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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.

NebulousReveal
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picakhu
  • 4,994
  • Seems like you have enough info to show $f \le x$; seems like this should also be strict on $(0,1]$. Also $f = |f|$ so $f^2 = |f|^2 \ge |f|^3 = f^3$. Using this and playing around with $\int |f| \ge |\int f|$ you should be able to come up with something. – solbap Jan 22 '11 at 19:31
  • I know one way to solve this problem, but it's really not pretty. You start with a slight generalization of http://math.stackexchange.com/questions/18548/combinatorial-reasoning-for-a-famous-identity that will allow you to prove an analogous statement for sequences, then prove it for Riemann sums. But I think there is a nicer solution; this is a Putnam problem, I think. – Qiaochu Yuan Jan 22 '11 at 19:34
  • I am a bit confused that this is a putnam problem, thats a bit too much for an university advanced placement question don't you think?

    http://www.science.nus.edu.sg/undergraduates/MA1102R%20APC%20Test%20Sample.pdf

     – picakhu Jan 22 '11 at 19:43
  • $f(x) \approx f(a)+f'(a)(x-a) < f(a)+(x-a)$ for $a \in (0,1)$. – NebulousReveal Jan 22 '11 at 19:48

1 Answers1

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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 $t \in (0,1)$.