Rightarrow f(x) for $$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 $$f(1)<1 \Rightarrow 1 = f(f(1)) and if $$f(1)>1 \Rightarrow 1=f(f(1))>f(1)$$, both of which are absurd. Thus $$f(1)=1.$$

• Andrew, how did you manage to prove $f(0) = 0$, $f(1)=1$, and that $f$ is strictly increasing? – amsmath Apr 19 at 17:04
• @amsmath I added the explanation. – AndrewC Apr 19 at 17:24
• Nice reasoning, Andrew. – amsmath Apr 19 at 17:30
• Have you tried applying Cauchy Scwartz to $f'(x)$ and $g(x)=x^{\sqrt{2}}$? – N. S. Apr 19 at 17:47
• @N.S. Yes, but the inequality I obtained wasn't strong enough. Would you mind providing your proof? – 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 $$\int_0^1(f'(x))^2\,dx\ge 1 > \frac{30}{31}$$.