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One of my pals asked me to look at this question

Let $f: [0,1] \to \mathbb{R}$ be differentiable. Suppose that $f(0) = 0$ and 0 < f'(x) < 1 for all $x \in (0, 1)$, where f'(x) is the derivative of $f$. Prove that

$ \left(\int_{0}^{1}f(x) \ dx \right)^{2} \geq \int_{0}^{1}(f(x))^{3} \ dx$

I don't really have a clue as to where to start. Any ideas will be appreciated.

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    f(x) \approx f(a)+f'(a)(x-a) < f(a)+(x-a) for $a \in (0,1)$.2011-01-22

1 Answers 1

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This Problem is given in the book

  • Problems in Mathematical Analysis - III by Kaczor and Nowak

The solution is as follows:

Set $F(t)= \biggl(\int\limits_{0}^{t} f(x) \ \text{dx}\biggr)^{2} - \int\limits_{0}^{t} (f(x))^{3} \ \text{dx}, \quad t \in [0,1]$

Then F'(t)=f(t) \cdot \biggl(2 \int\limits_{0}^{t} f(x) \ \text{dx} - (f(t))^{2}\biggr) and if $G(t)= 2 \int\limits_{0}^{t} f(x) \ \text{dx} - (f(t))^{2}$, then G'(t)=2f(t) \cdot (1-f'(t)) \geq 0. Consequently, $G(t) \geq G(0)=0$, which gives F'(t) \geq 0. So, $F(t) \geq 0$,, and in particular $F(1) \geq 0$.

Moreover if$F(1)=0$, then $F(t)=0$ for $t \in [0,1]$ and therfore F'(t)=f(t)G(t)=0. This, in turn, implies G'(t)=2f(t) \cdot (1-f'(t))=0 and 1-f'(t)=0 for $t \in (0,1)$.