Let $\phi : [0,\infty) \to [0,\infty)$ be a continuous function such that $\phi (0) =0$. If $(\phi (t))^{2} \leq 2 + \int _{0}^{t} \phi (s) ds \ \ \ \forall t \geq 0$ then what is $\phi ( \sqrt{2})$?
I tried using derivative but it didn't help me much.Can anyone please suggest me how to start?
Note that there are some obvious solutions to $\phi(t)^2\leq 2+\int_0^t\phi$ such as the trivial solution $\phi=0$ or $\phi(t)=\min(t,1)$. So $\varphi(\sqrt{2})$ cannot be determined.
However, you can bound $\phi(\sqrt{2})$, e.g., $\phi(2)\leq(1+\sqrt5)/\sqrt2$. Note that $\psi(t)=\sup_{[0,t]}\phi$ also satisfies the integral inequality, and we have $\int_0^t\psi\leq\psi(t)t$. So $\psi(\sqrt{2})^2\leq 2+\sqrt{2}\psi(\sqrt{2})$ giving the bound above.
