Prove that $x$ is bounded.

Question

Suppose that $x: \mathbb R^+ \to \mathbb R^+$ and that

$$ \frac{d x}{d t} \leq x^2 $$ $$\int_0^{\infty} x(s) d s <\infty $$

Post edit:( from the valuable comments)

Prove that $x$ is bounded. (Note: this is not obvious since the integral constraint does not eliminate singularities like $1 / \sqrt{t}$.

As a hint I am thinking $x^2$ as $x(t) x(t)$ and thinking about how to solve $x^{\prime}=a(t) x(t)$ but didn't get any help so far.

Answer 1

Suppose $x:\mathbb{R^+}\rightarrow\mathbb{R^+}$ is of class $C^1$. Since $x\left(t\right)>0$ we have $$ \frac{x'(t)}{x(t)}\leq{x(t)} $$ for all $t\in\mathbb{R}$.

Then $$ \int_{0}^{s}\frac{x'(t)}{x(t)}dt \leq \int_{0}^{s}{x(t)}dt \leq \int_{0}^{\infty}{x(t)}dt = c <\infty $$ for all $s>0$. Since the integral on the left side equals $\ln{x\left(s\right)}$ we have $x\left(s\right)\leq{e^c}$ so that $x$ is bounded.

Answer 2

This is answer to the original formulation of the question, where $x:\mathbb R \to \mathbb R^+$.

In fact $x$ does not need to be bounded if indeed defined for all $t\in \mathbb R$. To see it, consider the function $$ x(t) = \begin{cases} t^2, & \text{ if } t\leq 0 \\ 0,& \text{ if } t \geq 0. \end{cases} $$ Its derivative is negative on $(-\infty,0)$ and zero on $[0,\infty)$, so the inequality $x'\leq x^2$ is satisfied trivially. Likewise the integral condition is satisfied trivially as $\int_0^\infty x(s)ds = 0$.