Question

Let g (x), $x\ge 0$, be a non-negative continuous function and let $F\left(x\right)={\int }_0^{x}f\left(t\right)dt$, $x\ge 0$. If for some c > 0, f(x) $\le$ c F(x) for all $x\ge 0$, then

• A. $f\left(x\right)=0,\forall x\le 1$
• B. $f\left(x\right)=0,\forall x\ge 1$
• C. $f\left(x\right)=1,\forall x\ge 1$
• D. $f\left(x\right)=-1,\forall x\ge 1$

Answer 1

Answer: (A)

$f\left(x\right)=0,\forall x\le 1$

$f\left(x\right)\ge 0$ for $x\ge 0$

$F\left(x\right)={\int }_0^{x}f\left(t\right)dt$ for all $x\ge 0$

As $c<0,f\left(x\right)\le cF\left(x\right)$ for $x\ge 0$

$f\left(x\right)=0\Rightarrow F^{\prime }\left(x\right)=f\left(x\right)$ for $x\ge 0$

Since $f\left(x\right)\le cF\left(x\right)$ since $F^{\prime }\left(x\right)=f\left(x\right)$

$\Rightarrow e^{-cx}{F}^{\prime }\left(x\right)-c{e}^{-cx}F\left(x\right)\le 0$ by multiplying by $e^{-cx}$ where $e^{-cx}>0$

When $x\to \infty ,e^{-cx}\to 0$

$\Rightarrow \frac{d}{dx}\left(e^{-cx}F\left(x\right)\right)\le 0$ for all $x\ge 0$

$\Rightarrow e^{-cx}F\left(x\right)$ is a decreasing function for $x\ge 0$

Let $g\left(x\right)=e^{-cx}F\left(x\right)$ be a decreasing function.

$\Rightarrow g\left(0\right)\ge g\left(x\right)$ for $x\ge 0$

$\Rightarrow 1\times F\left(0\right)\ge e^{-cx}F\left(x\right)$

Given:$F\left(0\right)={\int }_0^{0}F\left(t\right)dt=0$

$\Rightarrow e^{-cx}F\left(x\right)\le 0$

$\Rightarrow F\left(x\right)\le 0$ and $e^{-cx}\ge 0$

$\Rightarrow c>0$

$\Rightarrow cF\left(x\right)\le 0$

$\Rightarrow f\left(x\right)\le cF\left(x\right)\le 0$

Since $f\left(x\right)$ is non-negative $f\left(x\right)\le 0\Rightarrow f\left(x\right)=0$