Prove a function is bounded above I am considering the following question but really get lost. Can anyone give me a little hint? Many thanks!
Suppose two (Riemann) integrable functions $f$ and $g$, defined on $[0,+\infty)$, satisfy the following properties:
(i) $f(x), g(x)>0$ for all $x\in [0,+\infty)$;
(ii) There exist $M_0, M_1>0$ such that  $\int_0^x g(t)\mathrm dt\le M_0$ and 
$$f(x)\le M_1\bigg(2+\int_0^x g(t)f(t)\mathrm d t\bigg)\ln \bigg(2+\int_0^xg(t)f(t)\mathrm dt\bigg)$$
for all $x\ge 0$.
Prove:  $f$ is bounded above on $[0,+\infty)$.
 A: Let's introduce $I(x)=2+\int_0^x f(t)g(t)\, dt$. Then your inequality gives that
$$
\frac{I'(x)}{I(x)} \le M_1 g(x)\ln I(x) .
$$
Integrate from $a$ to $x$ to obtain that $\ln I(x)\le C(a) + M_1(\int_a^x g(t)\, dt) \ln I(x)$. If we now take $a$ so large that the integral becomes sufficiently small, then the claim follows from this.
A: Let
$$I(x)=2+\int_0^x f(s)g(s)\,\mathrm d s, \,\,x\in [0,+\infty).$$
Then it is easy to see that $I(x)$ is differentiable, $I'(x)=f(x)g(x)$, and $I(x)\ge 2$. Therefore, by
$$f(x)\le C_1\bigg(2+\int_0^xf(s)g(s)\,\mathrm ds\bigg)\ln\bigg(2+\int_0^xf(s)g(s)\,\mathrm ds\bigg)$$
we have 
$$f(x)\le C_1 \,I(x)\ln[I(x)].$$ 
Multiplying $g(x)$ on both sides and rearranging, we obtain 
$$\frac{I'(x)}{I(x)}\le C_1\,g(x)\,\ln[I(x)].$$
Since $C_0\ge \int_0^t g(x)\,\mathrm d x$ and $g(x)>0$ on $[0,+\infty)$, there must be some $a>0$ such that 
$$C_1\int_a^t g(x)\,\mathrm d x<\frac 12,\,\,\forall t>a.$$ 
Integrating both sides over $[a,t]$ ($t>a$), we arrive at
$$\ln[I(t)]-\ln [I(a)]\le C_1\int_a^tg(x)\,\ln[I(x)]\,\mathrm dx=C_1\ln[I(\xi_{t,a})]\cdot \int_a^t g(x)\,\mathrm d x\le \frac 12\ln[I(\xi_{t,a})],$$
where the middle equality follows from the Mean Value Theorem of integral ($\xi_{t,a}\in (a, t)$), an the last inequality is due to the condition. Then divide by $\ln [I(t)]$ on both sides to get 
$$1-\frac{\ln [I(a)]}{\ln[I(t)]}\le \frac 12\cdot\frac{\ln[I(\xi_{t,a})]}{\ln[I(t)]}\le \frac 12,$$
where the second inequality is due to the fact that $\ln[I(t)]\ge \ln[I(\xi_{t,a})]>0$ (note that $I(x)$ is increasing and  $a<\xi_{t,a}< t$). As a result, we can solve for all $t>a$ that
$$\ln[I(t)]\le 2\ln [I(a)].$$
which implies that  $I(t)\le [I(a)]^2$ on $(a,+\infty)$. Since $I(x)$ is continuous, we denote by $C_a$ the maximum of $I(x)$ on the compact set $[0, a]$ (clearly $C_a>0$). Then 
$$f(t)\le C_1\cdot \max\{[I(a)]^2, C_a\}\cdot \max\{2\ln [I(a)], \ln C_a\}$$ for all $t\in [0,+\infty)$. 
