5
$\begingroup$

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ a $C^1$ class function with $0 < f'(t) \leq 1$ for all $t \in [0,1]$ and $f(0)=0$ prove that $$\left(\int_{0}^{1} f(t) dt\right)^2 \geq \int_{0}^{1} f^3(t) dt, $$ where $f^3=f\circ f\circ f$.

My attempt: So I tried to use Cauchy-Schwarz, Integration by Parts, Mean value Theorem for Integrals and wasn't unable to connected the dots. Then after a while I think if I use the identity function I will have a contradition. So now I don't know if this statement is right or wrong. I appreciate any help. Thank you

$\endgroup$
  • 1
    $\begingroup$ With the identity function the equality holds... It is $1/4$ in both sides. $\endgroup$ – DiegoMath Jun 12 '18 at 18:00
  • 3
    $\begingroup$ in this case the problem is wrong. But is interesting think in the problem with cubic power function. $\endgroup$ – DiegoMath Jun 12 '18 at 18:09
1
$\begingroup$

I think the question was not about function composition because then $f(x)=x$ is a counterexaple. It does hold when you assume it is a third power however:

Notice that $f$ is increasing in $[0,1]$ (since $f'>0$) and $f(0)=0$ thus $f(x)\geq{}0,\forall{}x\in{}[0,1]$. Set $F(x)=\int_0^xf(t)dt$ then if $h(x)=F^2(x)-\int_0^xf^3(t)dt$ we have $h'(x)=2f(x)F(x)-f^3(x)=f(x)(2F(x)-f^2(x))$. Now set $g(x)=2F(x)-f^2(x)$, then $g(0)=0$ and $g'(x)=2f(x)-2f(x)f'(x)=2f(x)(1-f'(x))\geq{}0,\forall{}x\in{}[0,1]$ and thus $g(x)\geq{}0,\forall{}x\in{}[0,1]$. Now $h'(x)=f(x)g(x)\geq{0},\forall{}x\in{}[0,1]$ and thus since $h(0)=0$ and $h$ is increasing we have $h(1)\geq{}h(0)=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.