Let $f:\mathbb R \to \mathbb R$ be a $C^1$ function and suppose that $0<f'(t)\leq 1$ on $[0,1]$ and $f(0)=0$. Does the inequality $$\left(\int_0^1 f(t)dt\right)^2 \leq \int_0^1f^3(t)dt$$ hold? By $f^3(t)$ I mean $(f\circ f\circ f)(t)$.

This question appeared on a test I took but there was no information about what $f^3(t)$ means. If I suppose that $f^3(t)=(f(t))^3$ then the inequality is true and very simple to prove, but as far as I know $f^3$ is the composition of $f$ three times.

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    $\begingroup$ Your question is answered. However, I suspect "$f^3(x)$", much like "$\sin^2(\theta)$", means the indicated power of the output of the function. The composition meaning is still odd enough that CS papers define it when they use it (and frequently use variations, like $\mathrm{lg}_2$ to mean $\mathrm{lg} \circ \mathrm{lg}$). $\endgroup$ – Eric Towers Oct 8 '17 at 17:45

No, it does not hold. Take for example $f(t)=at$ with $0<a<\frac{1}{2}$: $$\left(\frac{a}{2}\right)^2=\left(\int_0^1 f(t)dt\right)^2 > \int_0^1 (f\circ f\circ f)(t)dt=\frac{a^3}{2}.$$

  • $\begingroup$ Oh, that was fast. Thank you very much. $\endgroup$ – Hugocito Oct 8 '17 at 15:34

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