Inequality of Caratheodory function and its primitive Let $f: \Omega \times \mathbb{R} \to \mathbb{R}$ be a Caratheodory function. Set $F(x,t):= \int_{0}^{t} f(x,s)ds$ and suppose that $|f(x,t)| \le |t|$ for a.a. $x \in \Omega$, all $t \in \mathbb{R}$ and $\lim_{|t| \to \infty}\frac{f(x,t)}{t}=0$ uniformly for a.a. $x \in \Omega$.
It follows from the second condition that we can find some constant $C$ such that $F(x,u) \le \frac{1}{2} u^{2}+C$ for all $(x,u) \in \Omega \times \mathbb{R}$.
What happens to inequality if we throw out the second condition? I do not see where it could go wrong.
Thanks in advance for any help.
 A: As far as I understand, you either need only the first condition, or the second one and a small condition of integrability of $f$.

*

*If $f$ is a Caratheodory function and $|f(x, t)| \leq |t|$ for a.e. $x \in \Omega$, all $t \in \mathbb{R}$, then for a.e. $x \in \Omega$ you have :

$$ |F(x, t)| \leq \int_0^{|t|} |f(x, s)| ds \leq \int_0^{|t|} |s| ds = \frac{1}{2} t^2 $$


*If $f$ is a Caratheodory function and $\lim_{|t| \to \infty} \frac{f(x, t)}{t} = 0$ uniformly in $x$ for a.e. $x \in \Omega$, it implies that there exists some $t_0 > 0$ such that for a.e. $x \in \Omega$, for all $|t| > t_0$, $|f(x, t)| < \frac{1}{2} |t|$. Now all you need is that $\mbox{ess sup}_{x \in \Omega} \int_{-t_0}^{t_0} |f(x, s)| ds < \infty$ so that, if you call this supremum $C$, you have the majoration you wish for when $|t| \leq t_0$ and if $|t| > t_0$ :

$$ |F(x, t)| \leq C + \int_{t_0}^{|t|} \frac{1}{2} s ds \leq C + \frac{1}{4} t^2 $$
Note that you don't need a convergence to $0$ (a convergence to any $\alpha \in (-1, 1)$ would be enough) ; if you really have a convergence to $0$ you may also state it with an arbitrary small constant in front of $t^2$ (instead of $\frac{1}{2}$. The condition I gave about the essential supremum is maybe not the most general one, but you need some uniform control like this to prevent the integral to blow up when $t$ is fixed and $x$ varies.
