Let $f:[0, 1] \rightarrow \mathbb{R}$ be a differentiable function such that:
${\color{gray} \bullet} \;\; f(0)=0$ and
${\color{gray} \bullet} \;\; 0< f'(x) \leq 1$
Prove that:
$$\int_{0}^{x}f^3(t)\, {\rm d}t\leq \left ( \int_{0}^{x}f(t) \, {\rm d}t\right )^2, \;\; x \in [0, 1]$$
Solution
We consider the function $\displaystyle g(x)= \left ( \int_{0}^{x}f(t)\, {\rm d}t \right )^2 - \int_{0}^{x}f^3 (t)\, {\rm dt}, \; x \in [0, 1]$ which is clearly differentiable with a derivative given by:
$$g'(x)= 2f(x)\int_{0}^{x}f(t)\, {\rm d}t -f^3(x)= f(x)\left [ 2\int_{0}^{x}f(t)\, {\rm d}t-f^2(x) \right ]$$
However, $f'(x)>0$ meaning that $f$ is strictly increasing. Hence for $x>0$ we have that $f(x)>f(0)=0$. The function $\displaystyle h(x)= 2\int_{0}^{x}f(t)\, {\rm d}t -f^2(x)$ is obviously differentiable and $h'(x)= 2f(x)[1-f'(x)] \geq 0$. Hence $h$ is strictly increasing so $h(x) \geq h(0)=0$.
The result follows since $g \geq 0$ as $g$ is stricly increasing.
${\color{gray} \bullet} \;\; f(0)=0$ and
${\color{gray} \bullet} \;\; 0< f'(x) \leq 1$
Prove that:
$$\int_{0}^{x}f^3(t)\, {\rm d}t\leq \left ( \int_{0}^{x}f(t) \, {\rm d}t\right )^2, \;\; x \in [0, 1]$$
Solution
We consider the function $\displaystyle g(x)= \left ( \int_{0}^{x}f(t)\, {\rm d}t \right )^2 - \int_{0}^{x}f^3 (t)\, {\rm dt}, \; x \in [0, 1]$ which is clearly differentiable with a derivative given by:
$$g'(x)= 2f(x)\int_{0}^{x}f(t)\, {\rm d}t -f^3(x)= f(x)\left [ 2\int_{0}^{x}f(t)\, {\rm d}t-f^2(x) \right ]$$
However, $f'(x)>0$ meaning that $f$ is strictly increasing. Hence for $x>0$ we have that $f(x)>f(0)=0$. The function $\displaystyle h(x)= 2\int_{0}^{x}f(t)\, {\rm d}t -f^2(x)$ is obviously differentiable and $h'(x)= 2f(x)[1-f'(x)] \geq 0$. Hence $h$ is strictly increasing so $h(x) \geq h(0)=0$.
The result follows since $g \geq 0$ as $g$ is stricly increasing.
Hi Tolaso.
ReplyDeleteAdditionally, prove that :
\(\displaystyle{\int_{0}^{x}f^3(t)\,\mathrm{d}t\leq \left(\int_{0}^{x}f(t)\,\mathrm{d}t\right)^2\leq \dfrac{x^4}{4}\,\,,\forall\,x\in\left[0,1\right]}\) .
Hello Vaggelis,
ReplyDeletehere is a solution to the second inequality since the first one was proved above.
Since $f'(x)\leq 1, \; \forall x \in [0, 1]$ we have that the function $g(x)=f(x) -x , \; x \in [0, 1]$ is strictly decreasing since $g'(x)=f'(x)-1 \leq 0$. Now, $g(0)=0$ hence $g(x)\leq 0 \Leftrightarrow f(x)\leq x$.
Integrating we have that:
$$\int_0^x f(t)\, {\rm d}t \leq \frac{x^2}{2}$$
Squaring both sides we get the wanted result and the exercise is complete.
P.S: Sorry for not replying earlier.