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Let $f$ be a function having a continuous derivative on $[0,1]$and with the property that $0

Can someone help me pls?

2 Answers 2

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From $f(x)-f(0)=f'(\zeta)(x-0)$ for $0<\zeta<1$, and $0

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    How does one prove that $\int_0^1 f^2 \le (\int_0^1 f)^2$? – 2017-01-07
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    This is obviously false for $f(x) = e^x$.. – 2017-01-07
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    @Open Ball Given that f(0)=0, but your function doesn't satisfy the condition(assumption). – 2017-01-07
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    @user123 alright, but MyGlasses puts it as if it's true for any function. Anyway, how do you prove it for a function satisfying $f(0) = 0$? – 2017-01-07
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    @OpenBall Sorry. I was absent. I think you are right and maybe this inequality is wrong. I post it in this _Stack Exchange_ now. – 2017-01-07
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We consider the differentiable function $G$ with $G(u)=(\int_{0}^{u}f(x)dx)^2-\int_{0}^{u}f(x)^3dx$ for $u\in [0,1]$.

It is $\ G'(u)=f(u)(2\int_{0}^{u}f(x)dx-f(u)^2),\ \forall u\in [0,1]$.

Now consider the differentiable function $F$ with $F(u)=2\int_{0}^{u}f(x)dx-f(u)^2$ for $u\in [0,1]$. Then:

$F'(u)=2f(u)(1-f'(u)),\ \forall u\in [0,1]$.

We have $f'(u)>0,\ \forall u \in[0,1]$, which means that $f$ is strictly increasing in $[0,1]$ and hence $f(u)\geq f(0)=0\ (1)$, when $u \in [0,1]$. It is also given that $f'(u)\leq 1,\ \forall u\in [0,1]$. Therefore

$F'(u)\geq 0,$ i.e. $F$ is increasing and so $F(u)\geq F(0)=0\ (2)$.

Since $G'(u)=f(u)F(u)$, $(1)$ and $(2)$ imply that $G'(u)\geq 0,\ \forall u \in [0,1]$, i.e. $G$ is increasing and so $G(1)\geq G(0)=0$. This proves our inequality.

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    @user123 We 're welcome. I 'll try to find an example for the equality. – 2017-01-07