differential inequality of continuous functions Let $u:[0,+\infty)\to (0,+\infty)$ be a continuous function such that $\int_0^\infty u(x)dx<\infty$. Suppose there exist $a,b>0$ such that $\frac{du}{dx}\leq u(a+bu)$. Prove that $\lim_{x\to\infty} u(x)=0$. How to solve this problem?
 A: By contradiction, assume there exists some $\epsilon>0$ such that $u(x_n)>\epsilon$ for some sequence $\{x_n\}$ such that $x_{n+1}-x_n>\tau$ for some $\tau>0$ we will provide later. $u$ is positive, and by the inequality we really only bound the slope from above, but it can drop with unbounded slope. We are interested then, in bounding from below the area beneath the graph of $u$ on the intervals $[x_n-\tau,x_n]$.
Due to the upper bound of the slope and the fact that $u(x_n)>\epsilon$, we have $u(x_n-\tau)>\epsilon - \tau\epsilon(a+b\epsilon)$. This is the crucial observation. This is true because on the interval $u'$ is bounded by $u(a+bu)$, and if it was less than this value, $u$ couldn't increase over the distance $\tau$ to rise to $\epsilon$ without violating the bound. You can prove this formally by using any ODE comparison theorem with the bound and $u(0)\leq\epsilon - \tau\epsilon(a+b\epsilon)$ and proving $u(\tau)<\epsilon$.
We pick $\tau$ such that $\epsilon - \tau\epsilon(a+b\epsilon)=\epsilon/2\implies \tau = 1/2(a+b\epsilon)>0$. Now we have that the graph of our function lies above $\epsilon/2$ for a distance of $\tau$ on an infinite sequence of (almost) disjoint intervals, therefore
\begin{align}
\int_0^\infty u(x)\ dx &\geq \sum_n \int_{x_n-\tau}^{x_n} u(x)\ dx \\
&> \sum_n \int_{x_n-\tau}^{x_n} \epsilon/2 \ dx \\
&= \sum_n \tau\epsilon/2 \ dx \\
&= \infty
\end{align}
which is a contradiction.
